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

Answer

D

Explanation

`A` will be ahead of `B` when `x_A ge x_B`
`40(t - 10) ge (0)t + (1)/(2) (2)t^2`
As `A` is `10 sec` late than `B`.
`rArr t^2 - 40t + 400 le 0`
`rArr (t - 20)^2 le 0`.