A bus starts moving with `1 ms^(-2)` acceleration. A man 48 m away from bus starts moving with 10 ms to catch the bus, then man will reach to bus after ........

Options

A. 5 seconds
B. 6 seconds
C. 7 seconds
D. 8 seconds

Answer

D

Explanation

Suppose, ,man reaches to bus at distance .d.. Equation of motion for bus,
In `d = v _(0)t + 1/2 at ^(2) , v _(0) =0`
`d = 1/2 at ^(2) " "…(1)`
and distance covered by man,
`48 +d =ut`
`d = 10 t -48" "…(2)`
`therefore` From equation (1) and (2),
`1/2 at^(2) = 10t -48`
`1 xx t ^(2) =20t -96" "[a=1]`
`t ^(2) - 20 t + 96 =0`
`therefore (t -12) (t-8) =0`
`therefore t =12 s or t =8s`
But 12 s is not given in options.
Thus, answer `t=8s`