# Locomotives on a collision track

1. Jan 22, 2013

### kopinator

As a high-speed passenger train, traveling at 88.0 km/h, rounds a bend, the engineer realizes that a locomotive has improperly entered the train's track from a sidetrack distance d= 420 m ahead. That locomotive is moving at 22.0 km/h. The engineer immediately applies the brakes to reduce train's speed to the speed of the locomotive ahead. What minimal magnitude of the resulting constant acceleration is needed to prevent a collision?

I thought the equation to use would be Vfinal^2=Vinitial^2+2a(Xfinal-Xinitial). I converted the numbers so the answer would be in m/s^2 but that didn't work and then I tried leaving the numbers so my answer would be in km/h^2 but that didn't work either. Any help?

2. Jan 22, 2013

### mathman

It is easier to work out if you work in the coordinate system of the locomotive. Therefore v starts at 66 km/hr, while d starts at -.42 km. The problem then is to end up with a final speed = 0 and a final distance = 0.

The equations:
0 = 66 - at
0 = -.42 +66t - at2/2

a = acceleration (I put - in the equation, a will be a positive number), t = time.

Note that the time is in hrs and the acceleration would be in km/hr2. I suggest you convert 66 to m/sec2 and use 420 m before you start.

Last edited: Jan 22, 2013