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A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2 m / s^{2}$ and the car has acceleration $4 m / s^{2}$ the car will catch up with the bus after a time of:

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$\sqrt{120} s$

15s

$\sqrt{110} s$

$10 \sqrt{2} s$

answer is C.

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Detailed Solution

$u_{car} = u_{bus} = 0 m / s$

$a_{Car} = 4 m / s^{2} a_{Bus} = 2 m / s^{2}$

$a_{CB} = a_{Car} - a_{Bus} = 4 - 2 = 2 m / s^{2}$

$S = ut + \frac{1}{2} {at}^{2}$

$200 = \frac{1}{2} \times 2 t^{2} (∴ u = 0) Hence, Car will meet the bus after time t = 10 \sqrt{2} \sec$

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