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 2m/s² and the car has acceleration 4m/s². The car will catch up with the bus after a time of:

Relative acceleration of the car with respect to the bus is

  4 − 2 = 2m/s²

Now initial distance is 200m

as the Bus is taken as a reference so it can be treated as a rest

So we can use s = at²/2 because the initial velocity is zero.

So time t = $\sqrt{2s/a}$= $\sqrt{(2\times 200)/2}$ = $10\sqrt{2 }sec$