A bus begins to move with an accelaration of ` 1 ms^(-1)` . A man who is `48m` behind the bus starts running at ` 10 ms^(-1)` to catch the bus, the man will be able to catch the bus after .

Options

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

Answer

d

Explanation

`a=1ms^(-2)`
Bus starts from rest and travelled distance .S. in time .t.
So, `S=0+(1)/(2)at^(2)`
`=0+(1)/(2)t^(2)=(1)/(2)t^(2)`
A man is standing 48m from bus, he also takes time to catch the bus with velocity v=10ms
`rArr (48+S)=vt`
`48+(1)/(2)t^(2)=10t`
`t^(2)-20t+96=0`
t=12 seconds
t=8seconds
The minimum time taken by man is 8 seconds.