Question

A man is 10 m behind the door of a train. The train starts accelerating at the rate of 2m/s2 and the man starts running and able to catch the train after3 sec. What is the acceleration of the man?

Answer 1

Answer:

3.11 m/s^2

Explanation:

acceleration of train = a_t = 2 m/s^2

acceleration of man = a_m = ?

time = 3 s

distance traveled by train door in 3 sec = d_t = a_t . t^2 = 2 . 3^2

                                                         = 18 metres

distance traveled by man in 3 sec = d_m = 10 + d_t = 28 metres

Therefore, a_m = d_m/t^2 = 28/3^2 = 28/9 = 3.11 m/s^2