# Colliding trains

1. Sep 11, 2013

### lep11

1. The problem statement, all variables and given/known data
Collision - The engineer of a passenger train traveling at 25 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of -0.1 m/s^2, while the freight train continues with constant speed. Take x=0 as the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place?

2. Relevant equations
v0passenger=25 m/s
x0freight=200m

3. The attempt at a solution
When the collision takes place, xpassenger=xfreight
v0passenger*t +½*a*t2=x0freight +vfreight*t

Solving this quadratic equation for t gives me t=22.54s AND t=177.459...s
I choose t=22.54s because "it happens earlier"
Now x0freight+vfreight*t= 200m+(15m/s)*22.54s≈540m

Last edited: Sep 11, 2013
2. Sep 11, 2013

### MostlyHarmless

Here's a tip, use relative velocities. i.e. If you the passenger train is moving at 10 m/s, what is the speed of the freight train relative to the passenger train?

3. Sep 11, 2013

### MostlyHarmless

Sorry, I may have been unclear, if the passenger train is moving at 10m/s what would the freight trains velocity by so that their relative velocities are still the same?

4. Sep 11, 2013

### haruspex

22.54 became 24.54 then 24.534?
Check your answer by calculating where each will be after 22.54 (or however many) seconds.

5. Sep 11, 2013

### lep11

Corrected now.

6. Sep 11, 2013

### lep11

But why did I get two solutions for t and which one to pick and why?

7. Sep 11, 2013

### haruspex

Looks right now.
There are two solutions because the equations don't 'know' that the trains cannot pass through each other, or that the passenger train's acceleration will cease when it comes to rest. So the second solution corresponds to the passenger train overtaking the freight train, going into reverse and meeting it a second time.

8. Sep 11, 2013

### lep11

Would this be considered a good way to solve the problem?

9. Sep 11, 2013

### Tanya Sharma

The quadratic which i am getting is t2-200t-4000=0.

Solving it,I get t=218s , -18s .

Or may be I am making some mistake...

Edit: The equation should be t2-200t+4000=0.

Last edited: Sep 11, 2013
10. Sep 11, 2013

### MostlyHarmless

There is no need for time, although perfectly acceptable. If you look at relative velocities, we can say the freight train is stationary and making the same reduction to the passenger train gives a velocity of 10m/s, then you can use the formula Vf^2=Vo^2-a(x-xo). Where x is the position of the freight train and xo is the position of the passenger train.

11. Sep 11, 2013

### haruspex

You do not know the collision speed. The collision is necessarily represented as the two trains being in the same place at the same time. That makes it something of a challenge to solve the problem without reference to time.

12. Sep 11, 2013

### haruspex

I believe you have a sign wrong.

13. Sep 11, 2013

### MostlyHarmless

Did you mean "not necessarily"?

Velocity it relative regardless, the given values are relative to the ground, if we instead work with relative to each other, and then take the equation I posted, which is derived from solving the position formula for t, there is no need to actually find a value for t. You just by pass that step and simply find whether the passenger has a velocity

Edit: I read your post again, I don't see why we care how fast the one train is going when/if it collides. The question only asks if they collide. So you simply have to decide whether it has stopped before it gets to a certain point.

Last edited: Sep 11, 2013
14. Sep 11, 2013

### haruspex

No. The fact of collision is defined by their being in the same place at the same time.
You do not know the relative velocity when they collide (without first solving the problem).

15. Sep 11, 2013

### MostlyHarmless

We don't need the velocity when they collide, to find out if they collide, but the calculations are one in the same. If you get a positive, non-zero velocity from the formula I posted, then that value IS the relative velocity of the passenger train.*when they collide *

16. Sep 12, 2013

### haruspex

Your equation included two velocities (presumably the initial and final relative velocities) and an unknown distance x. Since you do not yet know the collision velocity (i.e. the final relative velocity), you have two unknowns but only one equation.
I see in one of your posts you mentioned the (relative) velocity becoming zero. That does not give you when they collide - it gives you the point at which they are moving at the same speed, which is of little relevance.

17. Sep 12, 2013

### MostlyHarmless

If you define one train as stationary at x=0 and keep the same relative velocity for the other train you the, it becomes a simple does the train stop in x distance w/ a certain initial velocity and a certain acceleration. If the freight train is stationary, then the displacement is not unknown. It is 200m. With the collision point being x=0.

18. Sep 12, 2013

### haruspex

Ok, I see. You're only addressing part (a). I should have realised that. The problem with that approach is that it doesn't help much with part (b), does it?