Thanks! This method is very useful in more complicated scenarios, the problem being that typically people shortcut this problem using relative velocity thus never learning the 'formal' way to solve the problem they will need later on down the line.
You just hinted at it, but you could have solved the time used to catch up using a reference frame moving with the blue car, and solve for the red car to move 100 mt
thats how many people solve this problem in their head (including me). the issue arises when the objects start accelerating, at that point you need to go back to the full position equations. i find it is best to teach the full version from the beginning so people dont get confused later on with more complicated problems.
helpful video but if you're someone who's fairly comfortable with physics, using relative velocity makes this much simpler. Relative velocity of car 1 wrt car 2 = 25-20=5 m/s. Distance to cover = 100m time taken = 100/5 = 20secs Distance travelled by car 1 = 25x20 = 500m.
@@INTEGRALPHYSICS completely agree. but question solving is always trying to find the simplest approach to the solution. The complexity can always increase. Say for example, the acceleration is time dependent and not constant, then even the equations of motion at constant acceleration don't hold and you'll have to integrate. Although I definitely do believe that the method you used here is better for beginners because it gives more conceptual clarity, thus why I said "if you're someone comfortable with physics".
Why is it 5t=100? I didn't understand that part. (I try using the formula since i have another problem that involves acceleration and deceleration). Hope you can explain it to me.
U can solve it within 5 seconds in your brain. Relative velocity is 25-29=5 m/s and distance between them is 100. So time taken is relative velocity/ time = 100/5 = 20 secs
He just explained the position equation, if you solve this question having both car some acceleration it will be difficult so solve by relative instead it's easy by using the formula
25-20=5 t=100/5=20 second 😅😅😅😅😅😅😅Such question which you all consider hard problem , this question in our area anyone can solve in our india even those who tend buffalow 😅😅😅
A buffalo is walking at 2m/s. 100m behind the buffalo, a buffalo herder is chasing the buffalo at 2.5m/s. Calculate the time it takes the buffalo herder to catch the buffalo.
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Love that this is all color coded!
I hear that's important for some people. =)
Useful video!! My teacher always tell me to learn the hard way, so in exam that wouldn’t be easy as that.
Thanks! This method is very useful in more complicated scenarios, the problem being that typically people shortcut this problem using relative velocity thus never learning the 'formal' way to solve the problem they will need later on down the line.
Great Way of Explation and Moreover That Presentation is just Awesome !
Thanks a ton
Thank you so much ......from Iran
You just hinted at it, but you could have solved the time used to catch up using a reference frame moving with the blue car, and solve for the red car to move 100 mt
thats how many people solve this problem in their head (including me). the issue arises when the objects start accelerating, at that point you need to go back to the full position equations. i find it is best to teach the full version from the beginning so people dont get confused later on with more complicated problems.
@@INTEGRALPHYSICS We could argue that even the "full form" is not general enough since it assumes constant acceleration, but I see your point.
Very true!
How would you modify the equation if the time is staggered?
use t for the object that starts last, and use time is t+1 (or whatever the stagger is) for the first vehicle.
helpful video but if you're someone who's fairly comfortable with physics, using relative velocity makes this much simpler.
Relative velocity of car 1 wrt car 2 = 25-20=5 m/s.
Distance to cover = 100m
time taken = 100/5 = 20secs
Distance travelled by car 1 = 25x20 = 500m.
true, however if one of the objects is accellerating, that method leaves most people stranded.
@@INTEGRALPHYSICS completely agree. but question solving is always trying to find the simplest approach to the solution. The complexity can always increase. Say for example, the acceleration is time dependent and not constant, then even the equations of motion at constant acceleration don't hold and you'll have to integrate. Although I definitely do believe that the method you used here is better for beginners because it gives more conceptual clarity, thus why I said "if you're someone comfortable with physics".
I have a physics test on Friday let's Hope this helps otherwise I'm getting kicked from school 😭😭😭😭
Good luck.
😂😂😂watch my comment
Why is it 5t=100? I didn't understand that part. (I try using
the formula since i have another problem that involves acceleration and deceleration). Hope you can explain it to me.
25t-20t = 100
5t = 100
The Question is way too simple, will never get this question on an exam
People gotta start somewhere.
I have a test in 4 hours am I failing
Try one hour 😭😭
Reading this and it says the comment was 5 hours ago... Hope it went well.
@@INTEGRALPHYSICS I got a 30/35 points, but that was only 1 question wrong so I take that as a win
U can solve it within 5 seconds in your brain. Relative velocity is 25-29=5 m/s and distance between them is 100. So time taken is relative velocity/ time = 100/5 = 20 secs
He just explained the position equation, if you solve this question having both car some acceleration it will be difficult so solve by relative instead it's easy by using the formula
25-29=5 m/s ? can u explain how u will get it
The given velocity of car 1 and 2 is 25 and 20 respectively so that the relative velocity of two car is 25-20=5
Such question which you all consider hard problem , this question in our area anyone can solve in our india even those who tend buffalow 😅😅😅
25-20=5
t=100/5=20 second
😅😅😅😅😅😅😅Such question which you all consider hard problem , this question in our area anyone can solve in our india even those who tend buffalow 😅😅😅
A buffalo is walking at 2m/s. 100m behind the buffalo, a buffalo herder is chasing the buffalo at 2.5m/s. Calculate the time it takes the buffalo herder to catch the buffalo.