How Fast Must a Passenger Train Decelerate to Avoid a Crash?

[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

 

[tms]

My attempt at the question:
V=u+at 0=80+at a=-80/t

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.

I tried to find at what time their distances were equal using

That is the right approach.

 

[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)

 

[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