@wO Well here's the problem: at infinite speed, time (for you) will stop. In order to stop running you have to brake for an amount of time. You don't have time anymore because it stopped. So you're trapped at infinite speed forever. Also, you're everywhere at the same time.
@wO You should run only one more lap and make it look like you ran both laps in the same time you already took to complete first one. Average may work with any quantity but wont work with time as it is already spent.
For the people that are confused with the second riddle, I’ll show you why he is correct: Let’s call the average speed of each lap: V1=D/T1, and V2=D/T2. Distance is unchanged. The one many are using is Vmean=(V1+V2)/2, which is actually the MEAN SPEED for this case where you have two different speeds, independently of its value, you can find that Vmean=D/2(1/T1+1/T2), so you can actually solve that to find that if you want Vmean=2V1, then you must have V2=3V1. This means that if you run the same distance 3 times faster than the first lap, the mean speed doubles the speed of the original lap. The actual definition of average speed, involves the total distance which is 2D in the total time which is T. Is easy to tell that T=T1+T2. Now if you use this you have that Vavg=2D/(T1+T2), and here is the tricky part. Since you want this to be 2V1 then by solving you have: 2D/(T1+T2)=2*(D/T1), So you’ll arrive to T1=T1+T2,which can only be achieved if T2=0. This in short means that you run at an infinte speed in your second lap, which is physically imposible. Remember don’t confuse MEAN SPEED with AVERAGE SPEED, they are only the same when the speed remains constant during the motion. I hope that this helps. Grettings.
@Lorenzo Gateau This might make it more clear: When you double velocity, you can either double the distance you ran and keep the time constant, or halve the time you ran and keep the distance constant. (V=D/T) By running the second lap faster, regardless of the speed, you already ran double the distance. You went from 1 lap to 2 laps. This means that in order for the average speed to be 2 times what you did before, the time has to be constant (what I typed in the first paragraph). You would have had to run the second lap in 0 time 1st lap distance 1 lap + 1 lap 2 laps ________________. = 1st lap Velocity _____________________ = __________ = 2* 1st lap velocity 1st lap time 1st lap time+ 2nd lap time 1st lap time You can see there that you can't afford to run the second lap with any time at all. Now let's be clear. This problem is obviously *possible* but just without the constraint of running only 1 more lap. You could run 3 more laps in the time it took you to run the first lap, which means distance = 4 laps and time = 2 lap time, which is 2* the speed of the first lap. But you had to do MORE laps. By running just.1 more and having you velocity double, it is impossible.
@leipero You're talking average speed with respect to distance, while he's referring to average speed with respect time - in other words, what average speed actually *is*
The research group at Sydney Uni where I did my PhD wants to collect some data on these riddles to see if you learned anything. If you have 2 min, please complete this survey: ve42.co/Rresearch
@@axemurderbambi2995 and half jar would not work, you need something to move the mass around when it stops the first time around, which means you need the empty space to shift, shifting the centre of gravity
Another way to think of the track question is this: Imagine you and another person both start at the start line and agree to run a two lap race. You're out of shape and they're a very good runner, but you're slowly improving with their training and you like the challenge. You line up at the start line, do a countdown, and then start running. You notice that the other person quickly gets away, easily outrunning you. Once you get to the end of the bend you see they've reached the end of the straight, twice as far as you! You then decide to only focus on your running. Once you get onto the final straight of your first lap you're feeling pretty good, thinking to yourself that you can keep this pace up for the whole two lap race. As you're approaching the end of your first lap, you turn around and see them running you down! You and the other person cross the line at the exact same time, you having completed your first lap, or half the race, and them having completed the whole race of two laps. How fast do you need to go on your second lap to tie them at the finish line? Here it makes more intuitive sense that, since you both are at the exact same line with them having finished the two lap race while you've only completed one, you've run out of time to catch up to them to make your average speed twice that of your first lap, or 2V, which happens to be the very pace they were running because they're very fit and you're not. You would need to travel twice the distance in 0 time. This is why it's impossible (mostly because you're a fluffy couch potato 🛋️ 🥔 ... run more 😉😅).
Well that makes sense. If you're running against someone. But i you yourself are making two laps and can choose to run however fast or slow you like. Can't you then just run e.g 1 or 3km/h for the first round and 3km/h or 9km/h for the second round?
@@sammyruncorn4165 You can do the math to find out that it is impossible. Turn the track, instead, into a straight track that's 2 km long, with a line at 0 km, a line at 1 km, and a line at 2 km. We'll also use your numbers. You decide to run from the 0 km line to the 1 km line at a pace of 1 km/h, which takes you 1 hour. You now need to run to the 2 km line in such a way that your average speed is 2 km/h. Using your numbers, you run from the 1 km line to the 2 km line at 3 km/h, which is ⅓ of an hour or 20 minutes. You have travelled a total of 2 km over the time span of 1 ⅓ hours. That makes your average speed 1.5 km/h. Let's run from the 1 km line to 2 km line faster this time, let's say 60 km/h. You will get from the 1 km line to the 2 km line in 1 minute, or 1/60 of an hour. You have travelled 2 km in 1 and 1/60th of an hour for an average speed of 1.967 km/h. Let's do from the 1 km line to the 2 km line at the speed of light, which is 299,792.458 km/s. You travelled 2 km in 1 and 0.000 000 000 927th of an hour. That gives you an average speed of 1.99999999815 km/h. It still isn't a 2 km/h average speed, even when your second half is the speed of light. Are you starting to see that the question is a trick question where the math, if you plot it out with various speeds for the second half, involves an asymptote? You can approach a 2 km/h average speed, but you will never reach it. Here's what the question is really asking: can you run a set distance and then cover twice that distance, in no additional time, so that your average speed is twice that of whatever speed you ran over the original set distance? It doesn't matter what the original distance is as long as you have to then travel some multiple of that distance further and wind up with the same multiple faster than you travelled over the first distance. The distance could be 1.1 feet where I challenge you to travel the first foot at whatever speed you like and then travel the additional 0.1 feet so that your average speed is 1.1 times that of your first foot speed. Your denominator, which in this case is time, needs to remain the same. You travelled 1 km in 1 hour, so 1 km/1 hour. Now you somehow need to double your speed to 2 km/1 hour. You need to double your distance but you have no additional time to use up. The key is that the distance is doubled but so is the average speed. The distance could be tripled and the average speed tripled and the trick would remain the same.
yeah so unless I'm missing something the two balls would be spinning and moving exactly the same then when they are separated all of their momentum stays conserved as gravity slowly brings them back together again until they're right as when they started.
I don't understand what you're thinking here, the small ball separates from the big one but should remain in exact synchronization with it being that there is no resistance or anything else applied, then due to gravity the small ball and the big one both pull towards each other until they're stuck together again.
FYI: The flange on a train wheel is only necessary on extreme curves. On straight tracks and even gentle curve tracks, the taper on the wheel's tread creates a cone shape that expands towards the middle of the train. this has the characteristic of automatically centering the train wheels on the track as it travels. If the train wheels required the flanges to stay on the track, the friction would make it inefficient and impractical.
okay but even if theres only a taper, that taper basically works the way the flange does because some part will always extend futher than the point of contact
@@sophie6878 The difference is, the flange is a hard stop to keep the wheels on the track in an extreme condition, where the taper allows the wheel to track and auto-center without inducing unnecessary friction.
Also, the flange on the train wheel is NOT moving backwards. Think of a bulldozer track in place of the wheel of a car. When the track is on top, its velocity is twice the velocity of the dozer, and when it hits the ground its velocity is zero until it’s picked up again on the other side. Great. Now think of a bulldozer track, but on a rail, and with a flange to keep it on the rail. How fast does the flange move when the rest of the track is touching the rail? If it’s anything but zero, the track will tear itself apart, because because the flange is still attached and part of the track that isn’t moving.
The track run is the same as the bartender joke, first patron asks for a beer, second asks for 1/2 beer, 3rd 1/4... bartender stops them all and pours 2 beers and says, "you've got to know your limits. "
@@dharmani_youtube If there would be an infinite amount of people asking for beer in the same pattern then the amount of beer in total would go against 2 beer (but never exactly hitting 2 Beer) . Kinda hard to explain, look up what the limit of a funtion is if you want it explained better.
Simplifying problem 2 with no advanced algebra: The trick is that people intuitively think that 2 laps means 2 runs, where you clock each run. The solution is then obvious... But it's a trick question: he's not asking you to do 2 runs, he's asking you to do 1 run, that consist of 2 laps, where the total speed of the run is double the speed of the first lap in that run. Since the time it will take to complete the run will always be greater than the time it takes to complete the first lap of the run, the answer is that its impossible. It will always take more time to complete 2 laps than it takes to complete 1 lap in the same run. No need for any equation here. Common sense, trick question. So here is the simplified version of the question: Can you complete a whole run faster than it took you to complete half of the run? The answer is obviously no. Forget laps and averages and doubles and what have you. 😂
that the problem, he tells about laps all this question. And most right think of it - Average speed it is a speed of first lap + speed of second / 2. Problem is in his wrong question explanation. So how he ask question - the wright answer is 3*V.
@@pekvekIf the goal was the average Speed of both laps the goal you would still be wrong. Your avg speed for the second lap would have to be 4 times faster than the avg speed of the first lap. However it is half the time it took to run the first lap that is requested. If the first lap took 2 minutes it would of course be impossible to do them both in 1 minute total.
Easy explanation for the running track riddle: Let's say your speed is 1min for 1 lap. To double your speed over 2 laps, you would need to run 2 laps in 1min. Since you already took 1 min for the first lap it's impossible to complete another lap. This works the same no matter the speed.
But in my physics class, i learned that average velocity was just (initial velocity + final velocity)/2 We would get questions like, you travel with a velocity of 10m/s for 5 seconds, then a speed of 100 m/s for 3 seconds, what's your average velocity? This is why it's hard for me to comprehend that it's not possible
@@MrMR-sk8jm See, average speed is a single value of speed with which you could travel a given distance and still take same amount of time as in the case where you will travel with different velocities. Actually that is the essence of average.
@@MrMR-sk8jm Even in your example, (v1 + v2) /2 will give 55m/s but to travel 350m in 8 second with a single constant speed, you will have to travel with 350/8 = 43.75m/s. And this is your actual average speed.
@@MrMR-sk8jm first lap velocity is 1 minute/lap. How fast do you need to run the second lap so that your average velocity is twice your first lap? You need to attain an average of 30 seconds/lap. Since you’re only running the second lap, you’ve already reached your limit of 1 minute per 2 laps (by running 1 minute the first lap). If you were to run the second lap at 1 second/lap (60x the velocity), your total time would be 61 seconds in 2 laps, or an average of 30.5 seconds/lap.
Theoretical answer to the Running Question- Let the circumference of the ground be 'd' average speed = total distance/total time therefore let the velocity for first lap be v1 and velocity for second lap be v2 time for first lap be t1 and time for second lap be t2 t1= d/v1 t2= d/v2 total distance = 2d (circled the ground 2 times) average velocity = 2d/(d/v1 + d/v2) =[2d(v1v2)]/[d(v1 + v2)] (d gets cancelled out) =2v1v2/(v1+v1) now this velocity should be equal to 2v1 therefore 2v1=2v1v2/(v1+v2) = 2v1(v1+v2)=2v1v2 (2v1 gets cancelled from both sides) = v1 + v2 = v2 (v2 gets cancelled from both sides) = v1 = 0 but if v1 = 0 the time will become infinite, so practically this is not possible unless v2 = infinite But going deep into the terminology used He said "what will be the velocity" velocity is a vector quantity and it is equal to displacement/time while circling a ground we are coming back to our original position so displacement = 0 hence velocity in both the cases would be zero the correct terminology here would have been speed or angular velocity (omega) Thanks for reading.
Wow, just posted before going home from work. Didn't expect that. It was no problem at all. As long as people have fun learning, I'm happy to be part of it. ^^ And thanks for choosing my video. :)
5:20 Actually the flange isn't what keeps the train on the rails most of the time. The wheels are slightly conic. Feynman has a great video about this.
For those who found the explanation for the speed thing confusing, the average speed of the person is the total distance divided by total time, or Vav = (v1*t1 + v2*t2)/(t1 + t2) keeping in mind you are supposed to run only 2 laps, the total distance you have ran is 2d and total time is t1+t2 then Vav = 2d/(t1+t2) since v1=d/t1 and we need the average to be 2v1 then we need t2 to become 0 which is impossible !
Well I didn't (in general). I'm too dumb. Can't one just walk like 3km/h for the first one and then 9km/h for the second so it averages at 6km/h. The distance is the same for both lapses so doesn't matter? That's what I thought, but I'm probably too unintelligent to get it...
@@sammyruncorn4165 Okay, so the key difference is that a racetrack is a fixed distance. If you were driving on a highway, for instance, you could absolutely do that. Say you travel 30km/h for one hour. Now you’ve gone 30km, right? So in order to make your average speed 60km/h, you can absolutely travel 90km/h for the next hour to make the total average 60km/h. But remember, that means you will have gone 90km in the 2nd hour. On a track, you can’t change the distance traveled like that unless you run more laps. Take that same highway scenario, but imagine there is only 60km of total road. Once you drive the first hour, there is only 30km of road left, _so you cannot just drive 90km to boost the average._ Even if you drive at 90km/hr for the second hour, you will run out of road after 20 minutes, and your total travel time will have been 1.333 hours to go the 60km (which obviously isn’t a rate of 60km/hr). In fact, it turns out that no matter how fast or slow you run the first lap (or drive half the road), your time for the second will need to be 0 in order to double your average - which obviously isn’t possible. Long story short, if you were able to run 3 laps after the first in the same total time you had run the first, that would double your average speed. But with only one more lap, it physically can’t be done.
3:56 If you run at the speed of light, according to special relativity, in the runner frame of reference, it will take 0 time to go anywhere. So the average speed is indeed twice.
@@sajinssha Yes the speed of light is finite, and will be observed to be exactly the same speed in any frame of reference. Right up to the point that you actually REACH the speed of light. At that point, you can not "witness" the speed of light, as there is no light to witness... No light will ever be able to "catch" you. According to special relativity, if you were to travel at the speed of light, you could travel a distance of 5,000 light years, and exactly 0 time would pass for you. Meanwhile, 5,000 years would pass for us here on earth.
The running two laps - 2V = 2d/t problem is hard to grasp so I did some actual examples and I now see it. Let's say your first lap is at 1 mph. You cover .25 miles in .25 hours (15 minutes). Let's say your second lap is at 3 mph. You cover .25 miles in .083333 hrs (5 minutes) Your combined half mile time equals .25 + .08333 = .3333333 hrs (20 minutes). Velocity is distance/time so .5 / .333333 = 1.5 mph This is NOT twice your first velocity - we are looking for 2 mph. Skip ahead to running you second lap at 6,000 mph. In your second lap you cover .25 miles in 0.15 seconds (.000041666 hours). Your combined half mile time is .25 + .000041666 = .2500416666 Hours Divide distance over time (like before) .5 / .250041666 = 1.9996667 MPH - Not quite 2 MPH. As a matter of fact, you'll never reach 2.0 MPH no matter how fast you go. I hope that helped someone. :)
Another way to look at it thru algebra: T1=time to run/walk 1st lap T2=time to run 2nd lap. D1=D2 =distance around on lap=D. V1= Velocity of lap one. So V1=D/T1 or T1=D/V1. Vave=2xV1 our goal. Vave=2xV1=2D(total dist two laps)/(T1+T2) or 2V1=2D/(T1+T2) . Divide both sides by 2 and get V1=D/(T1+T2) but we already stated correctly V1=D/T1. The only way V1=D/T1=D/(T1+T2) is if T2=0, in other words the time to run lap two is zero and V2=Infinity. Impossible.
This one was bs. It's not some confusing riddle just a poorly worded one. I thought it asked a speed such that the average of the two was 2 v1, not the average speed over the total time ran. Would have been able to give a much better answer had the question been clear Oke I'm done ranting
Finally i can feel smart for a change! Riddle 2 was easy with simple math and times of laps, not speeds. If you run a lap in 30 seconds, how fast should you run the second lap in order for the average of the 2 laps to be 15 seconds? Well... you should run the second lap in 0 seconds, so.. there...
@wO But don't forget you must keep with restrictions set by the problem. 2 laps is all you've got. So the second half of your run must have the same distance of the first half. If your 'lap' is 1km long, you only have 1km more to cut that average in half. Can't do 3km.
@wO not really, how did solve the problem? If he's incompetent like you then maybe he'd believe that someone finishes the 2nd lap with the average of 2V1.
For the track riddle, Isn’t velocity displacement over time? Therefore, since the total displacement is zero because it is a loop and you end up back where you started, it doesn’t matter what the second lap is. 2x0=0. As long as you complete the second lap you have done it.
doesn't quite work like that, you still need to complete 2 laps. So if you don't run 2 laps, you don't meet the conditions of the question, so it doesn't work. Also, moving slowly doesn't work, because his explanation is scale-able.
Lots of surprising answers in this one... but nothing baffles me as much as the fact it took me since last week to realise why you called those riddles "revolutionary". Damn puns...
The bike question also depends upon the friction between the back tire and the floor. With insufficient friction, it is not possible to move the bike forward by pulling backward. The same would apply to pulling backwards on the flange of a train wheel.
I don't think so. See, for the bike to roll forward the pedal must always be travelling forward with respect to the ground. If the pedal actually moves back, the bike will skid back, with its back wheel rotating normally, as when it is moving forward. If the friction betwen the tire and the ground is too great and does not allow it, you won't be able to move the pedal backward. The bike will be standing still.
For the people who are confused by the second riddle, just use numbers! Assume the following: - The lap distance "d" is 10 meters. d = 10 m - Your speed in the first lap "V1" is 1 meter per second. V1 = 1 m/s - You have to run the 2 laps at an average speed "Vavg" of 2 meters per second Vavg = 2V1 = 2×1 = 2 m/s Remember that V = d/t So: t1 = d/V1 = 10×1 = 10 s t(total) = 2d/Vavg = 2×10/2 = 10 s Notice that the time needed for both laps (10 s) equals the time that you spent in the first lap alone (10 s)!!! So basically, you have to complete the second lap in 0 seconds to be able to achieve Vavg = 2V1 !!! Which is impossible of course. Unless you can teleport..
But he didn't say anything about time... only speed. This is what's confusing to me. I listened to the riddle closely. He only said the speeds needed to average out to 2x the original speed. So if I walk at a pace of 1 mile an hour and then the second lap at a pace of 3 miles an hour, my average speed would be 2 miles an hour. There was zero mention of time in the original riddle.
@@bunny_0288 _"So if I walk at a pace of 1 mile an hour and then the second lap at a pace of 3 miles an hour, my average speed would be 2 miles an hour."_ Of course not. Your average speed would be 1.5mph for the two laps.
@@Stubbari Your mean speed would be 2 mph. 1+3=4 4/2=2..... However, I now realize he wasn't asking for the mean speed which is why that isn't the correct answer. It was a poorly worded riddle in the original video that was misinterpreted by many people.
Veritasium, I was wondering what you think about the idea of running the track (problem 3) on an inside lane. say it's 1k long. then the next lap, you run on an outer lane that is slightly longer, say 1.01k. then all you need to solve it is to run .01 times faster in the outer lane to finish in the same time and thus making 2v1. it may be a trick, but it's the same as the fog on the spinning floor. and still plausible. what do you think?
+Veritasium Have you done anything on the latent heat of fusion/vaporization yet? Lots of people don't realize that it takes energy to convert the water into steam, even if the water is at 100C, or to convert ice into water, even if it's at 0C. Could also include how salt actually gets rid of ice, and how the whole system's temperature lowers even though there's the same amount of latent energy present.
For the track question, you never specified whose reference frame we're measuring the time in. So in the reference frame of the runner, if he moves at the speed of light on the second lap, it will appear to him as if no time passed by in the second lap.
Is there a way to run the track backwards (ie in the opposite direction as the first lap) at, I dunno, half speed of the first lap so that you.... never mind
3:56 : "Even if you when the speed of light..." Actually, from your reference frame going the speed of light, no time will have passed going the second half of the distance. It really depends on the reference frame of the measurement.
I think instead of having no time pass, you meant no distance pass because even having your reference frame move at the speed of light, that still takes up time.
And even if you ran the first lap faster the average velocity of the two laps would be 2x the velocity of the first lap, since 2x0=0. Seriously every time he said velocity in this it's nails on a chalkboard to me. Speed, not velocity omg.
I'm still confused by the "answer" to that "riddle". Nowhere in the question does time come into play. If run 1 mph, then I run 3 mph, my average speed is 2 mph. Twice the speed of the first lap.
@@TylerDollarhide nope, because then suppose the lap is 3 miles long, you ran a total of 3+3=6 miles in 1+3=4 hours. making your average velocity 6/4 or 1.5mph
@@TylerDollarhide The average speed of the total journey = total distance / total time taken. Let d = circumference of a lap, and t = time it takes to do the first lap. Note that the average speed of the first lap is v = d/t. Now, the issue is that you MUST take t amount of time to run the first lap, then you need to run the second lap in whatever time is left (call it T). So, average speed of the total journey = 2*d / (t +T). But he is saying that you need to choose T so that 2*v1 = 2*d/t = 2*d/(t+T). But then T = 0. So you will need to "instantaneously" jump the last d distance. This means the speed of the second lap must be infinitely fast to spend no time covering it. Another way to see this problem is to draw a plot. Call the y-axis distance, and the x-axis time. If you draw a line from (0,0) to (t,d) (representing the first lap), then a line from (0,0) to (t, 2*d), you will find that the final side of that triangle goes vertical, leaving no additional time to get from d to 2*d. I don't know why he didn't draw the plot, it's much easier to see.
@@TylerDollarhide Time did come into play. From what you said, 1 mph and 3 mph, the "miles per hour" part already gave you the time component of this problem. 1 and 3 does indeed work if the laps are independent of each other, because now you are confusing the average speed and mean speed. Average and mean are very similar things but absolutely not the same.
You're right that the flanges are to keep the train on the track; but only a last resort, they're not designed to contact it - they'd be moving relative to it so they would wear and waste energy. It's the fact that the wheels' rolling surfaces are slightly wider on the inside that keeps it on the track normally: so the train is lowest when it's central and gravity corrects any deviation. Incidentally, it's the conical wheels which mean trains don't need differential gears. In a curve, the train is thrown out, so the outer wheel contacts the rail at a point where its diameter is greater (and vice versa)
At 6:22, there is an incorrect statement. The explanation is as simple as "the net force is backwards so the bike goes backwards." The complication is that the pulling force isn't the net force. The net force is the total of all of the forward friction and the backward pulling force. Whichever force is stronger will be the direction of the net force.
+Veritasium I am posting this here as I was particularly bugged that no one seems to have addressed the importance of torque and rotational dynamics in answering the bike pulling problem. The trochoid may provide a geometrical solution, but it has little predictive use if one had to calculate the acceleration of the bike with respect to the pulling force and gear ratio. As per net force, I'd say it has little to do with the problem at all. Assuming an infinite coefficient of static friction, regardless of the "net force", the bicycle would remain stationary; in this system, motion is only possible if the state of the system permits the rolling action of the wheels. Torque must be given consideration as there would otherwise be no way for the bike to move even kinematically. Regardless of the shape of the trochoid, there must be a condition at the point of rest where equilibrium dictates the acceleration of the bike; the derivative of the trochoid at that rest point would yield zero and no indication of the expected direction of rotation. Newton's "Second Law of Rotation" must be adhered to. In Armchair Explorers' video response in /watch?v=S1yX_LTqtms, I posted a comment mathematically deriving equations relating the _net torque_ to the dimensions of the bike and the applied force; I didn't have the resources to fashion a response video then. In the replies to that comment, I rearrange my equations and derive the same formula shown at 7:41. Just from first year first term university physics, I immediately saw it as a "basic" rotational dynamics and torque equilibrium problem. Torque is perhaps quite complex to most, but as you have in the past addressed misconceptions about translational forces, it might be an excellent educational opportunity to address misconceptions about the dynamics of rotation. Otherwise, these have been very thought provoking "revolutionary" riddles and I admire your work.
+Dan T The friction force is immensely important here, though the friction coefficient is not. Static friction is a useful force because it can have any value up to a limit. Since it opposes relative sliding, we just attempt the sliding and static friction pops up as a constraint force to prevent it. As long as we don't ask for to much static friction force, it's okay. Torque balance is certainly important to determine how much force is being exerted. The pedal arm length, gear ratio, and wheel radius all play a part.
+Jeff S To assume perfect rolling in this system "independent" of the applied force, we must also assume the parameters governing static friction (coefficient of static friction and normal force) to be as large as possible. An infinite coefficient of static friction simply sets for our model a limitless supply of a reactionary force to keep the wheel kinematically constrained to perfect rolling, assuming that sliding/kinetic friction is out of the question for possible solutions; in our case, it provides a limitless supply of counterclockwise torque moment upon the rear wheel proportional to the applied force at the pedal. As per torque, I'd argue that it is _much more important_ when discussing any form of motion at all involving rotational elements; "static friction" is but a variable constraint applying an opposing torque to the torque imparted from the pedal. The bike _will not move_ unless the torque moments imparted by static friction and the applied force are unbalanced. _net torque = sigma_tau = tau_1 + tau_2_ where angular _acceleration = alpha = I * sigma_tau_ . Torque balance is not about determining "how much force" as opposed to whether the bicycle wheels will undergo any rotational acceleration leading to translational motion at all. If you would review my analysis from my comment in /watch?v=S1yX_LTqtms, I calculate how these torques play out and eventually arrive upon the same relationship shown at 7:41 of this video. This is all from university physics.
I had the Force/tourque/radius explanation in my mind for the bike gearing problem. Which is a correct explanation, no doubt. But I really liked the purely kinematic explanation that you showed! Kinematic solutions are elegant!
What have the ping balls got to do with the first one? It's just air, and you get the same effect from without using them, just your container was smaller so it was a bias demonstration.
There might be a small difference. The honey sticks to the walls of the cylinder and the behaviour could change over time as it rolls down the slope compared to the initial state of no honey at the top of the cylinder and the bottom half full of honey, where as the ping pong balls always behave as spherical bubbles.
You need to keep the air bubbles together. In honey, air would float to the top and release/break through the surface. The only way to get this to work is to have the honey continually push the air bubbles down, which float back up by themselves.
I'm an aeronautical engineer, i graduated from the air force academy. Most of your videos are right but what you said about running the 2 laps is VERY VERY debatable if you consider that we move through time
Love trying to figure it out. But - failed, miserably. 1 out of 4. but I think I actually won since I was really thinking and trying to figure it out - not a normal activity for 70 yo.
Veritasium , this time it wasn't clear , concise and easy to understand . It did challenge my early beliefs and sort of intrigued my interest for rotational dynamics Actually , that ping pong ball , that's something I can't grab by books , It gave me a big BONER
The flanges don’t actually play that much of a role in keeping the train on the track. In fact, most of the time (read: in normal operation), they don’t touch the rail at all. It’s the geometry of the rolling part of the wheel (it’s cone-shaped) that does the trick.
Pretty much. Just to make sure the train doesn’t get derailed if there’s a track defect or in case of an emergency situation (like something bumping into a train sideways).
Not only for safety. In turns there is a cant so when train turns to the right, right rail is lower them left. Without flanges, it could slip from the track if it would have to slow down on that part.
Even being non intuitive, the lap riddle is a very common high school kinematics intro question. average speed= total distance travelled /total time taken Let time taken for Lap 1:t1 Lap 2:x And d be the distance of 1 lap let d/t1 be v1 avg speed=2v1=2d/t1=(d+d)/(t1+x) This leaves t1+x=t1 Obviously x=0 s therefore speed in lap 2 must be d/0 =infinity.
The track riddle can be done! You gave us an object to run around, not a distance. Assuming the track is a standard 400m, if you spend your second lap running in a 5.2m wide x 2m long zig-zag, you could effectively run 1200m on your second lap, adding making it take just as long to run thrice as fast, equating to a 2v average.
The answer to the running track question really depends on how you choose to interpret the question. Q: How fast do you have to run the 2nd lap to get an average speed of 2x v1? If you choose to average over time, there is no answer. If you choose to average over distance, the answer is 3x v1. If you are tasked with finding _the_ answer, then 3x v1 would certainly have to be seen as correct.
EternalSilverDragon That’s what I was saying. Such a riddle can be answered by simply doing it in your head. But this particular video was filled with nothing but cheap shot riddles. There was no way in holy hell we could have come up with the answers that is channel wanted us to come up for these particular riddles.
I agree with both of you. I wasn't expert enough to disgree, but I felt strongly that his explanation was just silly. I knew for a fact that we could get a reasonable average, if we assumed that there was no need for acceleration...because the time cost was the exact solution to a grade 11 Physics problem that was introduced to my class in high school. That being said, I still don't understand why his interpretation is what it is. It sounded like he was limiting the amount of time of the second lap to 0 seconds, which seemed so arbitrary.
@@eugenetswong because it was on.ly two laps. that is the key part to this riddle. if it was three laps then it would be possible. You have zero seconds in your send lap yet that is impossible as even the speed of light would take even a nano second which is more than zero which is why you cant double the average speed.
@@briandarlage9815 that sounds like what Derek says, and what we've been objecting to. Giving a person 0 seconds to do the second lap seems very arbitrary to me. Why do we get 0 seconds?
But physically speaking, you do not average speed over distance. V1 + V2/2 is incorrect in this case because time is the decisive factor. I actually did the maths on my phone, let me show you. (x/t + x/y) /(t + y) = 2x/t 2t + 2y = 1 + t/y 2y - 1 = t (1-2y/y) Let t be the time in which you run the first lap. X is the distance you run and y is the time in which you run the second lap. The last equation is as follows. 2y - 1 = t (1-2y)/y -1 = t/y So you have to run the second lap in the negative amount of time that you ran the first lap in. Negative time is physically impossible. And I disagree that the video had cheap tricks. They were all sensible and I personally got all of them right minus the two ping pong balls in the honey because that's way too specific of a guess.
You got the wrong answer for 3. :D The correct answer is 0. Lay down and have a nap. Don't go anywhere. Twice zero is still zero. You live at the starting line now.
You also die at the starting line as you have just imprisoned yourself for an eternity for such a silly reason, you cannot stop now or all previous time spent is wasted, hopefully some kind stranger will come by and provide you with food daily and if lucky they will even put up some form of shelter around you. But can we truly call that living? Is being stuck in one spot for eternity truly worth the results? For some maybe but not me, i choose to live my life even if it means i am a failure.
YES!! I said gearing needs to be low enough AND that the pedal will only pull back/bike move forward until the crank rotation causes the "acting" force vector to reduce with angle until equilibrium is reached. Then consider friction and the fact that the string has changed angle such that you are also pulling the bike downwards thereby slightly increasing this. The extra crank angle inflicted to pass static euilibrium (overcome friction) may further affect the second euilibrium crank angle that will result when skidding.
Yes thankyou finally someone with a brain, I don't know how many comments i had to go through to see someone with an actual right answer. His original question never specified the laps were the same distance, thus it is NOT impossible.
Aris Nikoletopoulos In fact you wouldn't be dobuling your distance, you would be changing you trayectory. The distance is a rect line between point A and point B. If you ran to the middle, went back to point A and then ran back to point B, the distance would be the same as if you ran in zig-zag from point A to point B.
Mateo Costa the distance between them will remain the same, of course; the distance you travel however will increase. Velocity is calculated using the distance traveled, not the distance between your start and finish point, or am I wrong??
no, speed is calculated using distance traveled. (s=d/t) velocity is calculated using displacement (v=x/t) however, because this track is circular - displacement is actually zero (start point and end point are the same) and therefore average velocity is also zero (v =0/t). which is why in this video he corrected velocity to speed. but then people want to try be 'smart' by arguing semantics and say 'oh well you never said lap distance is constant'. okay fine, well done, he didn't say distance is constant but that's clearly what he meant. if you zig zag in a 100m sprint, do you get to call it a 300m sprint? no. end of the day, the question is supposed to have distance fixed. if distance isn't fixed then youre asking a different question. so now that the clarification has been made, can you solve it?
Awesome video as always. I've just recently started my own science channel and I'm happy to say I found inspiration watching your channel and many others. Thank you and keep up the great work!!
Steam locomotives also have a rod called an eccentric rod which helps "change gear" by increasing or decreasing the movement of the piston inside the cylinder. The rod always moves backwards because of how the piston works
@@RR67890 you know, wheel flange either dont move backward relative to the ground. Question was wrong explained. One little area on flange which position on the wheel you need to change in your imagination to say it moves backward is not a part of the train.
@@Guapter I think you've misunderstood the explanation. You seem to be referring to the contact point, where the wheel doesn't move with relation to the ground. However, with a train wheel, the flange is a larger, concentric circle that must match the rotational speed. The following image isn't perfect, but I think it can help get on the correct path to understanding: Imagine drawing a line from the center of the wheel, through the contact point with the rail and to the flange's edge. This line functions a bit like a lever with the pivot point being the contact point between the wheel and the rail. When you move the center of the wheel forward, the pivot point remains stationary and the point at the flange moves in the opposite direction to the center of the wheel.
1 - The rod can't always move backwards. It would have to move forward as well. 2 - It would always need to be moving backwards with respect to the ground while the train is moving forwards, which isn't possible.
I consider the track problem to be a trick question. Your solution treats the two runs as one run with a slow half and faster half. Alternatively I, and I am sure many others thought of it as two separate runs in which case it is perfectly possible to find a solution without having to exceed the speed of light. This I suppose is due to the different perceptions between the pure analyst and the empiricist. The bike problem again depends on initial conditions, which were not specified. The compound vessel with its viscous liquid and variable geometry is only one of many different configurations which would all exhibit analogous behaviour. What this whole exercise illustrates very well is the importance of experiment as opposed to the ancient belief that real world problems can be resolved totally by the intellect. Aristotle certainly believed this and was accordingly guilty of passing on many erroneous explanations of real world phenomena [ example : the falling body problem ]. This is one of the reasons why physicists can be dangerous if given jobs better suited to engineers. An engineer is always ready to run a trial before committing to a final solution.
You are calculing your average value like: Avg Value = ( Value 1 + Value 2 )/2. The problem is that speed is the product of dividing distance and time (d/t), so the average velocity is calculated like: Avg Velocity = Total Distance / Total Time, so in that 2nd problem if you wanted the average velocity to be equal to 2 times Velocity 1, then you had: 2 V1 = (d+d)/(t1+t2)=(2d)/(t1+t2) and V1 = d/t1 so 2 times V1 equals 2d/t1, so: 2d / t1 = 2d/(t1+t2) For making that equation equal, you need to make t2 = 0, thats why its impossible to make the average velocity equal to 2 times velocity 1 (V1).
@@HAHA_468 Well, saying 2*V1 = d/(1/2)t1 is basically saying that you can double V1 by running the distance of V1 in half the time, but it doesnt say anything about the average speed, if you wanna desenvolupate the average speed into other things you need to start from Vavg = total distance / total time. if you start from another equation then your results arent "talking" about the average speed but about obvious things like "To double the speed you need to run the same distance but half the time".
I'd like to point out a technical mistake you did at 5:23. Flanges in train wheels are used only for switches on the railway. The reason why the train doesn't fall off the rail is because of the truncated cone shape of the wheels which makes them act as an automatic centerer :)
The other part of it is he said ground. While the tracks sit above the ground. Thus the wheels are still moving forward in respect to the ground because they don't go UNDER the ground they go under the track on top of the ground.
+Michael Brune If the track and the ground are both stationary (they are), then ANYTHING which is stationary with respect to the track is also stationary with respect to the ground.
no1unorightnow lol is that your gut speaking or do you actually know about it. Because I'm a civil engineer and I've been taught what this guy said. so unless it's just you trying to apply common sense here, if you actually have some practical knowledge about the wheels then do tell.
The lap trick is not impossible. Imagine you’re in a car, and you finish your first lap of 1km distance in 60sec. That means that the speed of your first lap was 60km/H. So you have to adjust your speed during the second lap to end up with an average speed of 120km/H at the end of the 2 laps. Then, you just drive at 180km/H during your second lap and your average speed over the 2 laps will be 120 km/h. Thats twice the speed of your first lap
Since you've run two 1-kilometre laps, your total distance was 2 kilometres. The first lap took one minute, and since you went three times as fast the second time, the second lap took 20 seconds. So your total distance is 2 km and your total time is 80 seconds. Average speed = total distance / total time = 2000 m / 80 s, so your average speed was 25 metres per second, or 90 kilometres per hour, which is only 1.5 x the speed of your first lap.
At 4:10 , u can say, in a sense, that the condition for the riddle can be fulfilled if a person runs first lap with initial speed(v) v--->(lim) 0. So the condition will be fulfilled for any velocitu v2 with which we run the second lap
@@NMTCG I doubt you can do the first lap backwards because when I tried to calculate it, I got a concrete v1 = 0 with no other answers (i.e. exactly what I'd do if someone was forcing me to get off my ass and do a lap)
None of these suggestions work. You need to go around the track to complete a lap which requires you to have a velocity greater than 0. -V is the same as a forward V, all - tells you is the direction
suppose you travel in train from one city to another which are 500km apart and the train takes you 5 hours in total then what will be the average speed of train according to you ? (Definitely the train would not be running at 100kmph all the time, sometimes it may move with higher speed and some times it may move with lower speed, but what matters to you to get average speed is the total time taken and total distance travelled)
If you are taking a loop track, you'd have to measure speed, not velocity, thus if you changed your total path length (i.e. by running backwards, then forward, then backwards, then forwards, or zigzagging) then you could plausibly solve that riddle.
A very small correction: The train wheels flanges are not intended to keep the train on the tracks, and they usually don´t make contact with them. They are used only when going over a set of points (BE) or a switch (AE). The train is kept on the tracks (even on curves) by the special shape of their top surface and the conicity of the wheels. Interesting video, btw!
For the bike riddle, more illustrative might be the case of a spool of string with raised, circular ends (standard gear, not extra-low gear). For all such systems, there exists an angle
if you reframe the track one in terms of time, it becomes a bit more intuitive: you want to run the second lap such that the average time across both laps is half the time of the first lap. But if, say, you walked that first lap at five minutes, then youd need to get the average for the two laps to be two and a half minutes. Which it already is before youve even started the second lap, so you need to do the second lap in zero seconds, or infinitely fast, regardless of how fast or slow you did the first lap.
The ping pong balls are irrelevant. The (lower-density) air inside them is what’s important. The jar half-filled with honey is close enough to be pretty much equivalent. I think it should count as a correct answer.
Sort of yeah, but the ping pong balls creates a vessel for the air which moves with the honey easier that just air, as it'll easier affect the position of the honey than the restricted air in the ping pong balls, although i agree, it should count as the correct answer, as it's has a veeeery similar effect, which would be hard to differentiate without seeing the content.
I feel like if our guess was 10w40 with hollowed out, sealed eggshells, it would still be wrong because the eggs are not shaped like spheres. While the ping pong balls offer more resistance to the fluid dynamics, a more viscous substance would yield an even more similar result. From the perspective of visual observation, indistinguishable!! I fart in your general direction on that one.
@@wilfdarr You are wrong. The shape of the balls is irrelevant. Replace them with balls the same density as the honey. It's the liquid and the air that matters here.
@@thetruth45678 Well, since you went there, No you're wrong. The tennis balls don't simply displace the honey but displace some of the honey VERTICALLY which forces the C of G higher in the cylinder while maintaining the total weight of the system, which you could not do by just adjusting the honey/air ratio. This high C of G is what makes the movement so dramatic. Good approximation, but not the same. Maybe next time try words like 'I think', 'I can't see how that would be' or 'Why wouldn't X work just as well?' rather than 'You're Wrong'.
I'm not buyin' the track riddle solution. 2d/t1 is the same as d/[(1/2)t1]. It's possible to run the same distance in half the amount of time. There's no need to "weight" the velocities in the average. Run the first lap in time t1, run the second lap in time one-third of t1. I think this can be shown pretty easily on a position vs time graph. You'll have one slope m1 for the first lap from y=0, t=0 to y=d, t=t1, then triple the slope between y=d, t=t1 and y=2d, t=(4/3)t1 for a slope of m2 The average of these two intervals. The average of these slopes will give a velocity twice as large (steep) as that of the initial interval.
I thought that you need to plot a velocity-time graph to find the average velocity? it should look like two bars next to each other where the height = the gradient from your graph
Great video, but I believe it is actually the taper on the train wheel surfaces (which at not 90 degrees) that keep the train centered in the track, not the safety lip. The lip is just a backup safety measure.
If the flanges did come in to contact, you'd hear awful screeching noises. The flanges do keep a model train on the track, but that's just because the speeds and accelerations are so slow that the screeching noises go unnoticed, and most likely, it isn't worth the effort to machine exact scale models of a real train wheel.
I have to say that 2, 3 and 4 are trick questions: 2. The question asked was what speed would need to be run, not velocity. The two are not the same. 3. The part of the train that moves backwards would be the pushrod pushing the wheels on the backstroke, which answers the question precisely. The flange is part of the wheel, which you called, 'the flange part of the wheel', not the flange part of the train. Technically it would be a part of a part. 4. Using any bike introduces inconsistent characteristics that should be taken into account.
For the 2nd riddle, you could run zig-zags across the lanes for the second lap such that you run 3 times the distance than the first lap, but also at 3 times the speed. The net result will be 2 laps with an average speed of twice the first lap. You're welcome.
I definitely thought the running around the track one was just a verbal typo that didn't get caught in editing. It would have been a lot more clear to say "the value of" instead of directly saying to compare speed and time.
I tell the track riddle differently. It's my favorite math trick question. You have a hill 1/2 mile up and 1/2 mile down. If you go up the hill at 30 mph, how fast do you have to down the hill to average 60 mph..many people say 90 knee jerk. But as you know you'd have to instantaneously be at the bottom. Your time is up at the top.
Yea but it's not a math problem, it's a communication problem. If i run a lap at 1m/s average speed and then a second at 2m/s, i would say i doubled my average speed. But if you consider it the same time series you only have a increase in average speed by 1/3. So if you ask an athlete for his average speed over a set distance such as 100m sprint for a given practise day, he will average the average speeds not average his total time spend on the track.
@@StillTryingMyBest Or those people were just confused as to why it is supposed to be a riddle. I think the question would be better if it included some kind of hint that the two laps are to be considered as one single measurement. But maybe you can't understand, because it was so obvious to you
In simple terms the leverage applied through the gears on a regular unmodified bike is not enough to overcome the force required to make the wheel turn. Aka you have to pull backwards harder than the tire can push forwards to make the wheel spin.
Nice video, cool riddles. That said, I have an argument about the flange: the flange is going forwards with the speed of the train: the part of the flange that goes backwards is not the same at all times so there's no part of the traind that is *always* going backwards - otherwise, there would not be much train left after a while.
Exactly. Veritasium is wrong on this one. By increasing the distance by a factor of 3, you get 4 laps in twice the time, which reduces to 2V1. If v1 = 1 lap per minute, and v2 = 3 laps per minute, then total is 4 laps in 2 minutes, which is the same as 2 laps in 1 minute (exactly twice v1). It's 100% doable, but Veritasium goofed big time on this one.
Porglit no... you cant add speeds like that to get 4 laps, because you only need to run one more lap. what you did is equivalent to adding fractures by simply summing the numerators apart and the denominators apart... this way 1/2+1/3=2/5
2d/t is the result you want. t is defined as the time it takes to do the first lap. How are you going to double distance but keep t the same? t1 + t2 = t1 For that to be true t2 has to equal 0, so you can't do it on the next lap.
I have a question- If a person is about to enter a black hole (he has not entered yet but is very close to the entrance point) would an observer from far away from the black hole see a image of that person entering the black hole like he is paused. If yes then would not there be sort of 2 existence of the same person somehow. One is what we see at the entrance of the black hole and the other that person is expireincing. Clear my doubts anyone
Does someone else thinks that the bicicle explanation is incomplete? Please, correct me if I'm wrong, but from my point of view it should also depend on if the tires are able to spin with or without sliding.... so... with a lower ratio.... the bike would always move forward or not move at all, just depending on the tires glide
Because the distinction exists. Don't be mad just because you weren't right - be happy that you came close to the right answer and now realize how you were wrong
I don't realize how I was wrong. I don't see how it makes a difference to the movement of the cylinder. He said it was close, I think it was actually the same. I don't think I was wrong, or at least I'm not ready to concede that I was until I see it properly explained why it matters that there were ping pong balls in it.
I agree. If the correct answer is not air, but ping pong balls, tell me what the difference is in that scenario. The little bit of plastic? Not good enough.
If you are running a circular track, your average velocity would be zero for any lap, since the first and second half vectors would cancel each other out. Thus to get an average of 2 times zero would mean that you could literally run the second lap at any speed slower or faster or even the same as the first lap and still get an average that is twice the first lap. Anything times 0=0, including 2.
@@steves9388 Sorry, but if he's too "dumb" to present a riddle hinging on semantics to "fool" the audience in a proper way (it's a video, he can edit stuff and not just make a split second text overlay), I am perfectly fine with anybody giving "not intended" solutions.
We have 2(t1)(v1) = {(t1)(v1) + (t2)(v2)) / 2 Which, under normal circumstances where (t1)(v1) = (t2)(v2), can only be solved where (tn)(vn) = 0 or infinity. However, if we consider speed from the perspective of the runner and include time dilation, then a(t1)(v1) != b(t2)(v2) where a != b, and a and b are factors of time dilation. 2a(t1)(v1) = {a(t1)(v1) + b(t2)(v2)} / 2 Otherwise our rule holds that (t1)(v1) = (t2)(v2), therefore: 2a = (a + b) / 2 3a = b Therefore, where the time dilation of the first run is 3 times the time dilation of the second, e.g. first lap at ~0.943c and second lap at ~0.001c We have a total average speed twice that of our first lap.
we can actually run the second lap with twice the distance of the first. If we run the first lap in the innermost strip of track and the second in the outermost strip with having the same time.
I see the error of my ways, I was calculating the average speed per lap, not the total speed. Put simpler (I think his phrasing was confusing): 0 = overall 1 = first lap d/t1 = V1 2d/t0 = 2V1 => t0 = t1 => t2 = 0 You can't run a lap in 0 seconds.
Paul George had a great answer. But here is the same explanation using only algebra: T1=time to run/walk 1st lap T2=time to run 2nd lap. D1=D2 =distance around one lap=D. V1= Velocity of lap one. So V1=D/T1 or T1=D/V1. V(average)=2xV1 our goal. V(ave)=2V1=2D(total dist two laps)/(T1+T2)total time or 2V1=2D/(T1+T2) . Divide both sides by 2 and get V1=D/(T1+T2) but we already stated correctly V1=D/T1. The only way V1=D/T1=D/(T1+T2) is if T2=0, in other words, the time to run lap two is zero and V2=Infinity. Impossible. Unless as Mark Kmiecki said, you refuse to run and V1=0. But if you refused to run because you know the goal is impossible, by refusing to run you prove yourself wrong because at V1=0 2V1=V(ave)=0 is true and you just sucked yourself into a cosmic paradox.
The bicycle riddle relies basically on the same principle as the wind powered car that can go faster than the wind, straight downwind. The gearing can multiply the force, and apply it in the opposite direction, or in the same direction.
3:58- even if you went the speed of light, If you ran with the speed of light, you would have infinite time because of the time dilation, So that makes it possible.... ❤️
The answer you have given is the average of two separate velocities, but you have forgotten they have different denominators (the laps are taking different times to run). With the equation for velocity (distance / time) we know that the distance is the same for both laps (i.e. 2d). Therefore to make this double the speed of lap 1 you need to keep the denominator the same i.e. Lap 1 = d/t so 2d/t can only be double the speed of the 2nd lap if no additional time is added into the equation. That is why adding in the speed of light (and faster) will always add some time that will make it just under double the speed of the first lap alone.
Vavg is using V1 as reference, so in order to complete the challenge you have to run *both* races within *t1* , amount of time that is only enough to do the first race. It's just a pedantic trick question.
Okay, so let me try to understand it in a different way. I run first lap at 5 KMpH and second lap at 15 KMpH. What's the average speed with which I completed one lap? If the answer is not 10 KMpH then please let me know what it is.
I can't get my head around a better way to ask the second riddle. The wording really made me want to just answer 3V1 but I can't think of a better way to word it. If I were to propose this riddle to a friend, I know full well it would spark an argument about how averages/means are calculated and how those words are defined. I think, instead of asking about speed, it's better to ask it in the context of time. So maybe something like this: "Let's say it takes you some time to run around a track. How long would it take you to run a second lap so that your average speed is doubled?" This is the best I've got, because the riddle is not asking for speed, it is asking specifically for time. I think this better emphasizes the meaning of the riddle without immediately thinking about just running three times faster.
Okay so I tested this on people around me, it encouraged them to think about trying to travel so fast that the average time for each lap is halved, meaning twice the distance in the same amount of time as the first lap, so the second lap has to be done in zero time, and is thus impossible.
@wO Don't worry man, I thought the same thing for a while. Even did the math on paper and everything. The mistake you are making is the same mistake that I made; while it is possible to reach an average velocity of 2V1, you have to run 3 more laps after the first lap to reach that average velocity. This is evident in your math, where you have to run 4km in 2 minutes to reach that average velocity. (first lap is 1km in 1 minute, second "lap" is 3km in 1 minute) However, in the question, you are only allowed to run 2 laps. If each lap is 1km, then you are only allowed to run 2km total. It's impossible to reach an overall average velocity of 2V1 when you can only run 2 laps. You have to be able to run 4 laps (total distance 4km assuming 1km per lap) to reach the desired average velocity, which is prohibited by the question.
@@nazimakhatoon8406 time is in account, but is a result of formula. So, however you are modifying speed, you have still the same distance. Now, to be still in the formula, whenever you are raising speed, you are at the same moment decreasing the time and vice-versa. Point is, that if you are counting average from TWO numbers, you are basically dividing by 2 the total distance and as your speed is result of distance divided by time, you always need to spend some time on the track, hence you always need to do t1 + t2. So, in formula d1+d2 / t1+t2, where d1+d2 = 2*d. And now, if you would need to have 2 as a result (v=d/t), you would need to have exactly t1+t2 = t1. And once again, that´s not possible, because t2 is always > 0, in a fact, it´s possible to make it only if you make 2nd round in exactly 0 time.
The flange on a train wheel is not STRICTLY necessary for it to not fall off the track. It's only necessary on sharper corners, when you hear that metal screeching sound. Most of the time the train stays on the tracks based on the conical wheel shape. The answer is still good though, since there's always a part of the cone riding below the level of the contact point, *just barely* though.
A train wheel flange DOES NOT stop the wheel falling off the track, it is only a safety feature. The wheels are CHAMFERED which is a very clever way of making the train self steer on corners - the outside wheel runs on the bigger part of the chamfer while the inside on the smaller part of the chamfer on corners. The flange never touches the track in normal situations.
Pulling the bike forward with the rubber duck tied, and then replaying the video backwards is the coolest way I have seen someone prove a point.
but pulling the string doesn't cause the pedals to go the other way like in the reverse clip... or am I missing something?
@wO you're not supposed to run at the two speeds for one minute, it's one LAP at each speed, That's how the hell it's impossible.
@wO Well here's the problem: at infinite speed, time (for you) will stop. In order to stop running you have to brake for an amount of time. You don't have time anymore because it stopped. So you're trapped at infinite speed forever. Also, you're everywhere at the same time.
@wO You should run only one more lap and make it look like you ran both laps in the same time you already took to complete first one. Average may work with any quantity but wont work with time as it is already spent.
coolest yet the cutest.
What if you refuse to run. Then, both laps are run at zero velocity. 2X0=0.
Mark Kmiecik illuminati confirmed
But can 0 be interpreted as the double of another 0 ? Maybe we should use a -0 that makes more sense x)
Only the lazy can achieve the impossible.
Peter Petyr roflmao
Then v2 is not much higher than v1 as he asked
For the people that are confused with the second riddle, I’ll show you why he is correct:
Let’s call the average speed of each lap: V1=D/T1, and V2=D/T2. Distance is unchanged.
The one many are using is Vmean=(V1+V2)/2, which is actually the MEAN SPEED for this case where you have two different speeds, independently of its value, you can find that Vmean=D/2(1/T1+1/T2), so you can actually solve that to find that if you want Vmean=2V1, then you must have V2=3V1.
This means that if you run the same distance 3 times faster than the first lap, the mean speed doubles the speed of the original lap.
The actual definition of average speed, involves the total distance which is 2D
in the total time which is T. Is easy to tell that T=T1+T2.
Now if you use this you have that
Vavg=2D/(T1+T2), and here is the tricky part. Since you want this to be 2V1 then by solving you have:
2D/(T1+T2)=2*(D/T1), So you’ll arrive to T1=T1+T2,which can only be achieved if T2=0.
This in short means that you run at an infinte speed in your second lap, which is physically imposible.
Remember don’t confuse MEAN SPEED with AVERAGE SPEED, they are only the same when the speed remains constant during the motion. I hope that this helps. Grettings.
Oh😅
Didn’t read what you said at all but I know that you put a ridiculous amount of effort into this comment so I liked
You’re awesome
@Lorenzo Gateau
This might make it more clear:
When you double velocity, you can either double the distance you ran and keep the time constant, or halve the time you ran and keep the distance constant. (V=D/T)
By running the second lap faster, regardless of the speed, you already ran double the distance. You went from 1 lap to 2 laps. This means that in order for the average speed to be 2 times what you did before, the time has to be constant (what I typed in the first paragraph). You would have had to run the second lap in 0 time
1st lap distance 1 lap + 1 lap 2 laps
________________. = 1st lap Velocity _____________________ = __________ = 2* 1st lap velocity
1st lap time 1st lap time+ 2nd lap time 1st lap time
You can see there that you can't afford to run the second lap with any time at all.
Now let's be clear. This problem is obviously *possible* but just without the constraint of running only 1 more lap. You could run 3 more laps in the time it took you to run the first lap, which means distance = 4 laps and time = 2 lap time, which is 2* the speed of the first lap. But you had to do MORE laps. By running just.1 more and having you velocity double, it is impossible.
@leipero You're talking average speed with respect to distance, while he's referring to average speed with respect time - in other words, what average speed actually *is*
Oh, of course
It's honey and ping-pong balls
It's so obvious
Is it clover honey or dandelion honey? Would eucalyptus honey work?
Yes
It is so “obvious”
hahaha lol
If you think abot it there is no real difference between the balls and just air
I remember the honey trick being used in some murder trick for some detective show I saw
The research group at Sydney Uni where I did my PhD wants to collect some data on these riddles to see if you learned anything. If you have 2 min, please complete this survey: ve42.co/Rresearch
Veritasium I'll be glad to participate.
+veritasium Done. Derek, you should pin this post.
I learnt that I nailed all 4, damn I'm bad! :p :p
I think it's a social experiment to see if people are honest to themselves, but to ask if we learned anything is still not deceptive
You should pin this to the top
"You guessed it was honey. It was actually honey....and two ping-pong balls!"
Oh, that Derek. What a character.
You could replace the ping-pong balls with air bubbles and have the same result!
@@scottsheffield6474 There's no way that is possible.
Brandon Ho so basically a half of jar of honey. Or does it have to be compressed air?
@@axemurderbambi2995 and half jar would not work, you need something to move the mass around when it stops the first time around, which means you need the empty space to shift, shifting the centre of gravity
Ronas Khoury thank you! Yeah, that makes sense.
4:08 "This may seem like a trick question, but actually, it's a trick question"
lol yeah
It’s not a trick question, it’s an inaccurate response to that question. He simply isn’t correct. Sorry
@@duplicantlivesmatter9083 why isnt he correct ?
@@robinschulz458 because you can decrease the time to get a faster velocity. 2V=D/(half) T. a.k.a. he math bad
@@his_mum no think about it...
Another way to think of the track question is this:
Imagine you and another person both start at the start line and agree to run a two lap race. You're out of shape and they're a very good runner, but you're slowly improving with their training and you like the challenge.
You line up at the start line, do a countdown, and then start running. You notice that the other person quickly gets away, easily outrunning you. Once you get to the end of the bend you see they've reached the end of the straight, twice as far as you! You then decide to only focus on your running.
Once you get onto the final straight of your first lap you're feeling pretty good, thinking to yourself that you can keep this pace up for the whole two lap race. As you're approaching the end of your first lap, you turn around and see them running you down! You and the other person cross the line at the exact same time, you having completed your first lap, or half the race, and them having completed the whole race of two laps.
How fast do you need to go on your second lap to tie them at the finish line?
Here it makes more intuitive sense that, since you both are at the exact same line with them having finished the two lap race while you've only completed one, you've run out of time to catch up to them to make your average speed twice that of your first lap, or 2V, which happens to be the very pace they were running because they're very fit and you're not. You would need to travel twice the distance in 0 time.
This is why it's impossible (mostly because you're a fluffy couch potato 🛋️ 🥔 ... run more 😉😅).
this actually made me understand it thank you
Well that makes sense. If you're running against someone. But i you yourself are making two laps and can choose to run however fast or slow you like. Can't you then just run e.g 1 or 3km/h for the first round and 3km/h or 9km/h for the second round?
@@jeremyxavier1939 You're welcome. 👍
@@sammyruncorn4165 You can do the math to find out that it is impossible.
Turn the track, instead, into a straight track that's 2 km long, with a line at 0 km, a line at 1 km, and a line at 2 km. We'll also use your numbers. You decide to run from the 0 km line to the 1 km line at a pace of 1 km/h, which takes you 1 hour. You now need to run to the 2 km line in such a way that your average speed is 2 km/h.
Using your numbers, you run from the 1 km line to the 2 km line at 3 km/h, which is ⅓ of an hour or 20 minutes. You have travelled a total of 2 km over the time span of 1 ⅓ hours. That makes your average speed 1.5 km/h.
Let's run from the 1 km line to 2 km line faster this time, let's say 60 km/h. You will get from the 1 km line to the 2 km line in 1 minute, or 1/60 of an hour. You have travelled 2 km in 1 and 1/60th of an hour for an average speed of 1.967 km/h.
Let's do from the 1 km line to the 2 km line at the speed of light, which is 299,792.458 km/s. You travelled 2 km in 1 and 0.000 000 000 927th of an hour. That gives you an average speed of 1.99999999815 km/h. It still isn't a 2 km/h average speed, even when your second half is the speed of light.
Are you starting to see that the question is a trick question where the math, if you plot it out with various speeds for the second half, involves an asymptote? You can approach a 2 km/h average speed, but you will never reach it.
Here's what the question is really asking: can you run a set distance and then cover twice that distance, in no additional time, so that your average speed is twice that of whatever speed you ran over the original set distance?
It doesn't matter what the original distance is as long as you have to then travel some multiple of that distance further and wind up with the same multiple faster than you travelled over the first distance. The distance could be 1.1 feet where I challenge you to travel the first foot at whatever speed you like and then travel the additional 0.1 feet so that your average speed is 1.1 times that of your first foot speed.
Your denominator, which in this case is time, needs to remain the same. You travelled 1 km in 1 hour, so 1 km/1 hour. Now you somehow need to double your speed to 2 km/1 hour. You need to double your distance but you have no additional time to use up. The key is that the distance is doubled but so is the average speed. The distance could be tripled and the average speed tripled and the trick would remain the same.
@@HungryTacoBoy thank you for the detailed explanation!
REVOLUTIONARY riddles... I finally caught the pun after 2 years of rewatching the video
?
@@anoijp8601 revolutionary as in, it revolves (goes in circles)
@@axxnub ooo thanks
Oh god same I feel so dumb because I didn't realize it.
So dumb
Came back for the bike.
it will land practically exactly in the same spot
yeah so unless I'm missing something the two balls would be spinning and moving exactly the same then when they are separated all of their momentum stays conserved as gravity slowly brings them back together again until they're right as when they started.
I don't understand what you're thinking here, the small ball separates from the big one but should remain in exact synchronization with it being that there is no resistance or anything else applied, then due to gravity the small ball and the big one both pull towards each other until they're stuck together again.
SmarterEveryDay awe
hey kid want a bike, it's in my van.
FYI: The flange on a train wheel is only necessary on extreme curves. On straight tracks and even gentle curve tracks, the taper on the wheel's tread creates a cone shape that expands towards the middle of the train. this has the characteristic of automatically centering the train wheels on the track as it travels. If the train wheels required the flanges to stay on the track, the friction would make it inefficient and impractical.
okay but even if theres only a taper, that taper basically works the way the flange does because some part will always extend futher than the point of contact
@@sophie6878 The difference is, the flange is a hard stop to keep the wheels on the track in an extreme condition, where the taper allows the wheel to track and auto-center without inducing unnecessary friction.
@@geoffschulz i know that but my point still stands.
that's interesting yes I can see the flanges would wear down fairly quickly if working hard all the time ,
Also, the flange on the train wheel is NOT moving backwards.
Think of a bulldozer track in place of the wheel of a car. When the track is on top, its velocity is twice the velocity of the dozer, and when it hits the ground its velocity is zero until it’s picked up again on the other side.
Great. Now think of a bulldozer track, but on a rail, and with a flange to keep it on the rail.
How fast does the flange move when the rest of the track is touching the rail? If it’s anything but zero, the track will tear itself apart, because because the flange is still attached and part of the track that isn’t moving.
The track run is the same as the bartender joke, first patron asks for a beer, second asks for 1/2 beer, 3rd 1/4... bartender stops them all and pours 2 beers and says, "you've got to know your limits. "
"I don't believe you. ."
Well, you've got to believe in something. . .
"OK. . . .I believe I'll have another beer🍺..."
Sorry dumb dumb here. Don't get it
@@dharmani_youtube If there would be an infinite amount of people asking for beer in the same pattern then the amount of beer in total would go against 2 beer (but never exactly hitting 2 Beer) . Kinda hard to explain, look up what the limit of a funtion is if you want it explained better.
I'm surprised people are so confused.
The "track run" question was the easiest question he asked!
@@gistfilm then answer it then
1:36 when you use the wrong formula but somehow stun yourself and the teacher by getting the right answer.
Relatable
That’s called a guess. Doesn’t work with big boy problems
But a good guess...
@@chrisgeorge84 what's wrong buddy. You got big boy problems?
@wO this is how my dumb brain thinks too and I want the answer to that too I want explanation.
I regret filling my bike tires with honey.
just add ping pong balls, it will fix itself
Now if you throw it off a balcony, will it move up or will it move down?
It will move to the left, clearly
Honey in bike tires would not do anything, as you are an outward force moving the pedals.
Filling your bike tires with honey is the only way to outrun a train rolling backwards uphill.
Simplifying problem 2 with no advanced algebra: The trick is that people intuitively think that 2 laps means 2 runs, where you clock each run. The solution is then obvious... But it's a trick question: he's not asking you to do 2 runs, he's asking you to do 1 run, that consist of 2 laps, where the total speed of the run is double the speed of the first lap in that run. Since the time it will take to complete the run will always be greater than the time it takes to complete the first lap of the run, the answer is that its impossible. It will always take more time to complete 2 laps than it takes to complete 1 lap in the same run. No need for any equation here. Common sense, trick question. So here is the simplified version of the question: Can you complete a whole run faster than it took you to complete half of the run? The answer is obviously no. Forget laps and averages and doubles and what have you. 😂
Your explanation is way clearer than what he says !!
that the problem, he tells about laps all this question. And most right think of it - Average speed it is a speed of first lap + speed of second / 2. Problem is in his wrong question explanation. So how he ask question - the wright answer is 3*V.
*Clearer question.*
u idiot, u run one lap speed x, u run 2nd lap twice that speed, easy, but not for dummies
@@pekvekIf the goal was the average Speed of both laps the goal you would still be wrong. Your avg speed for the second lap would have
to be 4 times faster than the avg speed of the first lap. However it is half the time it took to run the first lap that is requested. If the first lap took 2 minutes it would of course be impossible to do them both in 1 minute total.
Easy explanation for the running track riddle:
Let's say your speed is 1min for 1 lap. To double your speed over 2 laps, you would need to run 2 laps in 1min. Since you already took 1 min for the first lap it's impossible to complete another lap. This works the same no matter the speed.
*time is 1 min
But in my physics class, i learned that average velocity was just (initial velocity + final velocity)/2
We would get questions like, you travel with a velocity of 10m/s for 5 seconds, then a speed of 100 m/s for 3 seconds, what's your average velocity?
This is why it's hard for me to comprehend that it's not possible
@@MrMR-sk8jm See, average speed is a single value of speed with which you could travel a given distance and still take same amount of time as in the case where you will travel with different velocities. Actually that is the essence of average.
@@MrMR-sk8jm Even in your example, (v1 + v2) /2 will give 55m/s but to travel 350m in 8 second with a single constant speed, you will have to travel with 350/8 = 43.75m/s. And this is your actual average speed.
@@MrMR-sk8jm first lap velocity is 1 minute/lap. How fast do you need to run the second lap so that your average velocity is twice your first lap?
You need to attain an average of 30 seconds/lap. Since you’re only running the second lap, you’ve already reached your limit of 1 minute per 2 laps (by running 1 minute the first lap). If you were to run the second lap at 1 second/lap (60x the velocity), your total time would be 61 seconds in 2 laps, or an average of 30.5 seconds/lap.
Theoretical answer to the Running Question-
Let the circumference of the ground be 'd'
average speed = total distance/total time
therefore
let the velocity for first lap be v1 and velocity for second lap be v2
time for first lap be t1 and time for second lap be t2
t1= d/v1
t2= d/v2
total distance = 2d (circled the ground 2 times)
average velocity = 2d/(d/v1 + d/v2)
=[2d(v1v2)]/[d(v1 + v2)] (d gets cancelled out)
=2v1v2/(v1+v1)
now this velocity should be equal to 2v1
therefore 2v1=2v1v2/(v1+v2)
= 2v1(v1+v2)=2v1v2 (2v1 gets cancelled from both sides)
= v1 + v2 = v2 (v2 gets cancelled from both sides)
= v1 = 0
but if v1 = 0 the time will become infinite, so practically this is not possible unless v2 = infinite
But going deep into the terminology used
He said "what will be the velocity"
velocity is a vector quantity and it is equal to displacement/time
while circling a ground we are coming back to our original position so displacement = 0 hence velocity in both the cases would be zero
the correct terminology here would have been speed or angular velocity (omega)
Thanks for reading.
While writing avg. velocity,
you cancelled out a single 'd' with '2d'
Thanks that's just what needed
When you cancel a quantity from both sides of an equation, there is an implicit assumption that the quantity being cancelled is not zero.
Zenanov yeah only d would be canceld out. nigga please, learn maths first.
U should have subtracted 2v1v2 from both sides which would have left u with 2(v1)²=0.....
Nice, my video response was used. :)
I'm not in the description, but I'm happy to be part of the solution video. ^^
He should add you to the description. Your video answers all questions correctly and he even used part of it in his!
you should fight for your rights
everWonder - about the world? He should have added your video in the description...
Hey - sorry for the oversight. I very much enjoyed your video and will add your link.
Wow, just posted before going home from work. Didn't expect that.
It was no problem at all.
As long as people have fun learning, I'm happy to be part of it. ^^
And thanks for choosing my video. :)
5:20 Actually the flange isn't what keeps the train on the rails most of the time. The wheels are slightly conic. Feynman has a great video about this.
For those who found the explanation for the speed thing confusing,
the average speed of the person is the total distance divided by total time, or Vav = (v1*t1 + v2*t2)/(t1 + t2)
keeping in mind you are supposed to run only 2 laps, the total distance you have ran is 2d and total time is t1+t2 then Vav = 2d/(t1+t2)
since v1=d/t1 and we need the average to be 2v1 then we need t2 to become 0 which is impossible !
That's literally how he explained it. How does your comment make it less confusing for anyone who didn't get it the first time?
@@matrixphijr it did
That helped me a lot
Well I didn't (in general). I'm too dumb.
Can't one just walk like 3km/h for the first one and then 9km/h for the second so it averages at 6km/h.
The distance is the same for both lapses so doesn't matter?
That's what I thought, but I'm probably too unintelligent to get it...
@@sammyruncorn4165 Okay, so the key difference is that a racetrack is a fixed distance.
If you were driving on a highway, for instance, you could absolutely do that. Say you travel 30km/h for one hour. Now you’ve gone 30km, right? So in order to make your average speed 60km/h, you can absolutely travel 90km/h for the next hour to make the total average 60km/h. But remember, that means you will have gone 90km in the 2nd hour. On a track, you can’t change the distance traveled like that unless you run more laps.
Take that same highway scenario, but imagine there is only 60km of total road. Once you drive the first hour, there is only 30km of road left, _so you cannot just drive 90km to boost the average._ Even if you drive at 90km/hr for the second hour, you will run out of road after 20 minutes, and your total travel time will have been 1.333 hours to go the 60km (which obviously isn’t a rate of 60km/hr). In fact, it turns out that no matter how fast or slow you run the first lap (or drive half the road), your time for the second will need to be 0 in order to double your average - which obviously isn’t possible. Long story short, if you were able to run 3 laps after the first in the same total time you had run the first, that would double your average speed. But with only one more lap, it physically can’t be done.
@@sammyruncorn4165 Also thank you for literally making my point that OP didn’t explain jack squat lol.
No offense.
3:56
If you run at the speed of light, according to special relativity, in the runner frame of reference, it will take 0 time to go anywhere. So the average speed is indeed twice.
Speed of light is still finite. Not 0
@@sajinssha From a particular reference point, the time taken would be 0 due to relativity.
@@sajinssha Yes the speed of light is finite, and will be observed to be exactly the same speed in any frame of reference. Right up to the point that you actually REACH the speed of light. At that point, you can not "witness" the speed of light, as there is no light to witness... No light will ever be able to "catch" you. According to special relativity, if you were to travel at the speed of light, you could travel a distance of 5,000 light years, and exactly 0 time would pass for you. Meanwhile, 5,000 years would pass for us here on earth.
👏 of course you have to be massless
@@benwincelberg9684 What if my mass is greater than estimated mass of M87 black hole?
The running two laps - 2V = 2d/t problem is hard to grasp so I did some actual examples and I now see it.
Let's say your first lap is at 1 mph. You cover .25 miles in .25 hours (15 minutes).
Let's say your second lap is at 3 mph. You cover .25 miles in .083333 hrs (5 minutes)
Your combined half mile time equals .25 + .08333 = .3333333 hrs (20 minutes).
Velocity is distance/time so .5 / .333333 = 1.5 mph
This is NOT twice your first velocity - we are looking for 2 mph.
Skip ahead to running you second lap at 6,000 mph.
In your second lap you cover .25 miles in 0.15 seconds (.000041666 hours).
Your combined half mile time is .25 + .000041666 = .2500416666 Hours
Divide distance over time (like before) .5 / .250041666 = 1.9996667 MPH - Not quite 2 MPH.
As a matter of fact, you'll never reach 2.0 MPH no matter how fast you go.
I hope that helped someone. :)
Another way to look at it thru algebra: T1=time to run/walk 1st lap T2=time to run 2nd lap. D1=D2 =distance around on lap=D. V1= Velocity of lap one. So V1=D/T1 or T1=D/V1. Vave=2xV1 our goal. Vave=2xV1=2D(total dist two laps)/(T1+T2) or 2V1=2D/(T1+T2) . Divide both sides by 2 and get V1=D/(T1+T2) but we already stated correctly V1=D/T1. The only way V1=D/T1=D/(T1+T2) is if T2=0, in other words the time to run lap two is zero and V2=Infinity. Impossible.
This one was bs. It's not some confusing riddle just a poorly worded one. I thought it asked a speed such that the average of the two was 2 v1, not the average speed over the total time ran. Would have been able to give a much better answer had the question been clear
Oke I'm done ranting
Paul George I’ll be honest I was too lazy to read all of this
TY
THX
Finally i can feel smart for a change! Riddle 2 was easy with simple math and times of laps, not speeds. If you run a lap in 30 seconds, how fast should you run the second lap in order for the average of the 2 laps to be 15 seconds? Well... you should run the second lap in 0 seconds, so.. there...
@wO But don't forget you must keep with restrictions set by the problem. 2 laps is all you've got. So the second half of your run must have the same distance of the first half. If your 'lap' is 1km long, you only have 1km more to cut that average in half. Can't do 3km.
@wO
Bruh just acknowledged that you were wrong.
@wO not really, how did solve the problem? If he's incompetent like you then maybe he'd believe that someone finishes the 2nd lap with the average of 2V1.
@wO highly doubt that you can track me down considering how you failed to understand simple math concept.
@wO just take the L lol
For the track riddle,
Isn’t velocity displacement over time? Therefore, since the total displacement is zero because it is a loop and you end up back where you started, it doesn’t matter what the second lap is. 2x0=0. As long as you complete the second lap you have done it.
He corrected himself in the video description. Whenever he said velocity, he actually meant speed.
I'm looking forward to more riddles like these.
EEDude
i'm looking backwards!
wait... it depends...
2#you don't move...0X2=0 0+0=0 etc.
Lhr9scout10 or just move really slow...
doesn't quite work like that, you still need to complete 2 laps. So if you don't run 2 laps, you don't meet the conditions of the question, so it doesn't work.
Also, moving slowly doesn't work, because his explanation is scale-able.
Fine. Bring on the boring physics homework! My science will kick your dogma every time.
Lots of surprising answers in this one... but nothing baffles me as much as the fact it took me since last week to realise why you called those riddles "revolutionary". Damn puns...
Viniter you just made me realize.
Jamie Stivala me too xD
I still don't get it.Explain please??
Serene Shrestha
All the riddles have to do with revolution and rotation of some sort. So they’re...revolutionary, as veritasium put called it:)
And there is the true answer to all these riddles!
The bike question also depends upon the friction between the back tire and the floor. With insufficient friction, it is not possible to move the bike forward by pulling backward. The same would apply to pulling backwards on the flange of a train wheel.
I don't think so. See, for the bike to roll forward the pedal must always be travelling forward with respect to the ground. If the pedal actually moves back, the bike will skid back, with its back wheel rotating normally, as when it is moving forward. If the friction betwen the tire and the ground is too great and does not allow it, you won't be able to move the pedal backward. The bike will be standing still.
For the people who are confused by the second riddle, just use numbers!
Assume the following:
- The lap distance "d" is 10 meters.
d = 10 m
- Your speed in the first lap "V1" is 1 meter per second.
V1 = 1 m/s
- You have to run the 2 laps at an average speed "Vavg" of 2 meters per second
Vavg = 2V1 = 2×1 = 2 m/s
Remember that V = d/t
So:
t1 = d/V1 = 10×1 = 10 s
t(total) = 2d/Vavg = 2×10/2 = 10 s
Notice that the time needed for both laps (10 s) equals the time that you spent in the first lap alone (10 s)!!!
So basically, you have to complete the second lap in 0 seconds to be able to achieve Vavg = 2V1 !!!
Which is impossible of course.
Unless you can teleport..
But he didn't say anything about time... only speed. This is what's confusing to me. I listened to the riddle closely. He only said the speeds needed to average out to 2x the original speed. So if I walk at a pace of 1 mile an hour and then the second lap at a pace of 3 miles an hour, my average speed would be 2 miles an hour. There was zero mention of time in the original riddle.
@@bunny_0288 _"So if I walk at a pace of 1 mile an hour and then the second lap at a pace of 3 miles an hour, my average speed would be 2 miles an hour."_
Of course not. Your average speed would be 1.5mph for the two laps.
@@Stubbari Your mean speed would be 2 mph. 1+3=4 4/2=2..... However, I now realize he wasn't asking for the mean speed which is why that isn't the correct answer. It was a poorly worded riddle in the original video that was misinterpreted by many people.
Everybody who replied "my balls " to the 1st question got it right
What does that mean?
@@jf17thunder63 immature joke i made a year ago. Still find it funny tho
Utsav Manandharz Regret nothing
@@utsavmanandharz156 see you next year
@@utsavmanandharz156 lol 😂😆☠️ 🔥
I actually really like these types of videos, please do more!
I'm with you on that one. It's fun getting the juices flowing... in the head... not down south...
When I have more good ideas, I'll do more. It's tough to find problems that are interesting, that everyone hasn't heard already etc.
Veritasium i didnt understand the twice of the average speed of v1 riddle why was it wrong to do it mathematically?
Veritasium, I was wondering what you think about the idea of running the track (problem 3) on an inside lane. say it's 1k long. then the next lap, you run on an outer lane that is slightly longer, say 1.01k. then all you need to solve it is to run .01 times faster in the outer lane to finish in the same time and thus making 2v1. it may be a trick, but it's the same as the fog on the spinning floor. and still plausible. what do you think?
+Veritasium
Have you done anything on the latent heat of fusion/vaporization yet? Lots of people don't realize that it takes energy to convert the water into steam, even if the water is at 100C, or to convert ice into water, even if it's at 0C. Could also include how salt actually gets rid of ice, and how the whole system's temperature lowers even though there's the same amount of latent energy present.
For the track question, you never specified whose reference frame we're measuring the time in. So in the reference frame of the runner, if he moves at the speed of light on the second lap, it will appear to him as if no time passed by in the second lap.
Is there a way to run the track backwards (ie in the opposite direction as the first lap) at, I dunno, half speed of the first lap so that you.... never mind
Wait...if you run on the outside track on the first lap, then the inside track on the second lap then it is surely possible??
I'd like to see him try
I just watched the first video like "dang. Now I gotta wait a week." And then saw this in my recommended and realized it was uploaded 4 years ago haha
When it takes you 2 years to actually get the joke.
Which joke?
@@BESTofAlp "revolutionary"
all objects in the video "revolve"
@@cavemann_ ahhhh :D haha okay thanks 😄
3:56 : "Even if you when the speed of light..."
Actually, from your reference frame going the speed of light, no time will have passed going the second half of the distance. It really depends on the reference frame of the measurement.
Underrated comment. This deserves a spotlight.
@@MrLelopes yeah this is smart, but idk about you guys, I assumed that first picture of the train in the first video was the point of reference.
Well no time would pass for you while running at the speed of light, but for anyone stopping your time it will.
I think instead of having no time pass, you meant no distance pass because even having your reference frame move at the speed of light, that still takes up time.
Leandro Lopes looks like this guy didn’t understand a thing
technically speaking the velocity while running both the rounds was zero as the displacement was zero.
Prasad Sawant I love this comment.
Nice but he corrected that at 3:21 ;-)
Well you wouldn’t be running then..... youd be standing still
Thank you! I thought I was dumb
And even if you ran the first lap faster the average velocity of the two laps would be 2x the velocity of the first lap, since 2x0=0. Seriously every time he said velocity in this it's nails on a chalkboard to me. Speed, not velocity omg.
You actually can get your average to be 2V1 by running 3 times the speed but zig-zagging to reach the same time as the first lap
not quite what running a lap means but sure xd
I'm still confused by the "answer" to that "riddle". Nowhere in the question does time come into play. If run 1 mph, then I run 3 mph, my average speed is 2 mph. Twice the speed of the first lap.
@@TylerDollarhide nope, because then suppose the lap is 3 miles long, you ran a total of 3+3=6 miles in 1+3=4 hours. making your average velocity 6/4 or 1.5mph
@@TylerDollarhide The average speed of the total journey = total distance / total time taken. Let d = circumference of a lap, and t = time it takes to do the first lap. Note that the average speed of the first lap is v = d/t. Now, the issue is that you MUST take t amount of time to run the first lap, then you need to run the second lap in whatever time is left (call it T). So, average speed of the total journey = 2*d / (t +T). But he is saying that you need to choose T so that 2*v1 = 2*d/t = 2*d/(t+T). But then T = 0. So you will need to "instantaneously" jump the last d distance. This means the speed of the second lap must be infinitely fast to spend no time covering it. Another way to see this problem is to draw a plot. Call the y-axis distance, and the x-axis time. If you draw a line from (0,0) to (t,d) (representing the first lap), then a line from (0,0) to (t, 2*d), you will find that the final side of that triangle goes vertical, leaving no additional time to get from d to 2*d. I don't know why he didn't draw the plot, it's much easier to see.
@@TylerDollarhide Time did come into play. From what you said, 1 mph and 3 mph, the "miles per hour" part already gave you the time component of this problem. 1 and 3 does indeed work if the laps are independent of each other, because now you are confusing the average speed and mean speed. Average and mean are very similar things but absolutely not the same.
You're right that the flanges are to keep the train on the track; but only a last resort, they're not designed to contact it - they'd be moving relative to it so they would wear and waste energy. It's the fact that the wheels' rolling surfaces are slightly wider on the inside that keeps it on the track normally: so the train is lowest when it's central and gravity corrects any deviation.
Incidentally, it's the conical wheels which mean trains don't need differential gears. In a curve, the train is thrown out, so the outer wheel contacts the rail at a point where its diameter is greater (and vice versa)
Thank you. I came here to say the same thing
All us rail buffs will know that, but you might mention that it doesn't change the physics of the riddle.
that was my only nitpick... though I only learned that from watching a video with Feynman
Cool train info. Thanks.
At 6:22, there is an incorrect statement. The explanation is as simple as "the net force is backwards so the bike goes backwards."
The complication is that the pulling force isn't the net force. The net force is the total of all of the forward friction and the backward pulling force. Whichever force is stronger will be the direction of the net force.
good catch!
+Veritasium I am posting this here as I was particularly bugged that no one seems to have addressed the importance of torque and rotational dynamics in answering the bike pulling problem. The trochoid may provide a geometrical solution, but it has little predictive use if one had to calculate the acceleration of the bike with respect to the pulling force and gear ratio.
As per net force, I'd say it has little to do with the problem at all. Assuming an infinite coefficient of static friction, regardless of the "net force", the bicycle would remain stationary; in this system, motion is only possible if the state of the system permits the rolling action of the wheels. Torque must be given consideration as there would otherwise be no way for the bike to move even kinematically. Regardless of the shape of the trochoid, there must be a condition at the point of rest where equilibrium dictates the acceleration of the bike; the derivative of the trochoid at that rest point would yield zero and no indication of the expected direction of rotation. Newton's "Second Law of Rotation" must be adhered to.
In Armchair Explorers' video response in /watch?v=S1yX_LTqtms, I posted a comment mathematically deriving equations relating the _net torque_ to the dimensions of the bike and the applied force; I didn't have the resources to fashion a response video then. In the replies to that comment, I rearrange my equations and derive the same formula shown at 7:41. Just from first year first term university physics, I immediately saw it as a "basic" rotational dynamics and torque equilibrium problem.
Torque is perhaps quite complex to most, but as you have in the past addressed misconceptions about translational forces, it might be an excellent educational opportunity to address misconceptions about the dynamics of rotation.
Otherwise, these have been very thought provoking "revolutionary" riddles and I admire your work.
+Dan T The friction force is immensely important here, though the friction coefficient is not. Static friction is a useful force because it can have any value up to a limit. Since it opposes relative sliding, we just attempt the sliding and static friction pops up as a constraint force to prevent it. As long as we don't ask for to much static friction force, it's okay.
Torque balance is certainly important to determine how much force is being exerted. The pedal arm length, gear ratio, and wheel radius all play a part.
+Jeff S To assume perfect rolling in this system "independent" of the applied force, we must also assume the parameters governing static friction (coefficient of static friction and normal force) to be as large as possible. An infinite coefficient of static friction simply sets for our model a limitless supply of a reactionary force to keep the wheel kinematically constrained to perfect rolling, assuming that sliding/kinetic friction is out of the question for possible solutions; in our case, it provides a limitless supply of counterclockwise torque moment upon the rear wheel proportional to the applied force at the pedal.
As per torque, I'd argue that it is _much more important_ when discussing any form of motion at all involving rotational elements; "static friction" is but a variable constraint applying an opposing torque to the torque imparted from the pedal. The bike _will not move_ unless the torque moments imparted by static friction and the applied force are unbalanced. _net torque = sigma_tau = tau_1 + tau_2_ where angular _acceleration = alpha = I * sigma_tau_ .
Torque balance is not about determining "how much force" as opposed to whether the bicycle wheels will undergo any rotational acceleration leading to translational motion at all. If you would review my analysis from my comment in /watch?v=S1yX_LTqtms, I calculate how these torques play out and eventually arrive upon the same relationship shown at 7:41 of this video.
This is all from university physics.
I had the Force/tourque/radius explanation in my mind for the bike gearing problem. Which is a correct explanation, no doubt. But I really liked the purely kinematic explanation that you showed! Kinematic solutions are elegant!
Now I Understand why The Title is "REVOLUTIONary Riddles"
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What have the ping balls got to do with the first one? It's just air, and you get the same effect from without using them, just your container was smaller so it was a bias demonstration.
There might be a small difference. The honey sticks to the walls of the cylinder and the behaviour could change over time as it rolls down the slope compared to the initial state of no honey at the top of the cylinder and the bottom half full of honey, where as the ping pong balls always behave as spherical bubbles.
You need to keep the air bubbles together. In honey, air would float to the top and release/break through the surface. The only way to get this to work is to have the honey continually push the air bubbles down, which float back up by themselves.
I'm an aeronautical engineer, i graduated from the air force academy. Most of your videos are right but what you said about running the 2 laps is VERY VERY debatable if you consider that we move through time
does bolt measure his sprint times from his own perspective? 🧐
"... it does behave similar, and that's no coincidence; the mystery cylinder actually contains bees."
Love trying to figure it out. But - failed, miserably. 1 out of 4. but I think I actually won since I was really thinking and trying to figure it out - not a normal activity for 70 yo.
Wow - thank you for giving the riddles a shot! I think I got some of these wrong the first time I heard them.
Especially the bike riddle was tricky.
Veritasium , this time it wasn't clear , concise and easy to understand . It did challenge my early beliefs and sort of intrigued my interest for rotational dynamics
Actually , that ping pong ball , that's something I can't grab by books , It gave me a big BONER
You may not have won, but you definitely one.
Getting something wrong, is also a victory, a victory of better understanding.
The flanges don’t actually play that much of a role in keeping the train on the track. In fact, most of the time (read: in normal operation), they don’t touch the rail at all. It’s the geometry of the rolling part of the wheel (it’s cone-shaped) that does the trick.
good note!
What's the name of the program you used for train wheel simulation?
Ast A. Moore so are they just a safety measure?
Pretty much. Just to make sure the train doesn’t get derailed if there’s a track defect or in case of an emergency situation (like something bumping into a train sideways).
Not only for safety. In turns there is a cant so when train turns to the right, right rail is lower them left. Without flanges, it could slip from the track if it would have to slow down on that part.
Even being non intuitive, the lap riddle is a very common high school kinematics intro question.
average speed= total distance travelled /total time taken
Let time taken for
Lap 1:t1
Lap 2:x
And d be the distance of 1 lap
let d/t1 be v1
avg speed=2v1=2d/t1=(d+d)/(t1+x)
This leaves
t1+x=t1
Obviously x=0 s therefore speed in lap 2 must be
d/0 =infinity.
The track riddle can be done! You gave us an object to run around, not a distance. Assuming the track is a standard 400m, if you spend your second lap running in a 5.2m wide x 2m long zig-zag, you could effectively run 1200m on your second lap, adding making it take just as long to run thrice as fast, equating to a 2v average.
The answer to the running track question really depends on how you choose to interpret the question.
Q: How fast do you have to run the 2nd lap to get an average speed of 2x v1?
If you choose to average over time, there is no answer. If you choose to average over distance, the answer is 3x v1. If you are tasked with finding _the_ answer, then 3x v1 would certainly have to be seen as correct.
EternalSilverDragon That’s what I was saying. Such a riddle can be answered by simply doing it in your head. But this particular video was filled with nothing but cheap shot riddles. There was no way in holy hell we could have come up with the answers that is channel wanted us to come up for these particular riddles.
I agree with both of you. I wasn't expert enough to disgree, but I felt strongly that his explanation was just silly. I knew for a fact that we could get a reasonable average, if we assumed that there was no need for acceleration...because the time cost was the exact solution to a grade 11 Physics problem that was introduced to my class in high school.
That being said, I still don't understand why his interpretation is what it is. It sounded like he was limiting the amount of time of the second lap to 0 seconds, which seemed so arbitrary.
@@eugenetswong because it was on.ly two laps. that is the key part to this riddle. if it was three laps then it would be possible. You have zero seconds in your send lap yet that is impossible as even the speed of light would take even a nano second which is more than zero which is why you cant double the average speed.
@@briandarlage9815 that sounds like what Derek says, and what we've been objecting to. Giving a person 0 seconds to do the second lap seems very arbitrary to me. Why do we get 0 seconds?
But physically speaking, you do not average speed over distance. V1 + V2/2 is incorrect in this case because time is the decisive factor. I actually did the maths on my phone, let me show you.
(x/t + x/y) /(t + y) = 2x/t
2t + 2y = 1 + t/y
2y - 1 = t (1-2y/y)
Let t be the time in which you run the first lap. X is the distance you run and y is the time in which you run the second lap. The last equation is as follows.
2y - 1 = t (1-2y)/y
-1 = t/y
So you have to run the second lap in the negative amount of time that you ran the first lap in. Negative time is physically impossible.
And I disagree that the video had cheap tricks. They were all sensible and I personally got all of them right minus the two ping pong balls in the honey because that's way too specific of a guess.
You got the wrong answer for 3. :D
The correct answer is 0. Lay down and have a nap. Don't go anywhere. Twice zero is still zero. You live at the starting line now.
but it has to be 2 times faster than v1
Lil Schnoze 2 times 0 is still 0
You also die at the starting line as you have just imprisoned yourself for an eternity for such a silly reason, you cannot stop now or all previous time spent is wasted, hopefully some kind stranger will come by and provide you with food daily and if lucky they will even put up some form of shelter around you.
But can we truly call that living? Is being stuck in one spot for eternity truly worth the results? For some maybe but not me, i choose to live my life even if it means i am a failure.
But you haven't done any laps, so it doesn't work
What if he walks it backwards and makes NEGATIVE time!?
YES!! I said gearing needs to be low enough AND that the pedal will only pull back/bike move forward until the crank rotation causes the "acting" force vector to reduce with angle until equilibrium is reached. Then consider friction and the fact that the string has changed angle such that you are also pulling the bike downwards thereby slightly increasing this. The extra crank angle inflicted to pass static euilibrium (overcome friction) may further affect the second euilibrium crank angle that will result when skidding.
You had me at YES!!
that question about the rolling cylinder...
Could be there's also a weighty metal sphere inside. That would also work! :)
you could zig-zag during the second lap, thus doubling your distance traveled.
Yes thankyou finally someone with a brain, I don't know how many comments i had to go through to see someone with an actual right answer. His original question never specified the laps were the same distance, thus it is NOT impossible.
Aris Nikoletopoulos In fact you wouldn't be dobuling your distance, you would be changing you trayectory. The distance is a rect line between point A and point B. If you ran to the middle, went back to point A and then ran back to point B, the distance would be the same as if you ran in zig-zag from point A to point B.
Mateo Costa the distance between them will remain the same, of course; the distance you travel however will increase. Velocity is calculated using the distance traveled, not the distance between your start and finish point, or am I wrong??
Mateo Costa, that's (net) displacement, not distance.
no, speed is calculated using distance traveled. (s=d/t)
velocity is calculated using displacement (v=x/t)
however, because this track is circular - displacement is actually zero (start point and end point are the same) and therefore average velocity is also zero (v =0/t). which is why in this video he corrected velocity to speed.
but then people want to try be 'smart' by arguing semantics and say 'oh well you never said lap distance is constant'. okay fine, well done, he didn't say distance is constant but that's clearly what he meant. if you zig zag in a 100m sprint, do you get to call it a 300m sprint? no. end of the day, the question is supposed to have distance fixed. if distance isn't fixed then youre asking a different question. so now that the clarification has been made, can you solve it?
Awesome video as always. I've just recently started my own science channel and I'm happy to say I found inspiration watching your channel and many others. Thank you and keep up the great work!!
Steam locomotives also have a rod called an eccentric rod which helps "change gear" by increasing or decreasing the movement of the piston inside the cylinder. The rod always moves backwards because of how the piston works
But does it move backward in relation to the ground? Which was the question.
I think not.
@@RR67890 you know, wheel flange either dont move backward relative to the ground. Question was wrong explained. One little area on flange which position on the wheel you need to change in your imagination to say it moves backward is not a part of the train.
@@Guapter I think you've misunderstood the explanation. You seem to be referring to the contact point, where the wheel doesn't move with relation to the ground. However, with a train wheel, the flange is a larger, concentric circle that must match the rotational speed. The following image isn't perfect, but I think it can help get on the correct path to understanding: Imagine drawing a line from the center of the wheel, through the contact point with the rail and to the flange's edge. This line functions a bit like a lever with the pivot point being the contact point between the wheel and the rail. When you move the center of the wheel forward, the pivot point remains stationary and the point at the flange moves in the opposite direction to the center of the wheel.
1 - The rod can't always move backwards. It would have to move forward as well.
2 - It would always need to be moving backwards with respect to the ground while the train is moving forwards, which isn't possible.
I consider the track problem to be a trick question. Your solution treats the two runs as one run with a slow half and faster half. Alternatively I, and I am sure many others thought of it as two separate runs in which case it is perfectly possible to find a solution without having to exceed the speed of light. This I suppose is due to the different perceptions between the pure analyst and the empiricist.
The bike problem again depends on initial conditions, which were not specified. The compound vessel with its viscous liquid and variable geometry is only one of many different configurations which would all exhibit analogous behaviour.
What this whole exercise illustrates very well is the importance of experiment as opposed to the ancient belief that real world problems can be resolved totally by the intellect. Aristotle certainly believed this and was accordingly guilty of passing on many erroneous explanations of real world phenomena [ example : the falling body problem ]. This is one of the reasons why physicists can be dangerous if given jobs better suited to engineers. An engineer is always ready to run a trial before committing to a final solution.
3:25 that implies that your average speed is measured over *time*. If you were to take the average speed over distance it is definitely possible.
Speed (Velocity) = Distance / Time...If it didn't take time to cover distance, it'd be easily doable but that pesky time stops for no-one.
You are calculing your average value like: Avg Value = ( Value 1 + Value 2 )/2.
The problem is that speed is the product of dividing distance and time (d/t), so the average velocity is calculated like: Avg Velocity = Total Distance / Total Time, so in that 2nd problem if you wanted the average velocity to be equal to 2 times Velocity 1, then you had:
2 V1 = (d+d)/(t1+t2)=(2d)/(t1+t2) and V1 = d/t1 so 2 times V1 equals 2d/t1, so:
2d / t1 = 2d/(t1+t2)
For making that equation equal, you need to make t2 = 0, thats why its impossible to make the average velocity equal to 2 times velocity 1 (V1).
Huep JR why couldn’t you make V1*2 = d/(1/2)t1
@@HAHA_468 Well, saying 2*V1 = d/(1/2)t1 is basically saying that you can double V1 by running the distance of V1 in half the time, but it doesnt say anything about the average speed, if you wanna desenvolupate the average speed into other things you need to start from Vavg = total distance / total time. if you start from another equation then your results arent "talking" about the average speed but about obvious things like "To double the speed you need to run the same distance but half the time".
Huep JR thanks for the clarification 👍
I'd like to point out a technical mistake you did at 5:23.
Flanges in train wheels are used only for switches on the railway. The reason why the train doesn't fall off the rail is because of the truncated cone shape of the wheels which makes them act as an automatic centerer :)
The other part of it is he said ground. While the tracks sit above the ground. Thus the wheels are still moving forward in respect to the ground because they don't go UNDER the ground they go under the track on top of the ground.
+Michael Brune If the track and the ground are both stationary (they are), then ANYTHING which is stationary with respect to the track is also stationary with respect to the ground.
You are model
It's a little bit of both... conditions can vary, such that the cone shape doesn't do enough
no1unorightnow lol is that your gut speaking or do you actually know about it. Because I'm a civil engineer and I've been taught what this guy said. so unless it's just you trying to apply common sense here, if you actually have some practical knowledge about the wheels then do tell.
The lap trick is not impossible. Imagine you’re in a car, and you finish your first lap of 1km distance in 60sec. That means that the speed of your first lap was 60km/H. So you have to adjust your speed during the second lap to end up with an average speed of 120km/H at the end of the 2 laps. Then, you just drive at 180km/H during your second lap and your average speed over the 2 laps will be 120 km/h. Thats twice the speed of your first lap
Since you've run two 1-kilometre laps, your total distance was 2 kilometres. The first lap took one minute, and since you went three times as fast the second time, the second lap took 20 seconds. So your total distance is 2 km and your total time is 80 seconds. Average speed = total distance / total time = 2000 m / 80 s, so your average speed was 25 metres per second, or 90 kilometres per hour, which is only 1.5 x the speed of your first lap.
At 4:10 , u can say, in a sense, that the condition for the riddle can be fulfilled if a person runs first lap with initial speed(v) v--->(lim) 0. So the condition will be fulfilled for any velocitu v2 with which we run the second lap
one thing to consider is if your velocity is zero (you don't move). then next lap your velocity is zero and zero is double of zero. ;)
and another one would be if you do the first lap at the speed of light, the second one would be double as C is constant and 2C=C, no?
and another one if you do the first lap backward (velocity -1) and the second lap forward, the speed would be double? :D
when you go speed of light you do not have time, so question is who measuring time in this, runner or outsider?
@@NMTCG I doubt you can do the first lap backwards because when I tried to calculate it, I got a concrete v1 = 0 with no other answers (i.e. exactly what I'd do if someone was forcing me to get off my ass and do a lap)
None of these suggestions work. You need to go around the track to complete a lap which requires you to have a velocity greater than 0. -V is the same as a forward V, all - tells you is the direction
interesting answers, (thanks for using my video response)
thanks for making it!
Would make more sense if he said "Weighted Average Speed".
Arithmetic average can easily become 2V if V2 = 3V.
Average speed means total distance dividided by total time.
and it is quite obvious
suppose you travel in train from one city to another which are 500km apart and the train takes you 5 hours in total then what will be the average speed of train according to you ? (Definitely the train would not be running at 100kmph all the time, sometimes it may move with higher speed and some times it may move with lower speed, but what matters to you to get average speed is the total time taken and total distance travelled)
What are you talking about lol :P We all know what average speed is.
If you are taking a loop track, you'd have to measure speed, not velocity, thus if you changed your total path length (i.e. by running backwards, then forward, then backwards, then forwards, or zigzagging) then you could plausibly solve that riddle.
A very small correction: The train wheels flanges are not intended to keep the train on the tracks, and they usually don´t make contact with them. They are used only when going over a set of points (BE) or a switch (AE). The train is kept on the tracks (even on curves) by the special shape of their top surface and the conicity of the wheels. Interesting video, btw!
My mind exploded when I realized why "revolutionary" riddles! LOL
oh man
For the bike riddle, more illustrative might be the case of a spool of string with raised, circular ends (standard gear, not extra-low gear). For all such systems, there exists an angle
if you reframe the track one in terms of time, it becomes a bit more intuitive: you want to run the second lap such that the average time across both laps is half the time of the first lap. But if, say, you walked that first lap at five minutes, then youd need to get the average for the two laps to be two and a half minutes. Which it already is before youve even started the second lap, so you need to do the second lap in zero seconds, or infinitely fast, regardless of how fast or slow you did the first lap.
The ping pong balls are irrelevant. The (lower-density) air inside them is what’s important. The jar half-filled with honey is close enough to be pretty much equivalent. I think it should count as a correct answer.
Sort of yeah, but the ping pong balls creates a vessel for the air which moves with the honey easier that just air, as it'll easier affect the position of the honey than the restricted air in the ping pong balls, although i agree, it should count as the correct answer, as it's has a veeeery similar effect, which would be hard to differentiate without seeing the content.
I feel like if our guess was 10w40 with hollowed out, sealed eggshells, it would still be wrong because the eggs are not shaped like spheres. While the ping pong balls offer more resistance to the fluid dynamics, a more viscous substance would yield an even more similar result. From the perspective of visual observation, indistinguishable!! I fart in your general direction on that one.
The ping-pong balls aren't irrelevant since the displace the honey in a more irregular shape. It is a very good approximation though.
@@wilfdarr You are wrong. The shape of the balls is irrelevant. Replace them with balls the same density as the honey. It's the liquid and the air that matters here.
@@thetruth45678 Well, since you went there, No you're wrong. The tennis balls don't simply displace the honey but displace some of the honey VERTICALLY which forces the C of G higher in the cylinder while maintaining the total weight of the system, which you could not do by just adjusting the honey/air ratio. This high C of G is what makes the movement so dramatic. Good approximation, but not the same.
Maybe next time try words like 'I think', 'I can't see how that would be' or 'Why wouldn't X work just as well?' rather than 'You're Wrong'.
I'm not buyin' the track riddle solution. 2d/t1 is the same as d/[(1/2)t1]. It's possible to run the same distance in half the amount of time. There's no need to "weight" the velocities in the average. Run the first lap in time t1, run the second lap in time one-third of t1. I think this can be shown pretty easily on a position vs time graph. You'll have one slope m1 for the first lap from y=0, t=0 to y=d, t=t1, then triple the slope between y=d, t=t1 and y=2d, t=(4/3)t1 for a slope of m2 The average of these two intervals. The average of these slopes will give a velocity twice as large (steep) as that of the initial interval.
I thought that you need to plot a velocity-time graph to find the average velocity? it should look like two bars next to each other where the height = the gradient from your graph
Great video, but I believe it is actually the taper on the train wheel surfaces (which at not 90 degrees) that keep the train centered in the track, not the safety lip. The lip is just a backup safety measure.
If the flanges did come in to contact, you'd hear awful screeching noises. The flanges do keep a model train on the track, but that's just because the speeds and accelerations are so slow that the screeching noises go unnoticed, and most likely, it isn't worth the effort to machine exact scale models of a real train wheel.
I have to say that 2, 3 and 4 are trick questions:
2. The question asked was what speed would need to be run, not velocity. The two are not the same.
3. The part of the train that moves backwards would be the pushrod pushing the wheels on the backstroke, which answers the question precisely. The flange is part of the wheel, which you called, 'the flange part of the wheel', not the flange part of the train. Technically it would be a part of a part.
4. Using any bike introduces inconsistent characteristics that should be taken into account.
i really enjoyed it! :D one of the most valuable channel on youtube! Greets from Poland!
Train: Any gear/motor/engine that is rotating faster than the wheels involved in speed reduction/torque increase...
For the 2nd riddle, you could run zig-zags across the lanes for the second lap such that you run 3 times the distance than the first lap, but also at 3 times the speed. The net result will be 2 laps with an average speed of twice the first lap. You're welcome.
3:20 but the time doesn't matter the velocity needs to be doubled the time it takes doesn't change the velocity
He switched the puzzle for running the track. He asked the question based on velocity, but answered it based on time.
Velocity is distance over time. The distance doesn't change, therefore only by changing time can velocity change
I definitely thought the running around the track one was just a verbal typo that didn't get caught in editing. It would have been a lot more clear to say "the value of" instead of directly saying to compare speed and time.
I tell the track riddle differently. It's my favorite math trick question. You have a hill 1/2 mile up and 1/2 mile down. If you go up the hill at 30 mph, how fast do you have to down the hill to average 60 mph..many people say 90 knee jerk. But as you know you'd have to instantaneously be at the bottom. Your time is up at the top.
Yea but it's not a math problem, it's a communication problem. If i run a lap at 1m/s average speed and then a second at 2m/s, i would say i doubled my average speed. But if you consider it the same time series you only have a increase in average speed by 1/3.
So if you ask an athlete for his average speed over a set distance such as 100m sprint for a given practise day, he will average the average speeds not average his total time spend on the track.
@@StillTryingMyBest Or those people were just confused as to why it is supposed to be a riddle.
I think the question would be better if it included some kind of hint that the two laps are to be considered as one single measurement. But maybe you can't understand, because it was so obvious to you
In simple terms the leverage applied through the gears on a regular unmodified bike is not enough to overcome the force required to make the wheel turn. Aka you have to pull backwards harder than the tire can push forwards to make the wheel spin.
Nice video, cool riddles. That said, I have an argument about the flange: the flange is going forwards with the speed of the train: the part of the flange that goes backwards is not the same at all times so there's no part of the traind that is *always* going backwards - otherwise, there would not be much train left after a while.
in 3:45. why not use instead of 2d/t , d/(t/2)? you can't increase the distance but decrease the time it's possible.
If you've already run a lap, any more running will increase your total distance, not decrease it.
Exactly. Veritasium is wrong on this one. By increasing the distance by a factor of 3, you get 4 laps in twice the time, which reduces to 2V1.
If v1 = 1 lap per minute, and v2 = 3 laps per minute, then total is 4 laps in 2 minutes, which is the same as 2 laps in 1 minute (exactly twice v1). It's 100% doable, but Veritasium goofed big time on this one.
Porglit
no...
you cant add speeds like that to get 4 laps, because you only need to run one more lap.
what you did is equivalent to adding fractures by simply summing the numerators apart and the denominators apart...
this way 1/2+1/3=2/5
@a b:
Ah. You could eventually get up to 2V1, but it wouldn't be in just the next lap. Good point.
2d/t is the result you want. t is defined as the time it takes to do the first lap. How are you going to double distance but keep t the same? t1 + t2 = t1 For that to be true t2 has to equal 0, so you can't do it on the next lap.
I have a question-
If a person is about to enter a black hole (he has not entered yet but is very close to the entrance point) would an observer from far away from the black hole see a image of that person entering the black hole like he is paused. If yes then would not there be sort of 2 existence of the same person somehow. One is what we see at the entrance of the black hole and the other that person is expireincing.
Clear my doubts anyone
Does someone else thinks that the bicicle explanation is incomplete? Please, correct me if I'm wrong, but from my point of view it should also depend on if the tires are able to spin with or without sliding.... so... with a lower ratio.... the bike would always move forward or not move at all, just depending on the tires glide
I think the ping pong balls operated roughly similar to an air bubble. I don't see why the distinction is important.
Probably just that the ping pong balls always hold their shape so their behavior is more predictable
Because the distinction exists. Don't be mad just because you weren't right - be happy that you came close to the right answer and now realize how you were wrong
I don't realize how I was wrong. I don't see how it makes a difference to the movement of the cylinder. He said it was close, I think it was actually the same. I don't think I was wrong, or at least I'm not ready to concede that I was until I see it properly explained why it matters that there were ping pong balls in it.
I agree. If the correct answer is not air, but ping pong balls, tell me what the difference is in that scenario. The little bit of plastic? Not good enough.
balls might move in a different way than air bubbles because of friction, but that's just a guess, and rather a bad one.
If you are running a circular track, your average velocity would be zero for any lap, since the first and second half vectors would cancel each other out.
Thus to get an average of 2 times zero would mean that you could literally run the second lap at any speed slower or faster or even the same as the first lap and still get an average that is twice the first lap.
Anything times 0=0, including 2.
Hence the annotated correction from velocity to speed.
@@steves9388 Sorry, but if he's too "dumb" to present a riddle hinging on semantics to "fool" the audience in a proper way (it's a video, he can edit stuff and not just make a split second text overlay), I am perfectly fine with anybody giving "not intended" solutions.
We have 2(t1)(v1) = {(t1)(v1) + (t2)(v2)) / 2
Which, under normal circumstances where (t1)(v1) = (t2)(v2), can only be solved where (tn)(vn) = 0 or infinity.
However, if we consider speed from the perspective of the runner and include time dilation, then a(t1)(v1) != b(t2)(v2) where a != b, and a and b are factors of time dilation.
2a(t1)(v1) = {a(t1)(v1) + b(t2)(v2)} / 2
Otherwise our rule holds that (t1)(v1) = (t2)(v2), therefore:
2a = (a + b) / 2
3a = b
Therefore, where the time dilation of the first run is 3 times the time dilation of the second, e.g. first lap at ~0.943c and second lap at ~0.001c
We have a total average speed twice that of our first lap.
we can actually run the second lap with twice the distance of the first. If we run the first lap in the innermost strip of track and the second in the outermost strip with having the same time.
Oh look. It's Dirk from Veristablium!!
You mean Duke from the Vatican
Sudzy Soap what's Veritasblium?
Sudzy Soap is this supposed to be a variant of the buttercup cabbagepatch memes
Alain Gasco no, listen to the Podcast "Hello Internet"
It's Brady Haran of Numberfile fame, and CGP Grey, they call him that.
it's Draco from Veritaserum!
I see the error of my ways, I was calculating the average speed per lap, not the total speed.
Put simpler (I think his phrasing was confusing):
0 = overall
1 = first lap
d/t1 = V1
2d/t0 = 2V1
=> t0 = t1
=> t2 = 0
You can't run a lap in 0 seconds.
Actually it was well phrased, I got it right away, but yes, it may have sounded as a trick question.
Paul George had a great answer. But here is the same explanation using only algebra:
T1=time to run/walk 1st lap T2=time to run 2nd lap. D1=D2 =distance around one lap=D. V1= Velocity of lap one. So V1=D/T1 or T1=D/V1. V(average)=2xV1 our goal. V(ave)=2V1=2D(total dist two laps)/(T1+T2)total time or 2V1=2D/(T1+T2) . Divide both sides by 2 and get V1=D/(T1+T2) but we already stated correctly V1=D/T1. The only way V1=D/T1=D/(T1+T2) is if T2=0, in other words, the time to run lap two is zero and V2=Infinity. Impossible. Unless as Mark Kmiecki said, you refuse to run and V1=0. But if you refused to run because you know the goal is impossible, by refusing to run you prove yourself wrong because at V1=0 2V1=V(ave)=0 is true and you just sucked yourself into a cosmic paradox.
The bicycle riddle relies basically on the same principle as the wind powered car that can go faster than the wind, straight downwind. The gearing can multiply the force, and apply it in the opposite direction, or in the same direction.
"The string is always moving forward." So says science, and anyone who has ever ridden a bike with loose shoelaces.
3:58- even if you went the speed of light,
If you ran with the speed of light, you would have infinite time because of the time dilation,
So that makes it possible.... ❤️
No.. know your reference
3:45, you can run the second lap in half the time and that would give you d/0.5t1 which would make V2=3V1 and Vavg=2V1. Am I missing something here?
Exactly what i was thinking.
I'm surprised I had to scroll this far to find someone else who felt the same
The answer you have given is the average of two separate velocities, but you have forgotten they have different denominators (the laps are taking different times to run).
With the equation for velocity (distance / time) we know that the distance is the same for both laps (i.e. 2d). Therefore to make this double the speed of lap 1 you need to keep the denominator the same i.e. Lap 1 = d/t so 2d/t can only be double the speed of the 2nd lap if no additional time is added into the equation. That is why adding in the speed of light (and faster) will always add some time that will make it just under double the speed of the first lap alone.
Vavg is using V1 as reference, so in order to complete the challenge you have to run *both* races within *t1* , amount of time that is only enough to do the first race. It's just a pedantic trick question.
Okay, so let me try to understand it in a different way. I run first lap at 5 KMpH and second lap at 15 KMpH. What's the average speed with which I completed one lap? If the answer is not 10 KMpH then please let me know what it is.
I can't get my head around a better way to ask the second riddle. The wording really made me want to just answer 3V1 but I can't think of a better way to word it. If I were to propose this riddle to a friend, I know full well it would spark an argument about how averages/means are calculated and how those words are defined.
I think, instead of asking about speed, it's better to ask it in the context of time. So maybe something like this:
"Let's say it takes you some time to run around a track. How long would it take you to run a second lap so that your average speed is doubled?"
This is the best I've got, because the riddle is not asking for speed, it is asking specifically for time. I think this better emphasizes the meaning of the riddle without immediately thinking about just running three times faster.
Okay so I tested this on people around me, it encouraged them to think about trying to travel so fast that the average time for each lap is halved, meaning twice the distance in the same amount of time as the first lap, so the second lap has to be done in zero time, and is thus impossible.
Number 2 was a word trick. I made a trick question like that in the 1980's.
@wO Don't worry man, I thought the same thing for a while. Even did the math on paper and everything.
The mistake you are making is the same mistake that I made; while it is possible to reach an average velocity of 2V1, you have to run 3 more laps after the first lap to reach that average velocity. This is evident in your math, where you have to run 4km in 2 minutes to reach that average velocity. (first lap is 1km in 1 minute, second "lap" is 3km in 1 minute)
However, in the question, you are only allowed to run 2 laps. If each lap is 1km, then you are only allowed to run 2km total. It's impossible to reach an overall average velocity of 2V1 when you can only run 2 laps. You have to be able to run 4 laps (total distance 4km assuming 1km per lap) to reach the desired average velocity, which is prohibited by the question.
Run 1 lap in 6 minutes.
Then run lap 2 in (:$-'kcowkxm
@@AnhThuNguyen-zz2hm oh, that clears it up.
@@purge_bot2338 But why take distance into account and not time. Same dist and shorter time = Greater velocity
@@nazimakhatoon8406 time is in account, but is a result of formula. So, however you are modifying speed, you have still the same distance. Now, to be still in the formula, whenever you are raising speed, you are at the same moment decreasing the time and vice-versa. Point is, that if you are counting average from TWO numbers, you are basically dividing by 2 the total distance and as your speed is result of distance divided by time, you always need to spend some time on the track, hence you always need to do t1 + t2. So, in formula d1+d2 / t1+t2, where d1+d2 = 2*d. And now, if you would need to have 2 as a result (v=d/t), you would need to have exactly t1+t2 = t1. And once again, that´s not possible, because t2 is always > 0, in a fact, it´s possible to make it only if you make 2nd round in exactly 0 time.
4:03 Slight problem. 4/2 = 2/1, so you could just run 4x the distance in 2x the time.
He said only 2 laps
You cant run more distance though. Cuz in both cases you are running around the same track
The flange on a train wheel is not STRICTLY necessary for it to not fall off the track. It's only necessary on sharper corners, when you hear that metal screeching sound. Most of the time the train stays on the tracks based on the conical wheel shape.
The answer is still good though, since there's always a part of the cone riding below the level of the contact point, *just barely* though.
A train wheel flange DOES NOT stop the wheel falling off the track, it is only a safety feature. The wheels are CHAMFERED which is a very clever way of making the train self steer on corners - the outside wheel runs on the bigger part of the chamfer while the inside on the smaller part of the chamfer on corners. The flange never touches the track in normal situations.