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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, away from car A along the same direction. Find the time when car A overtakes car B.

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

The correct option is A 4 s
Relative velocity of A with respect to B is vAB,

At the start of the motion

vAB=vAvB=01=1 m/s

Relative acceleration of A with respect to B 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 where (x=12 m), initial relative velocity (vAB=1 m/s) and aAB=2 m/s2, we get

12=1t+12×2×t2
t2t12=0,
by solving for t we get
t=4 s, t=3 s
t=4 s As, time cannot be negative,

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