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(C) The particle starts from rest,so its initial velocity $u = 0$. After time $T$,the final velocity is $V$. The work done on the particle is equal to

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$\frac{1}{2} \frac{M V^{2}}{T}$

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(C) The particle starts from rest,so its initial velocity $u = 0$. After time $T$,the final velocity is $V$. The work done on the particle is equal to the change in its kinetic energy,according to the work-energy theorem. $W = \Delta K = \frac{1}{2} M V^{2} - \frac{1}{2} M u^{2} = \frac{1}{2} M V^{2} - 0 = \frac{1}{2} M V^{2}$. The average power delivered to the particle in time $T$ is defined as the work done divided by the time taken. $P_{	ext{avg}} = \frac{W}{T} = \frac{\frac{1}{2} M V^{2}}{T} = \frac{1}{2} \frac{M V^{2}}{T}$.

$A$ particle of mass $M$ starting from rest undergoes uniform acceleration. If the speed acquired in time $T$ is $V$,then the power delivered to the particle in time $T$ is:

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