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Question

Two cars, initially at a separation of 12 m, start simultaneously. First car "A", starting from rest, moves with an acceleration 2 m/s2, whereas the car "B" which is ahead, moves with a constant velocity 1 m/s along the same direction. Find the time when the car "A" overtakes car "B".

A
4 s
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B
7 s
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C
6 s
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D
8 s
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Solution

The correct option is A 4 s
I Method

Relative velocity of "A" with respect to "B", i.e vAB, at the start of motion is :
vAB=vAvB=01=1 m/s
Relative acceleration of "A" with respect to "B". i.e. aAB is :
aAB=aAaB=20=2 m/s2
Applying equation of motion for a single body (remember the second body is rendered stationary).
x=ut+12at2
12=1×t+12×2×t2
t2t12=0
t2+3t4t12=0
(t+3)(t4)=0
t=4 s

II Method
The car A will cover 12 m more than car B

Hence, displacement of A= displacement of B+12

Now equating the displacement by 2 nd equation of motion

12at2=vt+12
12(2)t2=1×t+12

t2t12=0
(t+3)(t4)=0
t=4 s

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