A bus starts from rest with an acceleration of 1 `m//s^(2)` A man who is 48 m behind the bus starts with a uniform velocity of 10 m `s^(-1)`. Then after how much time the man will catch the bus ?

Options

A. 12 s
B. 8s
C. 10 s
D. 4s

Answer

b

Explanation

Let them meet in t second. Distance coverede by bus in t second x=ut`(1)/(2) at^(2)=0+(1)/(2)1xxt^(2)`
Disatnce covered by man in t `s=10xxt`
`:. 10t=48+(1)/(2)t^(2)implies t=8` s.