For a particle starting from rest in a straight line, it is known that the speed is proportional to the square of acceleration at any moment.Then the time rate at which all the forces acting on it are doing work is proportional to

t^{2}

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t^{3}

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t^{4}

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t^{5}

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Solution

The correct option is B $t^{3}$
$\begin{matrix} A c c e l e r a t i o n i s p r o p o r t i a n a l t o t^{2} e . g ., a = k t^{2} H e r e, a i s t h e a c c e l e r a t i o n a n d K i s p r o p o r t i a n a l l y c o n s tan t \frac{V d V}{d t} = K t^{2} \Rightarrow \int V d V = K \int t^{2} d t \Rightarrow \frac{V^{2}}{2} = \frac{K t^{2}}{3} \Rightarrow V^{2} = \frac{2 K t^{2}}{3} - - - - - - - - - (1) w e k n o w, k i n e t i c e n e r g y = \frac{1}{2} m v^{2} = \frac{1}{2} m (\frac{2 K t^{3}}{3}) = \frac{m K t^{3}}{3} (f r o m e q u t a i o n (1)) H e r e i t i s c l e a r t h a t k i n e t i c e n e r g y i s d i r e c t l y p r o p o r t i o n a l t o t^{3} . \end{matrix}$
Hence,
option $(B)$ is correct answer.

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