# Kinematics involving two objects

1. Mar 31, 2016

### Ab17

1. The problem statement, all variables and given/known data
A passenger train is traveling at 27 m/s when the engineer sees a freight train 345 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 5.9 m/s.
(a) If the reaction time of the engineer is 0.36 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided? (m/s/s)
(b) If the engineer's reaction time is 0.73 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide? (m/s)
(c) For both reaction times, how far will the passenger train have traveled in the time between the sighting of the freight train and the collision? (km)

2. Relevant equations
Xf= Xi + vt + 0.5at^2

3. The attempt at a solution

(a) Xf1 = 27(t-0.36) + 0.5a(t-0.36)^2
Xf2 = 345 + 5.9t

Dont know what to do now. I am really confused

2. Mar 31, 2016

### jbriggs444

So you have two equations. One gives Xf1 (the final position of train 1) and one gives Xf2 (the final position of train2), both as a function of the unknown time to intercept and the unknown acceleration. That's two equations in two unknowns.

Without knowing how it's all going to work itself out, how about solving those two equations for t and a. See if any useful insights fall out.

3. Mar 31, 2016

### Ab17

Are the equations even right

4. Mar 31, 2016

### jbriggs444

The equations look right to me.

Personally, I would have approached the problem differently. I'm more comfortable using physical intuition rather than algebra. Transform to a frame of reference in which the front train is stationary see what acceleration is needed to attain a particular velocity over a particular separation distance. That becomes essentially a work = force times distance situation, easy to solve for velocity in terms of force, or for force in terms of velocity.