A bus starts from rest moves with an acceleration $1 \text{ m/ s}^{ 2 }$ . A man who is $48\text{m}$ behind the bus is running with $10\text{ m/s}$ . Calculate the minimum time (in seconds) taken by him to catch the bus.

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Bonus
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Can you explain why there are two time instances where he can board the bus?

The answer is 8.

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Let $x$ be the distance traveled by bus before the man board on it.Let

tbe the time taken for it.Displacement of man $= 48+x=10t$ ------

(1)Displacement of bus $= x= \frac { 1 }{ 2 } \times 1\times { t }^{ 2 }$ -------

(2)Using

(1)and(2),t=8sort=12sThere are two time instances because at first the man can overtakes the bus as his velocity is more initially.He could board the bus without overtaking.This corresponds to ${ 8 }^{ th }$ second. Then the bus accelerates and its can overtake the man.The man could also board the bus at this instance also. This corresponds to ${ 12 }^{ th }$ second.