Acceleration from Vi, Vf and Distance

[Dekoy]

Homework Statement

7. The engineer driving the Shinkansen (Japanese bullet train)
from Tokyo to Kyoto is having a nice day until he rounds a
bend and suddenly sees a slow-moving freight train on his
track. Both trains are traveling west, with speeds vs and vf
respectively. Of course, as soon as the Shinkansen engineer
sees the freight train, he applies the brakes; at that instant,
the distance between the two trains is D. Show that if a
collision is to be avoided, the magnitude of the Shinkansen's
acceleration must be at least

a =[(Vs-Vf)^2]/2D

Homework Equations

x(T)=Xi+ViT+(1/2)aT^2
D=VsT a=vT

The Attempt at a Solution

This is my attempt at a solution I got the same thing but I'm almost 99.99% certain my method isn't the correct one so any help will really help me my quiz is tomorrow heheh, so any hint please. Thanks in advance.

Since both trains move in the same direction the relative velocity Vr=Vs-Vf since they were @ different points @ T the distance between them @ all times =D1=Vr(T)-D and D1=(Vs-Vf)T-D if they meet D1=0 then 0=(Vs-Vf)T-D gives D=(Vs-Vf)T since D is position D=x(T)=(1/2)aT^2 solve for T on both equations T=D/(Vs-Vf) and T^2=2x(T)/a
then D/(Vs-Vf)=(2x(T)/a)^(1/2) from this (D^2)/(Vs-Vf)^2=2D/a so this gives
a=[(Vs-Vf)^2]/2D

 

[diazona]

since they were @ different points @ T the distance between them @ all times =D1=Vr(T)-D

That step isn't correct - you can't say that the distance between the trains at all times is equal to v_rt - D, because there is a nonzero acceleration.

 

[Dekoy]

Thanks 1 error to go many more to come hehe, I hope I'll get it before tomorrow