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

Question

Dr. Abubacker · Accepted Answer

(c) From work energy theorem, $\mathrm{\Delta}\ KE\  = W_{net}$
$K_{f} - K_{i} = \int_{}^{}{P\ dt}$
$\frac{1}{2}mv^{2} - 0 = \int_{0}^{2}\left( \frac{3}{2}t^{2} \right)dt\ or\ \frac{1}{2}(2)v^{2} = \frac{3}{2}\left\lbrack \frac{t^{3}}{3} \right\rbrack_{0}^{2} = 4$
$v = 2\ ms^{- 1}$

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

Explanation

Reviewed by Dr. Abubacker

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