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Question:

A car is standing 200m 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 :

15s

√120s

10√2s

√110s

Solution:

Car and bus distance = 200m
car ac = 4m/s²
bus aB = 2m/s²
ac with respect to bus = 2m/s² forward
So, S = ut + 1/2at²
200 = 0 + 1/2 × 2 × t²
t² = 200
t = 10√2s