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(C) Given,power $P = \frac{3t^2}{2}$.We know that power $P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v$.Substituting the given val

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$2$

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

Power supplied to a particle of mass $2\,kg$ varies with time as $P = \frac{3t^2}{2}\,W$. Here $t$ is in seconds. 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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