The video and what I see in the comments make it seem as though Einstein was fooled by the question. And yet at 11:31 he correctly calculated that there was no time available for the trip down. I would say that he got it!
It means he was fully going calculation mode and wanted to subtract t trip-tascent. He was never tricked. He had the right method throughout@@juncheok8579
@@juncheok8579 yeah exactly, that's the human error everyone has when encountering a problem like 7 * 8 you have to calculate it in your head for a few moments
@@juncheok8579 But he will have been aware that 45 was not the answer and that some calculation was needed instead. (Of course in this case no calculation is needed to see that time is already up after travelling exactly half of the distance at half of the required speed.)
6:46 ...I "invented" this problem a few years ago, trying to figure out why if I could ride my bike at 20 mph on the flat, why it was so hard to do so on a hilly course, if you get back all the energy you gained by going up the hill. I realized, that if the first half of the trip was up a hill you could only go up at 10 mph, you'd need to teleport down the hill to average 20 mph, and that allowed me to realize why the 20 mph assumption was wrong!
I like how asking someone very bad at math could actually have them solve it immediately! Me: "I want you to solve this problem: [...]" Friend (bad at math): "Impossible!" (as impossible for him) Me: "That's the correct answer!"
6:57 I've seen this one before. In order to average 30 mph for a 2 mile trip, the car has to make that trip in 4 minutes. But the car already took 4 minutes to travel the first mile, therefore the car would have to teleport the rest of the way! If you average a speed of x for the first half of any trip, it is impossible to average a speed of 2x for the full trip.
@@Sonny_McMacsson And then, there has been rumors that Einstein's spirit got cloned by magical equation sheets, so when he was proclaimed dead, his clone spirit quantum materialized a clone body and teleported underground into a base, where he would then change his name to "Doc Brown", and then he would then suffer a blow to the head which will cause him to use his theory of relativity to figure out the way to create a flux capacitor to make wormholes, as well as a hyperinfinipotent liquid known as "Time Juice" to power it, and then created a DeLorean Time Machine to the future, where he then drank a potion of immortality.
I've seen this before too, and my answer is that "average" is ambiguous for this - I'll stick with the arithmetic mean anyway, but if you average WRT distance instead of time you get that for the second mile the speed must have been 45mph. Remember - you can average anything WRT any variable you want. For what averaging speed WRT distance is (in the general case) that you integrate the (varying) speed WRT distance, then divide by total distance. When you're given averages for intervals that sum to the overall interval, that simplifies to a weighted average of those discrete intervals. When those smaller intervals are all equal, that simplifies to the sum-all-the-values-and-divide-by-the-number-of-values average - in this case (15+k)/2 = 30 which solves to k=45. Just because averaging WRT time is what you'd normally expect for problems involving speeds, doesn't mean it's the only possible option. People will say that speed is by definition an average given by distance per unit time, but they're wrong - speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously, and while it's natural to average WRT time when using a derivative WRT time (giving you a convenient integral-is-the-inverse-of-derivative simplification), you can average WRT any variable you like. ["speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously" is a slip-up, and unsurprisingly confusing. As mentioned in a much later comment, think in terms of a graph of speed WRT whatever - here, either time or distance, and averaging the height of the curve without caring what units are along the x-axis. The average is a speed irrespective - the units of the y-axis. The average WRT whatever based on integration is the integral of speed WRT whatever, divided by the total interval of whatever. The integral of speed WRT distance is in different units (distance squared per time) vs. the integral of speed WRT time (distance), but then you divide by the interval width along the x-axis in whatever units apply to that particular graph, and get back to speed either way. So WRT "but it's correct" below - well, no it wasn't (sigh) but that's an error in writing it out while half-asleep, not in the underlying principle.] It's not a popular answer, based on the last time I gave it, but it's correct - averages aren't even always arithmetic means (median, mode, geometric mean, harminic mean, many other kinds of mean) but, assuming the arithmetic mean, you can average over intervals of any kind you want, or you can even have averages of a list of numbers with no intervals specified (so assuming equal weight to each number) - again (15+45)/2 = 30. Personally, I'd consider my answer wrong normally (we generally know what is meant by "average speed" - ie. WRT time), but if someone asks an absurd question, I say exploit any ambiguity because ignoring conventional interpretation is the lesser absurdity. Averaging per distance travelled makes a lot more sense in reality when looking at fuel use - many road conditions vary with the interval of distance you're travelling on (others varying with time due to weather and traffic, others varying depending on the mood of the driver). Gallons per mile times miles = total gallons.
@@stevehorne5536 He explained why 45 is incorrect, you must go by total distance divided by total time to get average speed, even if the car drove 200 mph down the hill, the 4 minutes required for a 30 mph average speed are already "used up" so as additional seconds tick by, so the maximum possible average speed of the trip ticks down from that 30 mph target. After 8 minutes, it similarly becomes impossible to average 15 mph for the trip, because the required time threshold has already been passed. And as additional minutes pass from there, the maximum possible average speed gets lower and lower, no matter how fast the trip is completed.
The hill problem is something that I've often thought about as both a bicyclist and a runner. The problem is intuitive if you're talking about differences in speed over fixed amounts of time, but not distance. This is basically why dealing with hills, even on a round trip is more of a pain than just a level path.
When I was young I used to run such calculations while staring at my head unit on my bike. These days I just ride without caring much about the numbers.
@@mrosskne That would imply that the speed increase of going downhill is greater than the speed decrease of going uphill. Do you assert that to be true? I can't say it's wrong for sure, bit it strikes me as counterintuitive.
@@plentyofpaper It doesn't matter. The point of the problem is that you travel one speed for the first leg and a different speed for the second. It could be talking about a submarine or a spaceship. The problem's flavor is irrelevant.
@@mrosskne The point of the problem is that if an equal distance is traveled at 2 different speeds, the slower speed will be closer to the average than the faster speed. Because the slower trip takes longer. I honestly have no idea what you're trying to get at.
The sum of dice property is used to great effect in table-top games such as D&D. For instance, when rolling a character's stats, you roll 3 6-sided dice and sum them. The probability creates a bell curve that makes the midpoint between 1 and 18 the most common, which works out well at making 10 and 11 the most common stat, with either extreme (3 or 18) the least likely. So just as the "geniuses" are least common in reality, the rolling of a genius stat is also the least likely.
Just a correction there - it is not a bell curve. With 2 dice it's actually just an upside down V with no curves at all, just 2 straight lines. With 3 dice it's a bit more complicated, but still isn't a normal distribution. Also, the mid point with 3 dice is half way between 3 and 18, not 1 and 18.
@@asdfqwerty14587 He was referring to 3 dice, not 2, and he didn't say it was a normal distribution. He said it was a bell curve, which it is. This is an impressive amount of being wrong and simultaneously not responding to what the comment you're replying to actually said.
@@mrosskne A bell curve is the normal distribution. That is the dictionary definition of it. Google the definition of it and link on the front page will say that it refers to a normal distribution, including the one coming from a dictionary.
@@mrosskne Astonishing how bad your reading comprehension is and how wrong you are. Bell curve means normal distribution, not a "curve shaped like a bell". And the comment replies to what OP said. Corrects the mistake of it not being a bell curve and corrects the range to be from 3 to 18 not 1 to 18.
There's a story about someone giving John von Neumann this puzzle: Two trains 180 miles apart are on the same track going opposite directions toward each other, one traveling 10 miles/hour and the other 20 miles/hour. A fly starts on the front of one train and flies toward the other at 60 miles/hour. As soon as it reaches the other it turns around and flies back at that same speed to the first train. When it reaches that it reverses, and so on, until the fly finally is crushed when the two train collide. Question: how far does the fly fly? von Neumann instantly gives the correct answer. The person then said "Oh, you saw the trick! Most people just try to sum an infinite series!". von Neumann than said "I did sum the infinite series! There's a trick?"
Just so we're all on the same page, it's 360 miles, right? Of course I solved the answer in my head and I also summed the series. But the series was the sum of the two speeds giving me the relative speed between the trains (30mph), giving me a time to collision (6hr), and the total distance traveled by the fly (60mph x 6hr = 360miles).
I did it by combining the speeds, 30mph, 180miles, 6hrs, then how far the fly can get at 60mph. After solving, I realized, I could have solve as follows. Combined the speed, observe that the fly is traveling at double the speed, and note that it would cover double the distance, and arrived at 360mi
The third one does something to our brains, it seems obvious yet that's because we haven't accounted for time in the t/d/s triangle and it has already been spent. Thanks for explaining.
For the first puzzle, given the 2 results you can get a third result for r in terms of a and c only which comes to r = (a+c) + sqrt(2ac) or r = (a+c) - sqrt(2ac) (both values work as a,c > 0, so a+c > sqrt(2ac))
However, for the second solution the side lengths of the rectangle are a - sqrt(2ac) and c - sqrt(2ac) and at least one of them is guaranteed to be negative, which we probably don't want to accept.
Averaging speeds over the same distance requires taking a harmonic mean. 2/v_avg = 1/v1 + 1/v2 The reason is that you've defined the problem as saying the two legs of the trip cover the same _distance._ So to calculate it, the _distance_ needs to be in the denominator (since they're the same distance), just like when you add two fractions they need to have the same denominator. Since miles/hr has distance in the numerator, you need to flip it to put distance in the denominator before calculating the average. If the problem had been the car spends 1 hour driving 15 mph, and 1 hour driving v2 mph, then the time is equal for both legs, so time needs to go in the denominator. Which it already is for mph, so you can solve _that_ as v_avg = (v1 + v2)/2. And v2 = 45 mph would be correct. Same issue crops up with fuel efficiency in the U.S. (which uses miles per gallon). If you've got a car which gets 20 mpg and another car which gets 40 mpg and you drive both cars the same distance over the year, then the average mpg is the harmonic mean. Resulting in an average mileage of 26.7 mpg. OTOH, if you put 15 gallons in the 20 mpg car, and 15 gallons in the 40 mpg car (same gallons, instead of same distance), and drive both until their tank is empty, then the average mileage will be 30 mpg. Since the vast majority of driving you do is based on distance, not how much fuel you have, the rest of the world uses "liters per 100 km" to avoid this inversion issue (puts distance in the denominator).
None of this is necessary. If you want an average speed of 30 mph (~45 fps) over 2 miles (~10600 ft), the trip must take ~236 seconds. You traveled 5300 ft at 22.5 fps, so it took you 236 seconds. You have zero seconds to travel the remaining distance. If you want to reach an average speed of 30 mph, it will take you more than one mile of additional travel to reach it, no matter how fast you accelerate.
I agree with you...time has nothing to do with this problem. It's all about speed and distance. You go 15 for half and you have to go 45 the second half to average the SPEED. Who cares that one half takes 1/3 the time to complete. Speed is all that was asked and all that matters. Stick to the same unit of measure...don't bring in something that doesn't matter and use it to "prove" the correct solution wrong. If you are driving down the highway at 50 for a mile...and then for another mile you travel at 60...your average speed is 55. How long it took isn't the question.
the question isn't what is the average of the speeds of the dinamometer but what is the average speed of the trip. Also time always matters in speed since speed is composed by time. Speed is distance over time. Average speed by definition is distance over time
"I was going 70 miles an hour and got stopped by a cop who said, "Do you know the speed limit is 55 miles per hour?" "Yes, officer, I know but I wasn't going to be out that long..." -- Stephen Wright
Hah! That Einstein problem... I was getting annoyed with myself because I was getting an infinite speed/zero time for the descent and thought I was calculating it wrong. And then I thought, nah, it's a trick question 🙂
I was sure there was an answer, and so after confusing myself with the fact both require the same time, I watched the video, to find I was not actually mistaken.
Technically with time dilation it is possible and you wouldn't need to hit the speed of light because there is also very very very slight time dilation on the ascent as well. So it would be like (4 -(10^-32)) mins for the accent and 1O^-32 mins for the descent.
@@ericpaul4575 so I think you save 1.44 e-13 seconds on the ascent so I think you'd have to travel the equivalent of 6.944e12 miles per sec (without time dilation) I'm not sure how to convert that back to relative light speed (with time dilation) or I guess unwilling to figure it out.
So the first question, the answer is independent of a and c, which is not immediately obvious in the solution of the quadratic. I guess there is some interplay between a, b and c that would make the a and c terms cancel
The Feynman question has a problem in it. a, b, c are not independent of each other. That's why you get such different answers. Construct only a segment of length = a. From there if you draw a line straight up to the intersection with the circle, and from there strictly left, you can see that b and c are uniquely determined because of just fixing the a value.
@annanay007 if you fix only a, and "let b and c be free", and try to recreate the same picture from ONLY fixed a value, you will find out, that there would be anyway the only possible value for b and for c (because in the construction we have two lines with fixed right angles)
I actually came a variation of the Einstein problem decades ago (as a kid), so I already knew the answer to it. But it’s also a good teaching tool-the “average” isn’t always the arithmetic mean. Especially when dealing with complex units (a complex unit is one that combines 2 or more basic units).
I must be better at math than I thought. I had the b = r almost immediately. I've played craps in Vegas after reading a book on how to do it and already know that 11 is twice as likely as 12. I thought that the last one was a trick right away, too. It just didn't register exactly how it was tricky until the math was done. Another great video Talwalker!
They use this paradoxical question in 'Speed awareness courses' in the UK. I was allowed to leave when I answered "Easy just reduce your mass to a negative value and travel at 4C as you will arrive before you set off" it hurt the guy's head so much he knew he couldn't handle 4 hours with me 😂😂😂😂
No3 is a great example of something that happens all the time in industry and the media. I call it the average of averages and so many decisions are based on this flawed logic! This is a great way of pointing out that flaw.
I love your videos. They are beautifully designed and presented in an absolutely clear way. I'd just point out two mistakes: 1. The symbol for hour is just (upright or roman) h 2. Since time depends on the observer's perspective, going at light speed does solve the problem, assuming the one person interested in completing the trip on such average speed is the driver. From his perspective, going at light speed does take him to the 2-mile mark instantly, whereas an observer (e.g., someone who had an appointment with him) would be pissed due to him arrving a fraction if a second late.
@tezzerii that seems likely. Also, technically, a wheeled vehicle can only reach half light speed. This is because the highest speeds on a vehicle are those of the upper part of the wheels. If those where at light speed, their center would be at light speed divided by two.
I came up with 3 answers that ran through my mind in order after seeing the question and clicking on the video, so before i watch or read the comments i want to share/guess. 1. 60mph = 1 mile per min, 2miles = 2 min so to avg 30 mph you have 4 min to do it, @ 15mph it would take 8 min to go 2 mile sp if you use up 4 min on the 1st mile there is no time left. 2. 15 mph for half the trip + 45 for 2nd half to avg the 30. 3. half the speed the avg of 30 for 1st half you will have to double it to 60 for 2nd half. edit nevermind not going to watch it, once they start mixing letters and numbers its too much for me.
Trivially r = b ( b - a)^2 + ( b - c)^2 = b^2 b ^ 2 - 2 ( a + c) b + a^2 + c ^2 = 0 [ b - ( a + c) ] ^2 = 2 a c b = a + c + √ ( 2 a c) , a + c - √ ( 2 a c) , Hereby radius is r = b = a + c + √ ( 2 a c) , or r = b = a + c - √ ( 2 a c) , Second problem Uphill distance = downhill distance = D, say 2 D / 30 = D / 15 + D / v so it would require v = INFINITE
There are many variations of 'rate problems' that fool many people, this is a classic. Similar ones go something like, "If Bob can do a job in 2 hours, and Bill can do the same job in 3 hours..." Many folks will incorrectly combine the times vs rates.
Regarding Feynman's blunder, an interesting result is found when starting with the assumption that b=r, then expressing r in terms of a and c, by using the Pythagorean formula, then completing the square instead of using the quadratic formula: (r-c)**2 + (r-a)**2 = r**2 r**2 - 2cr +c**2 + r**2 -2ar +a**2 = r**2 r**2 - 2cr +c**2 -2ar +a**2 = 0 r**2 -2(a+c)r +a**2 +c**2 = 0 r**2 -2(a+c)r +a**2 +c**2 + 2ac = 2ac r**2 -2(a+c)r +(a+c)**2 = 2ac (r-(a+c))**2=2ac r=a+c+sqrt(2ac) Seeing as how a=r*versine(θ) and c=r*coversin(θ) (where "versine" is 1-cos and "coversine" is 1-sin), the above result would seem to imply the trig identity (1-ver-cvs)**2 = 2ver*cvs. But is that actually true? Let's check: (1-ver-cvs)**2 = (1 -(1-cos) -(1-sin))**2 = (-1+(cos+sin))**2 = 1 -2(cos+sin) +(cos+sin)**2 = 1 -2cos -2sin +cos**2 +2cos*sin +sin**2 = 2 -2cos -2sin +2cos*sin = 2(1-cos-sin+cos*sin) = 2(1-cos)(1-sin) = 2ver*cvs QED. New trig identity learned: (1-ver-cvs)**2 = 2ver*cvs
The first question felt so obvious that I was SURE I was doing it wrong, especially when you went into your explanation. Nope, turns out it was just obvious. ;)
7:59 i remember one video where you argued "half of two plus two" is 1/2 * 2 + 2 and not 1/2 * (2 + 2), because 1/2 (2 + 2) would be said as "half of *the sum* of two and two" or "half of *the quantity* two plus two" but look where we are now
Words aren't precise, that's why we don't use them for math. You can avoid ambiguity by saying "one half, multiply, open bracket, two, add, two, close bracket". But then you might as well just use the symbols.
@@mrosskne i just remember watching the "half of two plus two" video and they sounded very serious about it not being 1/2 * (2 + 2) and spent like half the video arguing so
@@mrosskne I believe the video being questioned was the use of the divide sign, AND juxtaposition of a(b+c), and whether people treat that as a*(b+c) verses (a*(b+c)), and was a lot of controversy because people learn to treat those differently, thus pointing out the problem with using such symbols. So to the point OP is making, half of 15 plus x, hasn't been explicitly stated as the sum of 15 and x, and should, by strict adherence to BEDMAS (or whatever you learned), be treated at half of 15, plus x.
looking at the car problem, to get a 30mph average, and you can only do 15mph for the 1st half of the trip you would have 0 time to do the 2nd mile, it takes the same amount of time to travel 2*D at 2*T as it does to travel 1*D at 1*T. this problem is deceptive in that it is really a time problem, not a speed problem.
When looking at the radius problem, I didn't notice the r=b shortcut at first. But when I looked at the quadratic solution, I immediately saw the 2ac - a² - c², and in my head converted it to -(a-c)², and now I had 2b²-(a-c)² and thus b² + b²-(a-c)² = b² + (a+b-c)(a+b+c), and slowly, it dawned me, that I missed something.
I have a weirdo way of thinking about the Einstein trick since I've seen it before, and it kinda works out by making the answer also a trick. So in some versions of the problem, it is vague about what average is being taken over and asks simply "what speed do you have to go for the total trip to average 30mph". This is generally assumed to be the average speed over the entire distance of the trip (2 miles). As the video shows, there's no way to get an average speed of 30 mph over the distance without traveling instantaneously. However, if you consider the average over time instead, then you end up back with 45 mph IFF you drive for another 4 minutes. Because it takes 4 minutes to go up at 15 mph, then another 4 minutes at 45 mph, you end up with an average speed of 30 mph. The distance you have traveled will end up being much further (4 miles here vs 2 in the original problem). If you reduce the time component out of this entirely (since it is constant), then it ends up being just the simple average of 15 and 45 like shown in the video, but it's important to clearly state the same amount of time as a constant.
My first thought was, "speed of light" The reason is simple, 15mph and to travel 1 mile, you have to spend the entire time as you would do with average for 2 miles. That is, there isn't time to spare!
The thumbnail question caught my attention, so I thought I'd do it out in my head before watching the video, but . . . . . This is infinity? Okay, so. . . 15 mph for 1 mile. Trip lasts another mile, 2 miles total. Average speed for total trip is 30 mph In my head I went "Okay, so 2 miles would be 1/15 of 30 miles. 1/15 of an hour is 4 minutes. So the total trip has to be 4 minutes" Then I looked at the first mile "Okay so that's also a rate of 1/15, so the first mile has to take 4 minutes. . . Wait a minute" Lemme do the problem another way just to be sure I'm doing it right (though, from the title I think I'm doing it correctly) We know the speed and distance for both the first mile and the total trip D/T=V. 1 mile / T{1} = 15 mph. 1 mile = 15 mph x T{1}. 1 mile / 15 mph = T{1}. 1/15 hour = T{1} 2 miles / T{2} = 30 mph. 2 miles = 30 mph x T{2}. 2 miles / 30 mph = T{2}. 2/30 = T{2}. 1/15 hour = T{2} The first mile takes the same amount of time as the total trip of two miles. The second mile must be completed instantaneously. The second mile has a velocity of infinity 1 mile / 0 hours = V{3}
I saw the average speed problem in a book many years ago, and I remember when I was 14 presenting this problem to my friends. They refused to believe that reaching the target average speed was impossible. Play confidently asserted the time had nothing to do with speed. One friend even claimed that his truck could accomplish it. I think they know better now. At least, I hope so.
I think the first problem is a good way of illustrating how you should always extrapolate what you're given to be true, into all of the other things that are also true as a result. I was able to get to the first "correct" answer, but the other solution can change how you look at stuff like this, which is the entire point of these types of videos.
Actually, if the 30mph is rounded to the nearest integer, this allows you to travel at an average of 29.5mph and still qualify. This works out at around 885mph for the downhill portion, which is less than 20% over the current land speed record. And so, of course, far less than infinity
@@kobeballer if you think specifying the level of precision changes the problem, please never consider becoming a scientist, or at the very least an engineer
@@mickdavies5647 lol so if my engineering manager asks me to design a part or process to 0.001 precision, then I can just present something to 0.01 and tell them "you shouldn't be in engineering if you think changing the precision changes the problem"? 😂😂
@kobeballer well, since you seem to think you can just consider the precision to be irrelevant, or ignore the fact it hasn't been provided, I pity both your manager and your clients
well because the other diagonal would also be b, it would be a straight line going from the center to the edge of the quarter circle, which means b is the radius of the circle. as for a and c, idk lol maybe somehow would be able to compute it in terms of r and b using trig functions.
For the ones who have been averaging the speeds: The cause of confusion here is that there are two possible physical quantities that this special case question can talk about.1. Equal distance segments: You cannot simply average the speeds. 2. Equal time segments: say, 15 miles per hour for 1 hour and 45 miles per hour for 1 hour (as an example), here you can simply average the two speeds.
You'd actually only need to be going ~18000mph along your decent when you factor in human reaction time were anyone timing you with a stop watch. The average human reaction time (~.2seconds) would then provide exactly enough gap for your car to reach the summit and then the base as they pressed the button to stop the timer at 4minutes. You'd also be going 1mile in .2seconds and would likely be the consistency of strawberry jam along the car's interior.
Haha! This was a dead giveaway to me but out of interest I subjected my family to this riddle and for sure my wife responded 45mph. Next my kid who is in second grade middle school gave the exact same answer, so I asked him if he could calculate what the resulting times for both distances would be for the given average speeds. For sure he did calculate it was 4 minutes for both but due to the nature of the question he then concluded that the additional mile had to be run in less than a minute and ended up answering 60mph. So next I asked him if he could use seconds instead of minutes and the big question mark started rising above his head - ehm 3600mph? Granted I don't really know if he already learned the concept of infinity slash division by zero, I think back then I already had it in first grade, but I believe he really did get it when I gave him the answer.
Translation: "If I have to be at work by 8am, and I'm only halfway there when the clock strikes 8am, how fast do I have to drive to get to work on time?" YOU CAN'T. You will be late to work no matter how fast you go.
the Einstein one is easier to visualize with different numbers: suppose you have an average 100 km/h to go 100km. What speed do you need to go through the next 100km to average 200km/h in the total trip? It's more immediately evident that a 100km trip at 100km/h would take 1h, and that a 200km trip at 200km/h would have to take 1h as well.
IMHO it is best to consider units. In the wrong solution the 2 must be miles (or else you woud clearly be assuming that the different distances were irrelevant, which is obviously wrong) and mph/m = 1/h which is not a speed.
When driving long distances over expressways where the speed limit is sometimes 70 mph or more, I TRY to have an average speed of about 65 or so. But it's really hard when you have to stop every 3 hours or so, even for just a few minutes (for gas, food, and the call-of-nature). Even brief stops quickly drops the 'average speed'.
You know, I actually thought about it using the second method you were talking about, with the idea of 4 minutes. I am not the best at math in my head for equations like this, but I did notice that it seemed like there wasn't enough time remaining to go that fast although I wasn't sure exactly why.
I didn't overthink the first one, it seemed obvious using trig that r=b. The second one is deeply obvious unless you think 6+5 and 5+6 are different. The last one takes a moment to realise the average speed is already sucked up by the ascent time. Sometimes overthinking or expectation stops the otherwise obvious from being apparent.
So Feynman's solution actually presents a restriction on viable values for a,b,c. If you instead use b for r in his equation and then solve for b you get that b=a+c-sqrt(2ac) so for any other value of b given a,c then the problem is unsolvable because the combination is invalid. For example if a=2,c=9 then we get b=17. If instead we are given instead b=19 then there is no possible rectangle with diagonal 19 such that to given quarter circle has the extra lengths a,c as given.
You got the wrong sign there. It should be PLUS sqrt not minus. There’s no way 2+9 minus a positive number equals 17 so you used it correctly in your first test.
That's made me feel great about my hitherto averagish brain. I got that Einstein-stumper straight away, just by looking at the thumbnail. Go brain! Go brain!
The speed for the second mile increases exponentially with every second spent driving under 30mph and hits infinity at 4min. If the car spent even a tiny fraction of a second driving faster, then a correct speed could be calculated to solve the puzzle. For example 15.0627mph for the first mile would leave 1 second at a speed of 3600mph to average out to 30mph for the whole trip.
Before watching, just looking at the thumb, I’m saying that this is impossible because to average 30mph the 2 mile trip must be done in 4 minutes, however it has already taken 4 minutes for the first mile at 15mph
Question in the thumbnail... I went directly to : How much time it take for the first part (4 minutes) and how much time do I have for the full trip (also 4 minutes) ?
I vaguely knew about the Einstein riddle but I never figured out why it truly was the case. Of course, students might have an easier time understanding it by considering that a single B on their record means they'll never have true straight As.
My "haven't watched the video yet" answer: Isn't the answer undefined (Basically infinity)? Speed is distance over time, to double the speed you need to double the total distance or halve the total time. We covered half the distance already, so the total distance is double, but we also used 100% of the time. We need to travel 1 mile in 0 time to double the average speed. 1÷0 mph. Undefined. The instinctual answer is probably 45, but that averages 22.5mph because time is the denominator and the time is not constant between the 2 legs.
Just look at the hill from a physics perspective instead of a math perspective. Due to significant figures you might make it if you go REALLY fast (for a car) on the descent: if the distances are 0.96 and 1.96 miles and the averages are 15.4 and 29.6 mph, but all written down with two significant figures, you would make it if you drove 2.6E+2 mph average, or accelerate with 1.6G if we don't account for stopping at the end (if I did all that correctly. Knowing my prowess with math, I probably made an error or two :P ). On the other hand, you could also be out of time before you even get to the top. Your mileage may vary...
Last question depends on if you are the observer or not, if you are not the observer you just need to instantly travel to the bottom. You can go at the speed of light if you are the one in the car and get to the bottom instantly. (assuming the car is massless.) You don't need to go faster than the speed of light you just need to travel at the speed of light if you are not the Observer.
Who is measuring the time spent on the decent? If you're travelling at the sped of light, time doesn't exist. As such, if you're the one timing the decent, the answer is the speed of light. If it's an external measurement, then yes - it's not possible.
Strictly the Leibniz problem is ambiguous: if you don't assume commutivity then you can interpret "a sum of" to be "a specific sum of" so that 5+6 is not the same sum as 6+5.
I got the b=r one instantly but that's probably because Feynman was primed for it to be a maths puzzle and tunnel visioned himself (I assume, I don't know anything about him outside that)
At that point you need to ask what clock you are going by. Because by the clocks of the rest of the universe it still takes you some time to travel that last bit, even at the speed of light.
There are a few problems with this. 1) If you're moving at/very close to the speed of light and you're talking about the frame of reference of the car, then the distance will not be 1 mile anymore. Relativity causes the lengths to change too, not just time. Even if you were talking about a wacky frame of reference the math would still work out the same and it would still be impossible. The only way you could come up with different answers is if you're talking about things like trying to calculate the speed of an object by taking the distance from 1 frame of reference and the time from a different frame of reference, which is a nonsense measurement which doesn't actually calculate anything useful (by that logic you could come up with literally any number, you could even say the car is travelling backwards if you wanted to). 2) If you're talking about the frame of reference of the object itself, then "the car isn't moving at all" - there's no such thing as measuring the speed of an object from its own frame of reference, because obviously the car is always at the same position relative to itself.
Thumbnail problem: Infinity. The car would have to instantly travel the remaining mile in order for it to average 30mph for the whole trip. Circle radius problem. The answer is just r = b. The two diagonals of a rectangle are of equal length. One diagonal is the radius. The other you've labeled b.
I refer to it as "a 5th-gear brain working on a 2nd-gear problem." If you think of the brain as a car engine, you can think of working on higher-end problems as being in a higher gear. Solving a simple problem with your brain in low gear is like trying to drive at a slow speed in a low gear; the engine handles it just fine. Trying to drive slow in a high gear, however, causes so few RPMs that the engine stalls and dies out, so it can't do it. ...A bit of a weird analogy, I admit, but it seems to get the point across.
My intuition said 60 mph, because if we go half the speed in the ascent, then we must double the speed in the descent. Alas, calculating it gave an overall speed of 24 mph. Very interesting problem!
I got the last puzzle right within seconds without calculations. I ride a bicycle a LOT, and I calculate my ride times by my average speeds. If I'm X miles from home, how fast do I have to go to get there in Y minutes? Yes, I call ahead and tell people I'm going to be late often. Why do you ask?
Used to do something similar to the driving problem when driving back and forth work to work out total mpg for the day when car only reports each journey. 😂 I know, but was quite mindful in a way. Anyway, this works similarly in that the average for the whole journey is the reciprocal of the average of the reciprocals of the two halfs. So, the initial calc would work if you were using 1/average speed instead of averag speed in esch case.
The speed would have to be infinite since a 30mph avg over 2 miles takes 4 minutes and so does a 15mph average over one mile. Since the allotted time was already all used up, the second mile would have to cover 1 mile in zero time.
The first one... if you draw the diagonal in the other direction then you get the radius is....B. The second one: Nope. Theres 1 way to get 12 and 2 ways of getting 11. (5-6) and (6,5)
I got the dice thing & I said the car would have to teleport instantaneously to the bottom. (I've heard a version of the last puzzle before, but I got the answer then too.) My answer to the 1st puzzle was a bit more complicated, I should have noticed r = b. I said r = sqrt (b^2- (r-a) ^2) +c & r = sqrt (b^2- (r-c) ^2) +a. This seems a valid answer to me, but what do you think?
There are several lines of thought to immediately get the answer to the Einstein puzzle. One of them is: If you you have a certain speed for a given distance and you want double the speed on double the distance then of course the time has to be exactly the same. This is true for any non-relativistic speed and any non-relativistic distance. If it's easier for you to calculate with certain numbers. Just imagine the first part of the hill to be exactly 15 retard units. Then the ascent takes 1 hour. The total distance would be 30 retard units and hence you also have 1 hour for the full distance to get the desired average speed. Which is of course impossible unless you are a massless particle...
The video and what I see in the comments make it seem as though Einstein was fooled by the question. And yet at 11:31 he correctly calculated that there was no time available for the trip down. I would say that he got it!
"Not until calculating did i notice..." implies he got tricked, even if it was for a few moments
It means he was fully going calculation mode and wanted to subtract t trip-tascent. He was never tricked. He had the right method throughout@@juncheok8579
@@juncheok8579 yeah exactly, that's the human error everyone has when encountering a problem like 7 * 8 you have to calculate it in your head for a few moments
@@juncheok8579 But he will have been aware that 45 was not the answer and that some calculation was needed instead. (Of course in this case no calculation is needed to see that time is already up after travelling exactly half of the distance at half of the required speed.)
I agree. I've heard it suggested that this problem stumped Einstein or that he couldn't solve it. But that's clearly not true.
I kept trying to do Einstein's question, getting to the divide by 0 part, and assuming I did something wrong and starting over for like an hour
6:46 ...I "invented" this problem a few years ago, trying to figure out why if I could ride my bike at 20 mph on the flat, why it was so hard to do so on a hilly course, if you get back all the energy you gained by going up the hill. I realized, that if the first half of the trip was up a hill you could only go up at 10 mph, you'd need to teleport down the hill to average 20 mph, and that allowed me to realize why the 20 mph assumption was wrong!
Therefore, Officer, it was impossible for me to be speeding.
I like how asking someone very bad at math could actually have them solve it immediately!
Me: "I want you to solve this problem: [...]"
Friend (bad at math): "Impossible!" (as impossible for him)
Me: "That's the correct answer!"
I drank six glasses of 5% ABV beer, you honour. But I only drank 95% of each glass.
Not impossible, simply highly improbable…
@@kjamison5951 No. The Improbability Drive is not enough, you need the Impossibility Drive (ref: Hitchhiker's Guide to the Galaxy).
I once thought that I made a mistake ... but I was wrong.
As a middle aged white man on the internet, I’ve never been wrong.
The correct expression is:
"I'm always right. I thought I was wrong once, but I was mistaken"
And that makes you right. Right?
6:57 I've seen this one before. In order to average 30 mph for a 2 mile trip, the car has to make that trip in 4 minutes. But the car already took 4 minutes to travel the first mile, therefore the car would have to teleport the rest of the way!
If you average a speed of x for the first half of any trip, it is impossible to average a speed of 2x for the full trip.
Einstein was figuring out HOW to do it, considering he's gonna need a wormhole.
@@Sonny_McMacsson And then, there has been rumors that Einstein's spirit got cloned by magical equation sheets, so when he was proclaimed dead, his clone spirit quantum materialized a clone body and teleported underground into a base, where he would then change his name to "Doc Brown", and then he would then suffer a blow to the head which will cause him to use his theory of relativity to figure out the way to create a flux capacitor to make wormholes, as well as a hyperinfinipotent liquid known as "Time Juice" to power it, and then created a DeLorean Time Machine to the future, where he then drank a potion of immortality.
Oh that’s why I was so confused. I kept using d=rt while looking at he thumbnail
I've seen this before too, and my answer is that "average" is ambiguous for this - I'll stick with the arithmetic mean anyway, but if you average WRT distance instead of time you get that for the second mile the speed must have been 45mph. Remember - you can average anything WRT any variable you want. For what averaging speed WRT distance is (in the general case) that you integrate the (varying) speed WRT distance, then divide by total distance. When you're given averages for intervals that sum to the overall interval, that simplifies to a weighted average of those discrete intervals. When those smaller intervals are all equal, that simplifies to the sum-all-the-values-and-divide-by-the-number-of-values average - in this case (15+k)/2 = 30 which solves to k=45.
Just because averaging WRT time is what you'd normally expect for problems involving speeds, doesn't mean it's the only possible option. People will say that speed is by definition an average given by distance per unit time, but they're wrong - speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously, and while it's natural to average WRT time when using a derivative WRT time (giving you a convenient integral-is-the-inverse-of-derivative simplification), you can average WRT any variable you like.
["speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously" is a slip-up, and unsurprisingly confusing. As mentioned in a much later comment, think in terms of a graph of speed WRT whatever - here, either time or distance, and averaging the height of the curve without caring what units are along the x-axis. The average is a speed irrespective - the units of the y-axis. The average WRT whatever based on integration is the integral of speed WRT whatever, divided by the total interval of whatever. The integral of speed WRT distance is in different units (distance squared per time) vs. the integral of speed WRT time (distance), but then you divide by the interval width along the x-axis in whatever units apply to that particular graph, and get back to speed either way. So WRT "but it's correct" below - well, no it wasn't (sigh) but that's an error in writing it out while half-asleep, not in the underlying principle.]
It's not a popular answer, based on the last time I gave it, but it's correct - averages aren't even always arithmetic means (median, mode, geometric mean, harminic mean, many other kinds of mean) but, assuming the arithmetic mean, you can average over intervals of any kind you want, or you can even have averages of a list of numbers with no intervals specified (so assuming equal weight to each number) - again (15+45)/2 = 30. Personally, I'd consider my answer wrong normally (we generally know what is meant by "average speed" - ie. WRT time), but if someone asks an absurd question, I say exploit any ambiguity because ignoring conventional interpretation is the lesser absurdity.
Averaging per distance travelled makes a lot more sense in reality when looking at fuel use - many road conditions vary with the interval of distance you're travelling on (others varying with time due to weather and traffic, others varying depending on the mood of the driver). Gallons per mile times miles = total gallons.
@@stevehorne5536 He explained why 45 is incorrect, you must go by total distance divided by total time to get average speed, even if the car drove 200 mph down the hill, the 4 minutes required for a 30 mph average speed are already "used up" so as additional seconds tick by, so the maximum possible average speed of the trip ticks down from that 30 mph target. After 8 minutes, it similarly becomes impossible to average 15 mph for the trip, because the required time threshold has already been passed. And as additional minutes pass from there, the maximum possible average speed gets lower and lower, no matter how fast the trip is completed.
The hill problem is something that I've often thought about as both a bicyclist and a runner.
The problem is intuitive if you're talking about differences in speed over fixed amounts of time, but not distance.
This is basically why dealing with hills, even on a round trip is more of a pain than just a level path.
When I was young I used to run such calculations while staring at my head unit on my bike. These days I just ride without caring much about the numbers.
The trip problem has nothing at all to do with hills, that's basically a red herring. The result is the same if you're on flat ground.
@@mrosskne That would imply that the speed increase of going downhill is greater than the speed decrease of going uphill.
Do you assert that to be true? I can't say it's wrong for sure, bit it strikes me as counterintuitive.
@@plentyofpaper It doesn't matter. The point of the problem is that you travel one speed for the first leg and a different speed for the second. It could be talking about a submarine or a spaceship. The problem's flavor is irrelevant.
@@mrosskne The point of the problem is that if an equal distance is traveled at 2 different speeds, the slower speed will be closer to the average than the faster speed. Because the slower trip takes longer.
I honestly have no idea what you're trying to get at.
The sum of dice property is used to great effect in table-top games such as D&D. For instance, when rolling a character's stats, you roll 3 6-sided dice and sum them. The probability creates a bell curve that makes the midpoint between 1 and 18 the most common, which works out well at making 10 and 11 the most common stat, with either extreme (3 or 18) the least likely. So just as the "geniuses" are least common in reality, the rolling of a genius stat is also the least likely.
Just a correction there - it is not a bell curve. With 2 dice it's actually just an upside down V with no curves at all, just 2 straight lines. With 3 dice it's a bit more complicated, but still isn't a normal distribution.
Also, the mid point with 3 dice is half way between 3 and 18, not 1 and 18.
@@asdfqwerty14587 He was referring to 3 dice, not 2, and he didn't say it was a normal distribution. He said it was a bell curve, which it is. This is an impressive amount of being wrong and simultaneously not responding to what the comment you're replying to actually said.
@@mrosskne A bell curve is the normal distribution. That is the dictionary definition of it. Google the definition of it and link on the front page will say that it refers to a normal distribution, including the one coming from a dictionary.
No, it's any curve shaped like a bell, sorry.
@@mrosskne Astonishing how bad your reading comprehension is and how wrong you are.
Bell curve means normal distribution, not a "curve shaped like a bell".
And the comment replies to what OP said. Corrects the mistake of it not being a bell curve and corrects the range to be from 3 to 18 not 1 to 18.
There's a story about someone giving John von Neumann this puzzle: Two trains 180 miles apart are on the same track going opposite directions toward each other, one traveling 10 miles/hour and the other 20 miles/hour. A fly starts on the front of one train and flies toward the other at 60 miles/hour. As soon as it reaches the other it turns around and flies back at that same speed to the first train. When it reaches that it reverses, and so on, until the fly finally is crushed when the two train collide. Question: how far does the fly fly?
von Neumann instantly gives the correct answer. The person then said "Oh, you saw the trick! Most people just try to sum an infinite series!".
von Neumann than said "I did sum the infinite series! There's a trick?"
Just so we're all on the same page, it's 360 miles, right?
Of course I solved the answer in my head and I also summed the series. But the series was the sum of the two speeds giving me the relative speed between the trains (30mph), giving me a time to collision (6hr), and the total distance traveled by the fly (60mph x 6hr = 360miles).
I did it by combining the speeds, 30mph, 180miles, 6hrs, then how far the fly can get at 60mph.
After solving, I realized, I could have solve as follows.
Combined the speed, observe that the fly is traveling at double the speed, and note that it would cover double the distance, and arrived at 360mi
Well, 180 miles covered at a combined speed of 30 mph takes 6 hours. Fly always travels at 60 mph during that time = 360 miles. Easy peasy.
Harder question would be, how many times does the fly turns?
@@Nobody-tu5wt As the 2 trains get closer and closer, the number of turns approach infinity if I'm not mistaken.
The third one does something to our brains, it seems obvious yet that's because we haven't accounted for time in the t/d/s triangle and it has already been spent. Thanks for explaining.
That last puzzle reminds me of trying to bump up my average speed during a bike ride. Always harder than you think!
I wish my Garmin calculated my mode average instead of my mean average. I'd still be slow, but my Strava activities would look better.
For the first puzzle, given the 2 results you can get a third result for r in terms of a and c only which comes to r = (a+c) + sqrt(2ac) or r = (a+c) - sqrt(2ac) (both values work as a,c > 0, so a+c > sqrt(2ac))
However, for the second solution the side lengths of the rectangle are a - sqrt(2ac) and c - sqrt(2ac) and at least one of them is guaranteed to be negative, which we probably don't want to accept.
Averaging speeds over the same distance requires taking a harmonic mean.
2/v_avg = 1/v1 + 1/v2
The reason is that you've defined the problem as saying the two legs of the trip cover the same _distance._ So to calculate it, the _distance_ needs to be in the denominator (since they're the same distance), just like when you add two fractions they need to have the same denominator. Since miles/hr has distance in the numerator, you need to flip it to put distance in the denominator before calculating the average.
If the problem had been the car spends 1 hour driving 15 mph, and 1 hour driving v2 mph, then the time is equal for both legs, so time needs to go in the denominator. Which it already is for mph, so you can solve _that_ as v_avg = (v1 + v2)/2. And v2 = 45 mph would be correct.
Same issue crops up with fuel efficiency in the U.S. (which uses miles per gallon). If you've got a car which gets 20 mpg and another car which gets 40 mpg and you drive both cars the same distance over the year, then the average mpg is the harmonic mean. Resulting in an average mileage of 26.7 mpg.
OTOH, if you put 15 gallons in the 20 mpg car, and 15 gallons in the 40 mpg car (same gallons, instead of same distance), and drive both until their tank is empty, then the average mileage will be 30 mpg.
Since the vast majority of driving you do is based on distance, not how much fuel you have, the rest of the world uses "liters per 100 km" to avoid this inversion issue (puts distance in the denominator).
...and because they use metric instead of (Hamburger*Eagle)/Shotgun^Footballfields.
None of this is necessary.
If you want an average speed of 30 mph (~45 fps) over 2 miles (~10600 ft), the trip must take ~236 seconds.
You traveled 5300 ft at 22.5 fps, so it took you 236 seconds. You have zero seconds to travel the remaining distance. If you want to reach an average speed of 30 mph, it will take you more than one mile of additional travel to reach it, no matter how fast you accelerate.
I agree with you...time has nothing to do with this problem. It's all about speed and distance. You go 15 for half and you have to go 45 the second half to average the SPEED. Who cares that one half takes 1/3 the time to complete. Speed is all that was asked and all that matters. Stick to the same unit of measure...don't bring in something that doesn't matter and use it to "prove" the correct solution wrong.
If you are driving down the highway at 50 for a mile...and then for another mile you travel at 60...your average speed is 55. How long it took isn't the question.
the question isn't what is the average of the speeds of the dinamometer but what is the average speed of the trip. Also time always matters in speed since speed is composed by time. Speed is distance over time. Average speed by definition is distance over time
Thank you, I needed that little confidence boost!
"I was going 70 miles an hour and got stopped by a cop who said, "Do you know the speed limit is 55 miles per hour?" "Yes, officer, I know but I wasn't going to be out that long..."
-- Stephen Wright
I think that would work in court. 😂😂
Also Stephen Wright:
Cop: Why were you going so fast?
Stephen: Because I pressed the gas pedal all the way down to the floor.
Hah! That Einstein problem... I was getting annoyed with myself because I was getting an infinite speed/zero time for the descent and thought I was calculating it wrong. And then I thought, nah, it's a trick question 🙂
Like reverse psychology, it's a reverse trick question. Ie: Not the trick you're thinking of!! 🤣
I was sure there was an answer, and so after confusing myself with the fact both require the same time, I watched the video, to find I was not actually mistaken.
Technically with time dilation it is possible and you wouldn't need to hit the speed of light because there is also very very very slight time dilation on the ascent as well. So it would be like (4 -(10^-32)) mins for the accent and 1O^-32 mins for the descent.
So what speed is that on decent?
@ashtonaughtband Except that you're in an old car that can only manage 15mph up the hill. =o)
@@ericpaul4575 so I think you save 1.44 e-13 seconds on the ascent so I think you'd have to travel the equivalent of 6.944e12 miles per sec (without time dilation) I'm not sure how to convert that back to relative light speed (with time dilation) or I guess unwilling to figure it out.
@@ashtonaughtbandyou just need to play with the Lorentz factor. I'll calculate it when I get up from my nap, if I remember
Yes, that's why the simple arithmetic problem didn't mention time dilation. Because you were supposed to consider time dilation. Christ.
So the first question, the answer is independent of a and c, which is not immediately obvious in the solution of the quadratic. I guess there is some interplay between a, b and c that would make the a and c terms cancel
The Feynman question has a problem in it. a, b, c are not independent of each other. That's why you get such different answers.
Construct only a segment of length = a. From there if you draw a line straight up to the intersection with the circle, and from there strictly left, you can see that b and c are uniquely determined because of just fixing the a value.
Did not understand. Can you please elaborate
@annanay007 if you fix only a, and "let b and c be free", and try to recreate the same picture from ONLY fixed a value, you will find out, that there would be anyway the only possible value for b and for c (because in the construction we have two lines with fixed right angles)
@@Serg_144 In other words, with that additional information Feynman's answer can be reduced to "b".
@@jmodified yep, he just didn't used it. They are functions b=b(a), c=c(a)
@@Serg_144 now finally understood. Thanks brother
"Never try! Trying is the 1st step to failure!" -- Grandmaster Ben Finegold
That means, Ben already knew the results before the match with Mittens
I actually came a variation of the Einstein problem decades ago (as a kid), so I already knew the answer to it. But it’s also a good teaching tool-the “average” isn’t always the arithmetic mean. Especially when dealing with complex units (a complex unit is one that combines 2 or more basic units).
I must be better at math than I thought. I had the b = r almost immediately. I've played craps in Vegas after reading a book on how to do it and already know that 11 is twice as likely as 12. I thought that the last one was a trick right away, too. It just didn't register exactly how it was tricky until the math was done. Another great video Talwalker!
They use this paradoxical question in 'Speed awareness courses' in the UK. I was allowed to leave when I answered "Easy just reduce your mass to a negative value and travel at 4C as you will arrive before you set off" it hurt the guy's head so much he knew he couldn't handle 4 hours with me 😂😂😂😂
No3 is a great example of something that happens all the time in industry and the media. I call it the average of averages and so many decisions are based on this flawed logic! This is a great way of pointing out that flaw.
Worked it out. Watched the video for confirmation. It’s good being intelligent 😊
I'm pleased that I realised the car question was impossible before I saw the answer.
I love your videos. They are beautifully designed and presented in an absolutely clear way. I'd just point out two mistakes:
1. The symbol for hour is just (upright or roman) h
2. Since time depends on the observer's perspective, going at light speed does solve the problem, assuming the one person interested in completing the trip on such average speed is the driver. From his perspective, going at light speed does take him to the 2-mile mark instantly, whereas an observer (e.g., someone who had an appointment with him) would be pissed due to him arrving a fraction if a second late.
Here's the thing though - it's an old car and can't manage more than 15mph up the hill ! So I reckon light speed is literally out of the question. =o)
@tezzerii that seems likely. Also, technically, a wheeled vehicle can only reach half light speed. This is because the highest speeds on a vehicle are those of the upper part of the wheels. If those where at light speed, their center would be at light speed divided by two.
It's an arithmetic problem, not a relativity problem. Get a clue.
It is a physics (kinematics) problem, @mrosskne. Make sure you are at least right before correcting others.
I feel smart when i instantly got the dice problem.
I came up with 3 answers that ran through my mind in order after seeing the question and clicking on the video, so before i watch or read the comments i want to share/guess.
1. 60mph = 1 mile per min, 2miles = 2 min so to avg 30 mph you have 4 min to do it, @ 15mph it would take 8 min to go 2 mile sp if you use up 4 min on the 1st mile there is no time left.
2. 15 mph for half the trip + 45 for 2nd half to avg the 30.
3. half the speed the avg of 30 for 1st half you will have to double it to 60 for 2nd half.
edit nevermind not going to watch it, once they start mixing letters and numbers its too much for me.
Trivially r = b
( b - a)^2 + ( b - c)^2 = b^2
b ^ 2 - 2 ( a + c) b + a^2 + c ^2 = 0
[ b - ( a + c) ] ^2 = 2 a c
b = a + c + √ ( 2 a c) ,
a + c - √ ( 2 a c) ,
Hereby radius is
r = b = a + c + √ ( 2 a c) ,
or r = b = a + c - √ ( 2 a c) ,
Second problem
Uphill distance = downhill distance
= D, say
2 D / 30 = D / 15 + D / v
so it would require v = INFINITE
There are many variations of 'rate problems' that fool many people, this is a classic. Similar ones go something like, "If Bob can do a job in 2 hours, and Bill can do the same job in 3 hours..." Many folks will incorrectly combine the times vs rates.
Regarding Feynman's blunder, an interesting result is found when starting with the assumption that b=r, then expressing r in terms of a and c, by using the Pythagorean formula, then completing the square instead of using the quadratic formula:
(r-c)**2 + (r-a)**2 = r**2
r**2 - 2cr +c**2 + r**2 -2ar +a**2 = r**2
r**2 - 2cr +c**2 -2ar +a**2 = 0
r**2 -2(a+c)r +a**2 +c**2 = 0
r**2 -2(a+c)r +a**2 +c**2 + 2ac = 2ac
r**2 -2(a+c)r +(a+c)**2 = 2ac
(r-(a+c))**2=2ac
r=a+c+sqrt(2ac)
Seeing as how a=r*versine(θ) and c=r*coversin(θ) (where "versine" is 1-cos and "coversine" is 1-sin), the above result would seem to imply the trig identity (1-ver-cvs)**2 = 2ver*cvs. But is that actually true? Let's check:
(1-ver-cvs)**2
= (1 -(1-cos) -(1-sin))**2
= (-1+(cos+sin))**2
= 1 -2(cos+sin) +(cos+sin)**2
= 1 -2cos -2sin +cos**2 +2cos*sin +sin**2
= 2 -2cos -2sin +2cos*sin
= 2(1-cos-sin+cos*sin)
= 2(1-cos)(1-sin)
= 2ver*cvs
QED. New trig identity learned: (1-ver-cvs)**2 = 2ver*cvs
the first puzzle actually made me do quadratic. After I went through the video I found out that I was just overthinking it LOL
The "famous right triangle theorem" made me LOL 😂😂
The first question felt so obvious that I was SURE I was doing it wrong, especially when you went into your explanation. Nope, turns out it was just obvious. ;)
It’s a trick question. The diagram is not an accurate representation of a quarter circle so there can be no radius
@bobross7473 diagrams in math questions are rarely to scale. That's not a trick, that's a convention
7:59 i remember one video where you argued "half of two plus two" is 1/2 * 2 + 2 and not 1/2 * (2 + 2), because 1/2 (2 + 2) would be said as "half of *the sum* of two and two" or "half of *the quantity* two plus two" but look where we are now
Words aren't precise, that's why we don't use them for math.
You can avoid ambiguity by saying "one half, multiply, open bracket, two, add, two, close bracket". But then you might as well just use the symbols.
@@mrosskne i just remember watching the "half of two plus two" video and they sounded very serious about it not being 1/2 * (2 + 2) and spent like half the video arguing so
ok. do you have an argument to contribute?
@@mrosskne search "half of two plus two" on their channel
@@mrosskne I believe the video being questioned was the use of the divide sign, AND juxtaposition of a(b+c), and whether people treat that as a*(b+c) verses (a*(b+c)), and was a lot of controversy because people learn to treat those differently, thus pointing out the problem with using such symbols. So to the point OP is making, half of 15 plus x, hasn't been explicitly stated as the sum of 15 and x, and should, by strict adherence to BEDMAS (or whatever you learned), be treated at half of 15, plus x.
looking at the car problem, to get a 30mph average, and you can only do 15mph for the 1st half of the trip you would have 0 time to do the 2nd mile, it takes the same amount of time to travel 2*D at 2*T as it does to travel 1*D at 1*T.
this problem is deceptive in that it is really a time problem, not a speed problem.
A speed problem is a time problem.
@@mrosskneit’s relative
When looking at the radius problem, I didn't notice the r=b shortcut at first. But when I looked at the quadratic solution, I immediately saw the 2ac - a² - c², and in my head converted it to -(a-c)², and now I had 2b²-(a-c)² and thus b² + b²-(a-c)² = b² + (a+b-c)(a+b+c), and slowly, it dawned me, that I missed something.
I have a weirdo way of thinking about the Einstein trick since I've seen it before, and it kinda works out by making the answer also a trick.
So in some versions of the problem, it is vague about what average is being taken over and asks simply "what speed do you have to go for the total trip to average 30mph". This is generally assumed to be the average speed over the entire distance of the trip (2 miles). As the video shows, there's no way to get an average speed of 30 mph over the distance without traveling instantaneously.
However, if you consider the average over time instead, then you end up back with 45 mph IFF you drive for another 4 minutes. Because it takes 4 minutes to go up at 15 mph, then another 4 minutes at 45 mph, you end up with an average speed of 30 mph. The distance you have traveled will end up being much further (4 miles here vs 2 in the original problem). If you reduce the time component out of this entirely (since it is constant), then it ends up being just the simple average of 15 and 45 like shown in the video, but it's important to clearly state the same amount of time as a constant.
Damn; you had me with the last one.
My first thought was, "speed of light" The reason is simple, 15mph and to travel 1 mile, you have to spend the entire time as you would do with average for 2 miles. That is, there isn't time to spare!
I love your videos
The thumbnail question caught my attention, so I thought I'd do it out in my head before watching the video, but . . . . . This is infinity?
Okay, so. . . 15 mph for 1 mile. Trip lasts another mile, 2 miles total. Average speed for total trip is 30 mph
In my head I went "Okay, so 2 miles would be 1/15 of 30 miles. 1/15 of an hour is 4 minutes. So the total trip has to be 4 minutes"
Then I looked at the first mile "Okay so that's also a rate of 1/15, so the first mile has to take 4 minutes. . . Wait a minute"
Lemme do the problem another way just to be sure I'm doing it right (though, from the title I think I'm doing it correctly)
We know the speed and distance for both the first mile and the total trip
D/T=V. 1 mile / T{1} = 15 mph. 1 mile = 15 mph x T{1}. 1 mile / 15 mph = T{1}. 1/15 hour = T{1}
2 miles / T{2} = 30 mph. 2 miles = 30 mph x T{2}. 2 miles / 30 mph = T{2}. 2/30 = T{2}. 1/15 hour = T{2}
The first mile takes the same amount of time as the total trip of two miles. The second mile must be completed instantaneously. The second mile has a velocity of infinity
1 mile / 0 hours = V{3}
and this is why traffic engineering differentiates between time average and distance average when it comes to average speed
I opted to change the relative speeds to 30 mph for the first half, and 60 mph average. Makes it easier to think in terms of miles/minute.
I saw the average speed problem in a book many years ago, and I remember when I was 14 presenting this problem to my friends. They refused to believe that reaching the target average speed was impossible. Play confidently asserted the time had nothing to do with speed. One friend even claimed that his truck could accomplish it. I think they know better now. At least, I hope so.
Thank you for this video!
I think the first problem is a good way of illustrating how you should always extrapolate what you're given to be true, into all of the other things that are also true as a result. I was able to get to the first "correct" answer, but the other solution can change how you look at stuff like this, which is the entire point of these types of videos.
Oh thank goodness. I clued in on r=b immediately, but thought I must have missed something.
Actually, if the 30mph is rounded to the nearest integer, this allows you to travel at an average of 29.5mph and still qualify. This works out at around 885mph for the downhill portion, which is less than 20% over the current land speed record. And so, of course, far less than infinity
Sure, if you change the problem, then you get a different answer...
@@kobeballer if you think specifying the level of precision changes the problem, please never consider becoming a scientist, or at the very least an engineer
@@mickdavies5647 lol so if my engineering manager asks me to design a part or process to 0.001 precision, then I can just present something to 0.01 and tell them "you shouldn't be in engineering if you think changing the precision changes the problem"? 😂😂
@kobeballer well, since you seem to think you can just consider the precision to be irrelevant, or ignore the fact it hasn't been provided, I pity both your manager and your clients
I've seen this riddle before and had hoped for a relativistic approach to the solution
well because the other diagonal would also be b, it would be a straight line going from the center to the edge of the quarter circle, which means b is the radius of the circle. as for a and c, idk lol maybe somehow would be able to compute it in terms of r and b using trig functions.
I came across the radius problem on an IQ test, many years ago. I solved it instantly. Take that Feynman!
Wow , so amazing technique of puzzle #1.
For the ones who have been averaging the speeds: The cause of confusion here is that there are two possible physical quantities that this special case question can talk about.1. Equal distance segments: You cannot simply average the speeds. 2. Equal time segments: say, 15 miles per hour for 1 hour and 45 miles per hour for 1 hour (as an example), here you can simply average the two speeds.
So If I drive 15 mi /hr for 1 mile and 45 mi/hr for the 2nd mile- what is my avg speed?
@@SunlendingSky Your question is outside of the described special case of equal time / distance... Use v=s*t to solve.
I use the time over distance problem everyday at work!
You'd actually only need to be going ~18000mph along your decent when you factor in human reaction time were anyone timing you with a stop watch. The average human reaction time (~.2seconds) would then provide exactly enough gap for your car to reach the summit and then the base as they pressed the button to stop the timer at 4minutes. You'd also be going 1mile in .2seconds and would likely be the consistency of strawberry jam along the car's interior.
Haha! This was a dead giveaway to me but out of interest I subjected my family to this riddle and for sure my wife responded 45mph. Next my kid who is in second grade middle school gave the exact same answer, so I asked him if he could calculate what the resulting times for both distances would be for the given average speeds. For sure he did calculate it was 4 minutes for both but due to the nature of the question he then concluded that the additional mile had to be run in less than a minute and ended up answering 60mph. So next I asked him if he could use seconds instead of minutes and the big question mark started rising above his head - ehm 3600mph? Granted I don't really know if he already learned the concept of infinity slash division by zero, I think back then I already had it in first grade, but I believe he really did get it when I gave him the answer.
Translation: "If I have to be at work by 8am, and I'm only halfway there when the clock strikes 8am, how fast do I have to drive to get to work on time?"
YOU CAN'T. You will be late to work no matter how fast you go.
If u go at infinite speed u can reach or just light speed
or it could still be 8am when you arrive.
What if the work place is in a timezone west of you? (could be as simple as a short drive crossing a border)
You are not thinking with portals
@@Hunni125 No, it could not.
the Einstein one is easier to visualize with different numbers: suppose you have an average 100 km/h to go 100km. What speed do you need to go through the next 100km to average 200km/h in the total trip?
It's more immediately evident that a 100km trip at 100km/h would take 1h, and that a 200km trip at 200km/h would have to take 1h as well.
5:15 "Imagine we roll one *dice."* Nope, I can't. But I can imagine rolling one *die.* 🎲
“Dice” is now totally accepted as a singular term.
@@ollyrukesapparently it's not accepted by everyone
@@jerkisoni like to eat one hamburgers every wednesday
Dr. Evil's son: This is what I'm talking about... YOU ALWAYS DO THAT!
@@xaigamer3129 Dice doesn’t have an s at the end of an existing word.
If r=b you could say r=b= sqrt of (r-c)squared + (r-a)squared then all terms are included.
IMHO it is best to consider units. In the wrong solution the 2 must be miles (or else you woud clearly be assuming that the different distances were irrelevant, which is obviously wrong) and mph/m = 1/h which is not a speed.
When driving long distances over expressways where the speed limit is sometimes 70 mph or more, I TRY to have an average speed of about 65 or so. But it's really hard when you have to stop every 3 hours or so, even for just a few minutes (for gas, food, and the call-of-nature). Even brief stops quickly drops the 'average speed'.
You know, I actually thought about it using the second method you were talking about, with the idea of 4 minutes. I am not the best at math in my head for equations like this, but I did notice that it seemed like there wasn't enough time remaining to go that fast although I wasn't sure exactly why.
I didn't overthink the first one, it seemed obvious using trig that r=b.
The second one is deeply obvious unless you think 6+5 and 5+6 are different.
The last one takes a moment to realise the average speed is already sucked up by the ascent time.
Sometimes overthinking or expectation stops the otherwise obvious from being apparent.
So Feynman's solution actually presents a restriction on viable values for a,b,c. If you instead use b for r in his equation and then solve for b you get that
b=a+c-sqrt(2ac)
so for any other value of b given a,c then the problem is unsolvable because the combination is invalid.
For example if a=2,c=9 then we get b=17. If instead we are given instead b=19 then there is no possible rectangle with diagonal 19 such that to given quarter circle has the extra lengths a,c as given.
You got the wrong sign there. It should be PLUS sqrt not minus. There’s no way 2+9 minus a positive number equals 17 so you used it correctly in your first test.
That's made me feel great about my hitherto averagish brain. I got that Einstein-stumper straight away, just by looking at the thumbnail.
Go brain! Go brain!
The speed for the second mile increases exponentially with every second spent driving under 30mph and hits infinity at 4min. If the car spent even a tiny fraction of a second driving faster, then a correct speed could be calculated to solve the puzzle. For example 15.0627mph for the first mile would leave 1 second at a speed of 3600mph to average out to 30mph for the whole trip.
In an old car that can only manage 15mph up the hill . . .
@@tezzerii Well the whole question doesn't make any sense since a car has to accelerate/decelerate in the first place.
If you average over distance rather than time, 45mph is correct. There's just the small problem that we never average speed over distance
Before watching, just looking at the thumb, I’m saying that this is impossible because to average 30mph the 2 mile trip must be done in 4 minutes, however it has already taken 4 minutes for the first mile at 15mph
Question in the thumbnail... I went directly to : How much time it take for the first part (4 minutes) and how much time do I have for the full trip (also 4 minutes) ?
Einstein got it right. Thats what the video said. Also, he had time left to make bagels!
I vaguely knew about the Einstein riddle but I never figured out why it truly was the case.
Of course, students might have an easier time understanding it by considering that a single B on their record means they'll never have true straight As.
one way to look at the last one is if you double the speed required for average to travel you would double the distance in the same amount of time.
My "haven't watched the video yet" answer:
Isn't the answer undefined (Basically infinity)? Speed is distance over time, to double the speed you need to double the total distance or halve the total time. We covered half the distance already, so the total distance is double, but we also used 100% of the time. We need to travel 1 mile in 0 time to double the average speed. 1÷0 mph. Undefined.
The instinctual answer is probably 45, but that averages 22.5mph because time is the denominator and the time is not constant between the 2 legs.
Ah thank God. I was thinking all the while “how is the radius not simply b?” Good thing it is. Was beginning to doubt myself.
6:32 thanks, i needvto hear that
Now I'm afraid to drive 30MPH.
Just look at the hill from a physics perspective instead of a math perspective. Due to significant figures you might make it if you go REALLY fast (for a car) on the descent: if the distances are 0.96 and 1.96 miles and the averages are 15.4 and 29.6 mph, but all written down with two significant figures, you would make it if you drove 2.6E+2 mph average, or accelerate with 1.6G if we don't account for stopping at the end (if I did all that correctly. Knowing my prowess with math, I probably made an error or two :P ).
On the other hand, you could also be out of time before you even get to the top. Your mileage may vary...
Last question depends on if you are the observer or not, if you are not the observer you just need to instantly travel to the bottom. You can go at the speed of light if you are the one in the car and get to the bottom instantly. (assuming the car is massless.)
You don't need to go faster than the speed of light you just need to travel at the speed of light if you are not the Observer.
Who is measuring the time spent on the decent?
If you're travelling at the sped of light, time doesn't exist. As such, if you're the one timing the decent, the answer is the speed of light.
If it's an external measurement, then yes - it's not possible.
Strictly the Leibniz problem is ambiguous: if you don't assume commutivity then you can interpret "a sum of" to be "a specific sum of" so that 5+6 is not the same sum as 6+5.
How is it ambigous? Rolling 11 is more likely and that can easily be tested.
I remember the car riddle from an old Encyclopedia Brown book.
I got the b=r one instantly but that's probably because Feynman was primed for it to be a maths puzzle and tunnel visioned himself (I assume, I don't know anything about him outside that)
ultimately, the equation becomes 1/30 = 1/30 + 1/(2v). In this, 1/(2v) must equal zero, which is not possible
It isn't necessary to travel at an infinite speed for the trip down. Just travel at the speed of light and time stands still.
At that point you need to ask what clock you are going by. Because by the clocks of the rest of the universe it still takes you some time to travel that last bit, even at the speed of light.
Love pedantry. ;)
Still, either way, the given problem has no practical solution..
The answer is C. Even though it isn't a multiple choice question.
In an old car which can only manage 15mph up the hill.
There are a few problems with this.
1) If you're moving at/very close to the speed of light and you're talking about the frame of reference of the car, then the distance will not be 1 mile anymore. Relativity causes the lengths to change too, not just time. Even if you were talking about a wacky frame of reference the math would still work out the same and it would still be impossible. The only way you could come up with different answers is if you're talking about things like trying to calculate the speed of an object by taking the distance from 1 frame of reference and the time from a different frame of reference, which is a nonsense measurement which doesn't actually calculate anything useful (by that logic you could come up with literally any number, you could even say the car is travelling backwards if you wanted to).
2) If you're talking about the frame of reference of the object itself, then "the car isn't moving at all" - there's no such thing as measuring the speed of an object from its own frame of reference, because obviously the car is always at the same position relative to itself.
Thumbnail problem: Infinity. The car would have to instantly travel the remaining mile in order for it to average 30mph for the whole trip.
Circle radius problem. The answer is just r = b. The two diagonals of a rectangle are of equal length. One diagonal is the radius. The other you've labeled b.
Circle radius is a trick question, there is no quarter circle because the two radii are unequal
@@bobross7473 - Perhaps. I've encountered other trick problems that are effectively "I purposefully drew this inaccurately in a misleading way.".
I refer to it as "a 5th-gear brain working on a 2nd-gear problem." If you think of the brain as a car engine, you can think of working on higher-end problems as being in a higher gear. Solving a simple problem with your brain in low gear is like trying to drive at a slow speed in a low gear; the engine handles it just fine. Trying to drive slow in a high gear, however, causes so few RPMs that the engine stalls and dies out, so it can't do it. ...A bit of a weird analogy, I admit, but it seems to get the point across.
my answer on intuition while he was reading the problem was, his trip would have to end at the top of the hill
My intuition said 60 mph, because if we go half the speed in the ascent, then we must double the speed in the descent. Alas, calculating it gave an overall speed of 24 mph. Very interesting problem!
If I do the second half of the trip at the speed of light, then time would stop for me.
I got the last puzzle right within seconds without calculations. I ride a bicycle a LOT, and I calculate my ride times by my average speeds. If I'm X miles from home, how fast do I have to go to get there in Y minutes? Yes, I call ahead and tell people I'm going to be late often. Why do you ask?
Insane! I was puzzled just like Einstein was!
Used to do something similar to the driving problem when driving back and forth work to work out total mpg for the day when car only reports each journey. 😂 I know, but was quite mindful in a way. Anyway, this works similarly in that the average for the whole journey is the reciprocal of the average of the reciprocals of the two halfs. So, the initial calc would work if you were using 1/average speed instead of averag speed in esch case.
i will literally confuse my friends with this question before our math exam saying that this is in our syllabus xdxd
The speed would have to be infinite since a 30mph avg over 2 miles takes 4 minutes and so does a 15mph average over one mile. Since the allotted time was already all used up, the second mile would have to cover 1 mile in zero time.
The first one... if you draw the diagonal in the other direction then you get the radius is....B.
The second one: Nope. Theres 1 way to get 12 and 2 ways of getting 11. (5-6) and (6,5)
Wrong.
There are 2 ways of getting 11: 5+6 and 6+5
@@Stubbari He said there were two ways. There was just a typo putting a hyphen instead of a comma. It was perfectly obvious what he meant.
@@Raven-Creations Yep, I see he edited the mistake.
I got the dice thing & I said the car would have to teleport instantaneously to the bottom. (I've heard a version of the last puzzle before, but I got the answer then too.) My answer to the 1st puzzle was a bit more complicated, I should have noticed r = b. I said r = sqrt (b^2- (r-a) ^2) +c & r = sqrt (b^2- (r-c) ^2) +a. This seems a valid answer to me, but what do you think?
There are several lines of thought to immediately get the answer to the Einstein puzzle. One of them is: If you you have a certain speed for a given distance and you want double the speed on double the distance then of course the time has to be exactly the same. This is true for any non-relativistic speed and any non-relativistic distance.
If it's easier for you to calculate with certain numbers. Just imagine the first part of the hill to be exactly 15 retard units. Then the ascent takes 1 hour. The total distance would be 30 retard units and hence you also have 1 hour for the full distance to get the desired average speed. Which is of course impossible unless you are a massless particle...