...in motion
A man can run on the ground with maximum speed of 6 m/s.
The person stands on the rear end of the roof of a train moving with constant velocity.
The length of the train is 120m.
The man begins to run at this maximum capacity towards the front end, reaches the front end, immediately turns back and runs towards the rear end.
Find the net displacement of the man w.r.t. ground by the time he reaches back to the rear end.
The velocity of the train is 10 m/s.
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1 solution
To cover full length of the train, he takes 120\6=20 s, both ways. The speeds relative to the ground are 10+6=16 forward as he runs to the front, 10-6=4 forward as he runs to the rear. So distance covered is 16 * 20 + 4 * 20=400m.
OR>>>Since he is covering same distance + and - with same speed, w.r.t .ground he covers no distance. So distance covered by the train is the total distance covered. But the whole running is for 20+20=40s. So distance covered is 10 * 40=400m.