A man is 9 m behind a train at rest. The train starts with an acceleration of 2 `"m.s"^(-2)` and simultaneously the man starts running. He is able to board the train somehow after 3 s. Find the acceleration of the man.

Explanation

While boarding the train the positions of the man and of the train must be the same. Let the acceleration of the man be a and the distance traversed by the train in 3 s be x.
We know, s = `ut+(1)/(2)`. Here u = 0 as the train as well as the man starts from rest.
So, for the train `x=(1)/(2)xx2xx(3)^(2) " "`or, x = 9 m
Thus the distance traversed by the man in this time = 9 + 9 = 18 m
Then , s = 18 m, u = 0 and t = 3s
`:. a = (2s)/(t^(2))=(2)/(3^(2))xx18 = 4 "m.s"^(-2)`.