Physics question

[Kelemvor]

Originally posted by: GML3G0
So here's the problem.

Train moving at 36 m/s hit's brakes and decellerates at a rate of 3 m/s^2 to avoid colliding with a train 100m in fron of it which is moving at a constant velocity of 11 m/s in the same dirrection. Will they collide?

I say yes, they will, but according to my teacher they won't...

Here's what I did:

You want the faster train to slow down to 11 m/s before it reaches the same distance as the other train.

So...

V(final) = V(initial) + at
11 = 36 + (-3)t
t= 8 and 1/3 s


Train 1

d = 36 ( 8 1/3) + 0.5 (-3) (8 1/3) ^2
d = 195.83333...

Train 2

d = 11 (8 1/3)
d = 91.666...

195.8333.... = 91.6666... + 100 ? No, so the first train overtook the first, so obviously they collided...

Teacher's reasoning was:

When you put the values in the distance formula, it takes 12s for the fast train to come to a halt, and then plug 12s into the distance formulas for both trains and the slower train went further. OK, but what about before that time????

That's why his method doesn't work...

Train 1: ||----------|1|--------|2|---|3|--|4|-|5|

Train 2: ||---------------|1|--|2|--|3|--|4|--|5|--|6|--|7|

So who is right?

To prove the Prof wrong, figure out at what time(t) they actually collide and have him plug those numbers into the formula to see what happens.

 

[PurdueRy]

You have an incorrect assumption, you assume that the first train will be gaining on the second train when its velocity is less, obviously this is not true...Sorry dude

You should have your final velocity for the first train be the speed of the second train. That way you can find out how close it gets before it starts losing ground again.

 

[PurdueRy]

Originally posted by: PurdueRy
You have an incorrect assumption, you assume that the first train will be gaining on the second train when its velocity is less, obviously this is not true...Sorry dude

You should have your final velocity for the first train be the speed of the second train. That way you can find out how close it gets before it starts losing ground again.

EDIT: My bad it looks like you did assume that. Let me look at it again

Didn't mean to quote myself....

I like your answer. I agree with you

 

[BigJ]

Originally posted by: GML3G0

Oh, and by the way, I wasn't trying to find the time in which they collided. I know it's not 8.333, I'm just saying if the faster train's position at that point was less, then they wouldn't have collided. I was just using that as a reference point. I know how to figure out the time of impact, I just wasn't looking for that. Anyways, thanks everyone.

Doing this you got lucky. If you had picked a point time after 10, the originally slower train would be distancing itself from the once faster train. Then you wouldn't have been able to prove anything.

Your best bet for collision problems is setting the displacement equations equal to each other, adjusting for starting positions.

 

[mugs]

Originally posted by: GML3G0

Originally posted by: mugs
Now the question is, can you convince your teacher that she's wrong without showing her this thread (which might piss her off)?

him*

My bad, I saw a physics teacher that was clueless about physics and I just assumed.

 

[bonkers325]

they will crash because train B (the constant velocity one) wont be able to distance itself away from train A fast enough.

edit: and your solution is wrong.

use v = v(initial) + a*t <----- you did this part correctly to find the time needed to slow down to 11m.s


then you need to use x = x(initial) + v(initial) * t + .5 * a * t * t

solve the distance travelled for the first train and you will find it travels 195.83 meters in 8.33 seconds. since the trains are 100m away, it will travel 95.83 meters (with train B as the frame of reference) - think of the initial position as -100meters.

now you find the time required for train B to travel 95.83 meters is 8.712 seconds, and bam you ahve a collision

 

[GML3G0]

Originally posted by: bonkers325
they will crash because train B (the constant velocity one) wont be able to distance itself away from train A fast enough.

edit: and your solution is wrong.

use v = v(initial) + a*t <----- you did this part correctly to find the time needed to slow down to 11m.s


then you need to use x = x(initial) + v(initial) * t + .5 * a * t * t

solve the distance travelled for the first train and you will find it travels 195.83 meters in 8.33 seconds. since the trains are 100m away, it will travel 95.83 meters (with train B as the frame of reference) - think of the initial position as -100meters.

now you find the time required for train B to travel 95.83 meters is 8.712 seconds, and bam you ahve a collision

How is my solution wrong? I proved that Train A overtook train B. I'm not trying to say what time they collided, I'm just tring to prove it passed it, and it did.

I know this method may not work all the time, but it was a quick way of proving they crashed. If this method proves they crashed, great, then it's 100% they crashed, but if this method tells me they didn't, then there's the possibility they crashed before, in which case I would have to fully work out the problem and set the distances equal and solve quadratically.

I know all this, I'm just trying to use a quick method to point out to my teacher that he had made a mistake...

 

[bonkers325]

i guess its wrong because on a test you'd get drilled for doing it the way you did. but thats just from my perspective

what you said is "right" but if you show that solution to your teacher then they may use it against you to show you what you did wrong in the solution and make you look like a total goof.