From the given information, we can see that the relative speed of the passenger train with respect to the freight train is V_passenger = 25 m/s - 15 m/s = 10 m/s.
Let's find the time (t) it takes for the passenger train to come to a stop.
v = u + at
0 = 25 + (-0.4)t
t = 25/0.4
t = 62.5 seconds
Now, let's find the distance (d) traveled by the passenger train during this time (t).
d = ut + (1/2)at^2
d = 25(62.5) + (1/2)(-0.4)(62.5^2)
d = 1562.5 - 781.25
d = 781.25 meters
Now, let's find the distance (d_f) traveled by the freight train during this time (t).
d_f = 15 * 62.5
d_f = 937.5 meters
Since the freight train was initially 200 meters ahead, we can find the remaining distance between the two trains by subtracting the distance traveled by the passenger train from the total distance traveled by the freight train.
Remaining_distance = 937.5 - 781.25
Remaining_distance = 156.25 meters
Therefore, the remaining distance between the two trains after the passenger train stops is 156.25 meters.
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 , 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.
1 answer
1 answer