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Power supplied to a particle of mass 2 kg varies with time is $P=\large\frac{3t^2}{2}$ watt, velocity of particle at $t=0$ is $v=0.$ The velocity of particle at $t=2s$ will be

\[(a)\;1\;m/s \quad (b)\;2\;m/s \quad (c)\;4\;m/s \quad (d)\;2 \sqrt 2\;m/s \]

1 Answer

From work energy Theorem
$\Delta KE=W_{net}$
$\large\frac{1}{2}$$mv^2=\int \limits _0^2 p dt$
Hence b is the correct answer.
answered Jul 19, 2013 by meena.p
edited Jun 1, 2014 by lmohan717

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