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Question

(D) The particle starts from rest,so its initial velocity $u = 0$. Given acceleration $a$ is uniform,the velocity $V$ at time $T$ is given by $V = aT$

Vedclass · Accepted Answer

$\frac{1}{2} \left( \frac{MV^2}{T} \right)$

Vedclass · Answer

(D) The particle starts from rest,so its initial velocity $u = 0$. Given acceleration $a$ is uniform,the velocity $V$ at time $T$ is given by $V = aT$,which implies $a = \frac{V}{T}$. The force acting on the particle is $F = Ma = M \left( \frac{V}{T} \right)$. Power $P$ delivered to the particle is defined as $P = F \cdot V$. Substituting the values of $F$ and $V$: $P = \left( M \frac{V}{T} \right) \cdot V = \frac{MV^2}{T}$. However,the instantaneous power at time $T$ is $P = F \cdot V = (Ma) \cdot (aT) = Ma^2T$. Since $a = V/T$,$P = M(V/T)^2 T = \frac{MV^2}{T}$. Wait,let's re-evaluate: The work done $W = \Delta K = \frac{1}{2}MV^2$. Average power $P_{avg} = \frac{W}{T} = \frac{MV^2}{2T}$. Looking at the options,option $D$ represents the average power delivered to the particle.

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

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