A passenger train is behind a freight train, both on the same set of tracks and

Question

A passenger train is behind a freight train, both on the same set of tracks and moving in the same direction. The freight train is moving at a constant speed of 27. 0 m/s and maintains that speed. The passenger train is originally moving at 32. 0 m/s when the engineer of the passenger train spots the freight train and applies the brakes. At the time the brakes are applied, the separation between the trains is 425. 0 meters, and the application of the brakes causes the passenger train to have a constant acceleration in the direction opposite to the train’s velocity.

A) If there is to be no collision, what must the minimum value of the magnitude of the acceleration of the passenger train?
B) Assuming this minimum value, how far will the passenger train travel before it stops, and what will be the minimum separation between the trains? How much time will it take the passenger train to stop?
C) Draw a single position-time diagram showing both trains from the initial spotting of the freight train to the stopping of the passenger train. This should be a carefully drawn graph with appropriate labels and scales.

Explanation / Answer

Velocity of Freight train=27m/s
Initial Velocity of Passenger train=32m/s
Initial separation=425m
Passenger train comes to a stop so its final velocity= zero
Let acceleration=-a m/s2

To prevent collision at minimum acceleration, the relative velocity should be zero and the separation between them is just zero. Both of these should occur at same time

Initial Velocity of Passenger train relative to Freight Train=(32-27)m/s=5m/s
Final Velocity of Passenger train relative to Freight Train=(0-27)m/s=m/s
Acceleration of Passenger train relative to Freight Train=(-a-0)=-am/s2
Distance covered by passenger train relative to freight train=425m

v2=u2+2as

(0)2=(5)2+2*(-a)*425

a=0.028m/s2

Minimum separation between two trains =0

a=0.028m/s2

Velocity when passenger train stops: 0m/s

v=u-at

0=32-at

t=32/0.028=1142.85s

v2=u2+2(-a)s

0=32*32-2*0.028*s

s=18285.71m

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