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MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    A particle moves along the\[x-\]axis as \[x=u(t-2s)+a{{(t-2s)}^{2}}\]. The initial velocity of the particle is

    A)  \[u-2a\]              

    B)  \[u-4a\]

    C)  \[2a-u\]              

    D)  \[2a-3u\]

    Correct Answer: B

    Solution :

                    \[x=u(t-2s)+a{{(t+2s)}^{2}}\] \[v=\frac{dx}{dt}=u+2a(t-2s)\] \[v=\frac{dx}{dt}=u+2a(t-2)\] \[\therefore \]Initial velocity\[v(0)=u-4a\]


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