# 2 trains headed for eachother

1. Aug 31, 2007

### Yapper

For phys 170 we do assignments on a website from the textbook edugen.wiley.com.

1. The problem statement, all variables and given/known data

A red train traveling at 72 km/h and a green train traveling at 144 km/h are headed toward each other along a straight, level track. When they are 870 m apart, each engineer sees the other's train and applies the brakes. The brakes slow each train at the rate of 1.0 m/s^2. Is there a collision?
If so, give (a) the speed of the red train, (b) the speed of the green train, and (c) the separation between the trains when they collide (0 m).
If not, give (a) the speed of the red train (0 m/s), (b) the speed of the green train (0 m/s), and (c) the separation between the trains when they stop.

2. Relevant equations
v0= initial velocity
v=v0+at
dx=(1/2)at^2 + v0t

3. The attempt at a solution
72km/h = 20 m/s 144 km/h = 40 m/s

Ok so I checked the stopping distances and the red train will stop after 20 seconds, 200 m farther down and after that 20 sec the green train will be (1/2)(-1)20^2 + 40(20) = 600m farther down the track, so I assumed the red train's velocity to be 0m/s since it stops before the collision.

Then for the green train I set 1/2(-1)x^2 +40x = 670, to see how long it would take for the green train to hit the stationary red train, and solved for x. I got 23.8755 seconds. Then solved for velocity 40-23.8755 =16.1245 m/s.
So I inputed a) 0 m/s b) 16.1245 m/s c) 0 m

The website said I was wrong and after many different tries it told me that the answer was:
(a)
Number: 3.127717305696
Units: m/s
(b)
Number: 23.127717305696
Units: m/s
(c)
Number: 0
Units: m

2. Aug 31, 2007

### rootX

haven't got anywhere..

but I am pretty sure that you need to use relative distance for these like questions.
So, find an equation for the distance between two trains. I got:
d(t) = 870-60*t+t^2 (using relative a and v)
and it says d(t) is min when d(t) = -30, and t = 30,
so that means they collided,

and d(t) = 0 when t = 24.52277442

according to that answer relative v is 20 when d btw them is 0,
but mine equations says relative v is 10.. so try finding some error in my way

Last edited: Aug 31, 2007
3. Aug 31, 2007

### Yapper

I did it first that way but then I realized that the red train stops before 24.522, so for those extra 4.552 seconds its moving backwards, which shouldn't be possible solely from the application of brakes. And based on the answer the time to collision is 40-23.127 which is like 16.something seconds which I don't get near at all. And isn't the relative velocity going to be 60-2t(until red train stops) since both trains are decelerating away from each other at 1m/s^s and relative velocity is 60m/s to start with.

Last edited: Aug 31, 2007
4. Aug 31, 2007

### rootX

umm.. I think that first approach is not so correct, because the collision occurs before the red train stops....
relatve a is 1-(-1)

5. Aug 31, 2007

### Yapper

according to them yes, but they don't show there work and I have no idea how they figure the trains hit before the red train stops seeing as how after 20 sec the green train has closed 600 m and the red train has closed 200m and stopped, and they started 870m apart...

((-1m/s^2)(20s)^2)/2 +20m/s*20s = 200 m ((-1m/s^2)(20s)^2)/2 +40m/s*20s = 600m 600+200=800 which leaves 70m between them at t = 20 sec

Last edited: Aug 31, 2007
6. Sep 1, 2007

### learningphysics

Yeah, I think something's wrong with the question... Yapper, I'm getting exactly the same answer as you...

My work:

I'm getting that the red train stops in 20s... which is at a distance of 200m... In that 20s, the green train travels 40(20)+1/2(-1)(20)^2 = 600m... so there's still a 70m gap between them... the collision still takes place, but the red train is stopped... so the velocity of the red train should be 0.

The green train travels a distance of 70m before the collision takes place... using v2^2 = v1^2 + 2as, I get the velocity of the green train as 16.1245 m/s when the collision takes place.

Last edited: Sep 1, 2007
7. Sep 1, 2007

### Yapper

K, thanks for all the help. Ill make sure to point out the problem to my prof.

8. Sep 1, 2007

### rootX

I guess mine relative distance equation is wrong because acc. to it, the green train starts traveling backwards after 20 s..
now, i know why it used to take so much time to solve these like problems.

thanks! - that other approach is much better than mine!! ^^

9. Sep 1, 2007

### rootX

new equation:
d(t) = 870-60t+t^2; t<=20 s
d(t) = 70-20(t-20)+0.5(t-20)^2 t>20