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. 6s
B. 5s
C. 3s
D. 8s

Answer

D

Explanation

`S_1 = 1/2 at^2 = 1/2 xx 1 xx t^2 = (t^2)/(2) m`
`S_2 = ut = 10xx t`
Now `10t = 48 + (t^2)/(2) rArr t^2 - 20t + 96 = 0`
(t-12) (t-8) =0
t = 8sec and t = 12 sec
least time = 8sec