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A body of mass 3 kg is moving along a straight line with a velocity of 24 ms^–1. When it is at a point 'P' a force of 9 N acts on the body in a direction opposite to its motion. The time after which it will be at 'P' again is,

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16 s

12 s

24 s

8 s

answer is B.

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Detailed Solution

S = 0 \Rightarrow S = ut - \frac{1}{2}\,\,a{t^2} \Rightarrow ut = \frac{1}{2}a{t^2}

$t = \frac{2 u}{a}, a = \frac{F}{m} \Rightarrow t = \frac{2 u}{\frac{F}{m}} = \frac{2 \times 24}{\frac{9}{3}} = 16 s e c$

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