A 150 m long train accelerates uniformly from rest.If the front of the train passes a raikway worker 50 m away from the station at a speed of 25m/s,what will be the speed of the back part of the train as it passes the worker?
Answers
Answered by
8
Acceleration = v / t
= 25 / 50
= 0.5 m/s^2
Velocity of back of the train
v’ = sqrt(2aS)
= sqrt(2 × 0.5 × 200)
[Here S = 200 m because back of the train travels 200 m to pass the worker]
= 10√2 m/s
= 14.14 m/s
= 25 / 50
= 0.5 m/s^2
Velocity of back of the train
v’ = sqrt(2aS)
= sqrt(2 × 0.5 × 200)
[Here S = 200 m because back of the train travels 200 m to pass the worker]
= 10√2 m/s
= 14.14 m/s
Answered by
9
correct answer is 50m/s
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