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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` |
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Answer» Correct Answer - 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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