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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} $

1 Answer

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 meena.p
Your answer is correct, no doubt in it but write the exact statement  before solving any problem.  You haven't explained  the frame of reference  between man and the bus.
Good Luck.

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