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(C) Given: Mass $m = 2\, kg$,Power $P = \frac{3t^2}{2}$,and initial velocity $v(0) = 0$.We know that power $P = F \cdot v = (m \cdot a) \cdot v = m \c

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(C) Given: Mass $m = 2\, kg$,Power $P = \frac{3t^2}{2}$,and initial velocity $v(0) = 0$.We know that power $P = F \cdot v = (m \cdot a) \cdot v = m \cdot v \cdot \frac{dv}{dt}$.Substituting the given values: $2 \cdot v \cdot \frac{dv}{dt} = \frac{3t^2}{2}$.Rearranging the terms: $v \cdot dv = \frac{3t^2}{4} \cdot dt$.Integrating both sides from $t = 0$ to $t = 2$ and $v = 0$ to $v = V$:$\int_{0}^{V} v \, dv = \int_{0}^{2} \frac{3t^2}{4} \, dt$.$\left[ \frac{v^2}{2} \right]_{0}^{V} = \frac{3}{4} \left[ \frac{t^3}{3} \right]_{0}^{2}$.$\frac{V^2}{2} = \frac{1}{4} \cdot (2^3 - 0^3) = \frac{8}{4} = 2$.$V^2 = 4$,which gives $V = 2\, m/s$.

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

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