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

$\begin{array}{1 1} 4.8\;s \\ 8\;s \\ 10\;s \\12\;s \end{array}$

Can you answer this question?

## 1 Answer

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Answer : $8\;s$
$10t =48+\large\frac{1}{2}$$+ 1 \times t^2$
or $t^2-20t+96=0$
or $t^2-8t-12t+96=0$
or $t(t-8) -12 (t-8) -12 (t-8)=0$
or $(t-12) (t-8) =0$
or $t=8s$ or $12\;s$
But we are interested in minimum time.
answered Aug 19, 2014 by

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