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Two bikes A and B start from a point. A ...

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 :

A

`20s`

B

`8s`

C

`10s`

D

`A` is never ahead of `B`

Text Solution

Verified by Experts

The correct Answer is:
D
`A` will be ahead of `B` when `X_(A)gtX_(B)`
`40(t-10)gt(0)t+(1)/(2)(2)t^(2)`
as `A` is `10sec. `late than `B`.
`rArr t^(2)-40t+4 lt0`
`rArr (t-20)^(2) lt0`
Which is not possible . So `A` will never to ahead at `B`.
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