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Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    The motion of a particle along a straight line is described by the function\[x={{(2t-3)}^{2}}\], where\[x\]is in metre and \[t\] is in second. Then, the velocity of the particle at origin is

    A)  0                                            

    B)  1

    C)  2                            

    D)         None of these

    Correct Answer: A

    Solution :

    Given that,\[x={{(2t-3)}^{2}}\] On differentiating w.r.t.\[x\], we get                 \[\frac{dx}{dt}=2(2t-3)(2)\] At origin,\[x=0\] \[\therefore \]  \[2t-3=0\]    \[\Rightarrow \]    \[t=\frac{3}{2}\] \[\therefore \]  Velocity\[=\frac{dx}{dt}=4\left( 2\times \frac{3}{2}-3 \right)=0\]


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