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Power supplied to a particle of mass 2 kg varies with time as $P = 3 t^{2} / 2$ . Here t is in second. If the velocity of particle at t = 0 is v = 0, the velocity of particle at time t = 2 s will be

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1 ms^-1

4 ms^-1

2 ms^-1

$2 \sqrt{2}$ ms^-1

answer is C.

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Detailed Solution

From work-energy theorem,

$ΔKE = W_{net} or K_{f} - K_{i} = \int Pdt$

$\begin{array}{l} or \frac{1}{2} {mv}^{2} - 0 = \int_{0}^{2} (\frac{3}{2} t^{2}) dt \\ or v^{2} = {|\frac{t^{3}}{2}|}_{0}^{2} or v = 2 {ms}^{- 1} \end{array}$

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