Question

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 $t=2 \, s$ will be

• A. $1 \,ms ^{-1}$
• B. $4 \,ms ^{-1}$
• C. $2 \sqrt{\frac{2}{3}} \,ms ^{-1}$
• D. $2 \sqrt{2} \,ms ^{-1}$

Answer 1

Answer: (C)

From work-energy theorem, $\Delta KE =W_{\text {net }}$
$K_{f}-K_{i} =\int P d t$
$\frac{1}{2} m v^{2}-0 =\int\limits_{0}^{2}\left(\frac{t^{2}}{2}\right) d t $
or $\frac{1}{2}(2) v^{2} =\frac{1}{2}\left[\frac{t^{3}}{3}\right]_{0}^{2}$
$v=2 \sqrt{\frac{2}{3}}\, ms ^{-1}$