A car is standing 600m behind a bus which is also at rest. The two start moving at the same instant but with different forward acceleration. The bus has acceleration 3 $m / \sec^{2}$ and the car has acceleration $6 m / \sec^{2}$ .The car will catch up with the bus after a time of

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5sec

10sec

10 $\sqrt{2}$ sec

20sec

answer is A.

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Detailed Solution

$\begin{array}{l} relative separation of car w . r . t bus = s_{CB} \\ U_{C} = U_{B} = 0 m / S u_{c b} = 0 m / s \\ r e l a t i v e a c c e l e r a t i o n o f c a r w . r . t b u s a_{C B} = 6 - 3 = 3 m / \sec^{2} \\ S_{C B} = U_{C B} + \frac{1}{2} a_{C B} t^{2} \\ \Rightarrow 600 = \frac{1}{2} 3 t^{2} \\ \Rightarrow \frac{1200}{3} = t^{2} \\ \Rightarrow t = 20 s \end{array}$