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At time $t = 0,$ a particle starts moving along the $x - a x i s$ . If its kinetic energy increases uniformly with time $t$ , the net force acting on it must be proportional to

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$\sqrt{t}$

$\frac{1}{\sqrt{t}}$

$t$

$c o n s \tan t$

answer is D.

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Detailed Solution

$K = c t$

Momentum, $P = \sqrt{2 m K} = \sqrt{2 m c t} = c^{'} t^{1 / 2}$

∴ Force, $F = \frac{d p}{d t}$

$= c^{'} \frac{d}{d t} t^{1 / 2}$

$= c^{'} \frac{1}{2} t^{- 1 / 2}$

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