Part B If so, where will it take place? The engineer of a passenger train travel
ID: 2126426 • Letter: P
Question
Part B
If so, where will it take place?
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 ahead on the same track. The freight train is traveling at 15.0 in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of , while the freight train continues with constant speed. Take at the location of the front of the passenger train when the engineer applies the brakes. (Figure 1) Will the cows nearby witness a collision? yes or no? If so, where will it take place?Explanation / Answer
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 be xf = 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 our equation 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 into the 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 comes out 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