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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 = 0 sec and A starts from the same point at t = 10s, then the time during the journey in which A was ahead of B is:

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20 s

10 s

8 s

A is never ahead of B

answer is D.

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Detailed Solution

A will be ahead of B when $x_{A} \geq x_{B}$

40(t-10) $\geq (0) t + \frac{1}{2} (2) t^{2}$

As A is 10 sec late than B.

$\Rightarrow t^{2} - 40 t + 400 \leq 0$

$\Rightarrow {(t - 20)}^{2} \leq 0$

Which is not possible. So A will never be ahead of B.

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