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Standard XII

Physics

Relative Position

Two cars A ...

Question

Two cars $A$ and $B$ are initially $100 m$ apart with $A$ behind $B$ . Car $A$ starts from rest with a constant acceleration $2 m / s^{2}$ towards car $B$ and at the same instant car $B$ starts moving with constant velocity $10 m / s$ in the same direction. Time after which car $A$ overtakes car $B$ is:

14.36 s

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16.18 s

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10 s

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16.54 s

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Solution

The correct option is B $16.18 s$
Distance traveled by car $A = S_{A}$
Distance traveled by car $B = S_{B}$
According to formula $S = u t + \frac{1}{2} a t^{2}$
$S_{A} = 0 t + \frac{1}{2} \times 2 t^{2}$
$S_{B} = 10 t + \frac{1}{2} \times 0 t^{2}$
Car $A$ will overtake car $B$ when
$S_{A} = S_{B} + 100$
Putting the value $S_{A}$ and $S_{B}$
$+ 2 = 10 t + 100$
$+ 2 - 10 t - 100 = 0$
After solving for $t$ , we get
$t = 16.18 sec$ possible
$t = - 6.10 sec$ not possible
So option B is correct.

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Two cars A and B are initially 100 m apart with A behind B. Car A starts from rest with a constant acceleration 2 m/s2 towards car B and at the same instant car B starts moving with constant velocity 10 m/s in the same direction. Time after which car A overtakes car B is: