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Q.
A body of mass $1 \,kg$ crosses a point $O$ with a velocity $60\, ms ^{-1}$. A force of $10\, N$ directed towards $O$ begins to act on it. It will again cross $O$ in
AMUAMU 2003
Solution:
From Newtons second law $F = ma $
$ \Rightarrow $ $ a=\frac{F}{m}=\frac{10}{1}m/s^{2} $
Since, the body returns to the point $O$ under a force, its displacement is zero. From equation of motion, we have
$ s=ut+\frac{1}{2}at^{2} $
$ 0=60\,t-\frac{1}{2}\times 10\times t^{2}$
$\Rightarrow \,t=12\,s $