A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts.
If `S=25.5` m and passenger keeps running, find the time in which he will catch the train:

Options

A. 5 sec
B. 4 sec
C. 3 sec
D. `2sqrt(2)` sec

Answer

C

Explanation

At time t, `X_(t)` and `X_(p)` are coordinates of train and passenger respectively.
`X_(t)=(1)/(2)a_(1)t^(2)` and `X_(p)=v_(P)t-S`
If passenger catches the train, ,brgt `X_(t)=X_(p)`
or `(1)/(2)a_(1)t^(2)=v_(P)t-S or t=(V_(p)-sqrt(V_(P)^(2)-2a_(1)S))/(a_(1))`
`=(10-sqrt((10)^(2)-2(1)(25.5)))/(1)=3` seconds