Two bikes `A and B` start from a point. A moves with uniform speed `40 m//s and B` starts from rest with uniform acceleration `2 m//s^2`. If `B` starts at `t = 10` and `A` starts from the same point at `t = 10 s`, then the time during the journey in which `A` was ahead of `B` is :

Options

A. `20 s`
B. `8s`
C. `10 s`
D. `A` is never ahead of `B`

Answer

D

Explanation

A will be ahead of B when `x_(A) gt X_(B)`
40(t-10) gt (0) t +`1/2(2) t^(2)`
as A is 10 sec. late than B.
`rArr t^(2)-40 t +400 lt 0` `rArr (t-20)^(2) lt 0`
Which is not possible. So A will never be ahead at B.