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JCECE Engineering JCECE Engineering Solved Paper-2011

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

    A) \[1\,\,m/s\] 

    B) \[4\,\,m/s\]

    C)  \[2\,\,m/s\]

    D)  \[2\sqrt{2}\,\,m/s\]

    Correct Answer: C

    Solution :

    From work energy theorem                 \[\Delta KE={{W}_{net}}\] or            \[{{K}_{f}}-{{K}_{i}}=\int{p\,\,dt}\]                 \[\frac{1}{2}m{{v}^{2}}-0=\int_{0}^{2}{\left( \frac{3}{2}{{t}^{2}} \right)dt}\]                 \[{{v}^{2}}=\left[ \frac{{{t}^{3}}}{2} \right]_{0}^{2}\]                 \[v=2\,\,m/s\]


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