Walking on a train: A train of proper length L and speed 3*C / 5 approaches a tu

Question

Walking on a train: A train of proper length L and speed 3*C / 5 approaches a tunnel of length L. At the moment the front of the train enters the tunnel, a person leaves the front of the train and walks (briskly) toward the back. She arrives at the back of the train right when it (the back) leaves the tunnel. a) How much time does this take in the ground frame? b) What is the person's speed with respect to the ground? c) How much time elapses on the person's watch? Please quickly explain the steps/principles used

Explanation / Answer

b) solving first part, we first imagine we are sitting in train so time taken by person to go to end will be same in which train covers L distance so 5L/3C,,,so speed of man will be same as that of train but when we see from ground frame his speed will appear to be 2*3C/5. a) time will be 5L/6C,, c) sitting inside train is like train is at rest and person moves L distance so time elapsed in his watch = 5L/6C

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