Question

The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The freight train is traveling at 15.0 m/sin the same direction as the passenger train. The engineer of the passenger train immediatly applies the brakes causing a constant acceleration of .100m/s^2 in adirection opposite to the trains velocity, while the freight train continues with constant speed. will there be a collision?

Answer 1

Okay so to start off let's describe the motion of both trains using the kinematics equation xf = xi + v*t + (1/2)*a*t^2.

The first train has a velocity of 25 m/s, an acceleration of -.100 m/s^2 and we will say it has a initial starting position of 0. so our equation would bexf = 0 + 25t -.05t^2

The second train has a velocity of 15 m/s, no acceleration, and because it is 200 meters away we will say it has a starting position of 200 meters. So ourequation would be xf = 200 + 15t

Since we want to know when they will collide we know that their final position will be at the same place. So we can substitute one of the equations intothe others, hence: 0 + 25t -.05t^2 = 200 + 15t

If we then put it into this form: .05t^2 -10t + 200 = 0 then we can solve for the time it took the trains to crash using the quadratic formula. It comesout to be 22.54 seconds. Then we plug that time into either of our two first equations and we get a distance of 538.1 meters.

Therefore, there will be a collision