Train Collision Problem - Calculating Head Start

[egrr10]

Homework Statement

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15.0 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 0.100 m/s^2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brakes.
If the speeds of the two trains and the deceleration of the passenger train remain as originally stated, how much of a head start would the freight train need in order to avoid collision?

Homework Equations

y=yo+vo*t+1/2*a*t^(2)

The Attempt at a Solution

I want to know if my set up is right.
I know then that yf(freight train) must be greater than yp(passenger) so knowing that they will collide at about 25 seconds, would I set
yf=yp and solve for distance?
yf=yo+vot
vp=vpot-(1/2)at^2
vo+(15m/s)=(25m/s)-(1/2)(-0.1m/s^2)

 

[HallsofIvy]

I think I would work with "relative" speed. In order to avoid a collision the passenger train does NOT have to come to a stop, relative to the track, but just slow to the same speed as the freight train, v= 0 relative to the freight train. The passenger train is traveling at 25.0 m/s and the freight train at 15.0 m/s so the passenger train is initially closing on the freight train at 20.0- 15.0= 10.0 m/s. At that point, the passenger train starts decelerating at .100 m/s^2. At that acceleration, how long will it take for the passenger train to slow to the same speed as the freight train- for its speed to be 0 relative to the freight train? How far, relative to the freight train, will it have gone in that time?

You can use $y= v_0t- (1/2)at^2$ to determine the distance traveled but since the acceleration is constant, you could also use the fact that the average speed will just be the average of the initial speed, 10 m/s relative to the freight train, and the final speed, 0, relative to the freight train- that is, whatever the acceleration is, the average relative speed will be 5 m/s.