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

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

Grade:12

1 Answers

Arun
25750 Points
6 years ago
Dear Pritam
 
 
Distance between car and bus = 200m
Acceleration of bus                   = 2m/s2
Acceleration of car                    = 4m/s2
Relative acceleration of car with respect to bus = 4m/s2 – 2m/s2 = 2m/s2
EXPLANTION- Imagine you are sitting in the bus and which starts moving in forward direction with an acc. of 2m/s2 and a car follows you with an acc. 4m/s2. Therefore acceleration of car with respect to you (that is as seen by you) would be 2m/s2.
therefore distance(S) = 200m , initial velocity = 0 (since at rest), relative acc. = 2m/s2
Applying second equation of motion 
S = ut + 1/2at2 
200 = 0 + 1/2*2*t2
t2 = 200
t = 10\large \sqrt2 sec
t = 14.14 sec
 
Regards
Arun (askIITians forum expert)

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