A bus start from rest with an acceleration of$\ 1\ ms^{- 2}$. A man who is 48m behind the bus starts with a uniform velocity of $10ms^{- 1}$. The minimum time after which the ma will catch the bus

Question

Dr. Abubacker · Accepted Answer

(b) $10t = 48 + \frac{1}{2} 	imes 1 	imes t^{2}$ or $t^{2} - 20t + 96 = 0$
or $t^{2} - 8t - 12t + 96 = 0$ or $t(t - 8) - 12(t - 8) = 0$
or $(t - 12)(t - 8) = 0$ or $t = 8$ s or 12 s
But we are interested in minimum time.

Q. A bus start from rest with an acceleration of$\ 1\ ms^{- 2}$. A man who is 48m behind the bus starts with a uniform velocity of $10ms^{- 1}$. The minimum time after which the ma will catch the bus

Explanation

Reviewed by Dr. Abubacker

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