Car A and B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a=4m/s2, while car B moves with a constant velocity v=1m/s. At time t=0, car A is 10m behind car B. The time when car A overtakes car B is:

# Car A and B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a=4m/s2, while car B moves with a constant velocity v=1m/s. At time t=0, car A is 10m behind car B. The time when car A overtakes car B is:

1. A

5.5 s

2. B

7.5 s

3. C

3.5 s

4. D

2.5 s

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### Solution:

Given,

uA = 0, uB = 1m/s, aA = 4 m/s2, aB = 0

Assuming car B to be at rest, we have,
uAB= uA - uB = 0-1 = -1 m/s
aAB= aA - aB = 4-0 = 4 m/s2

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^{2}$
$10=-\mathrm{t}+\frac{1}{2}\left(4\right){\mathrm{t}}^{2}$

Solving, we get,

t= 2.5 or t= -2

Since time never is negative, t = 2.5 s is the correct response.

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