# Linear acceleration question

1. Feb 12, 2013

### Woolyabyss

A passenger train, which is traveling at 80 m/s is 1500 m behind a goods train which is traveling at 30 m/s in the same direction on the same track. At what rate must the passenger train decelerate to avoid a crash?

My attempt at the question:
V=u+at 0=80+at a=-80/t
I tried to find at what time their distances were equal using
s=ut+1/2 (a)(t^2)
80t +1/2 (-80/t) t^2-1500=30t
Simplify and I got 10t=1500
t=150
sub value of t into original equation and you get a=-8/15 m/s^2
The back of my book says its 5/6 m/s^2
Any help would be appreciated

2. Feb 12, 2013

### tms

The passenger train does not have to stop, it just has to slow down to the speed of the freight train before crashing into it.
That is the right approach.

3. Feb 12, 2013

### pongo38

In your first equation you have used V=0, but to avoid a crash the passenger train only needs to decellerate to 30 m/s

(sorry, tms got there before me with same solution)

4. Feb 12, 2013

### Woolyabyss

Thanks I replaced 0 with 30 in the first equation and carried out the same method as before and got 5/6 m/s^2