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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    The relation between the time t and distance \[x\] is given by \[t=p{{x}^{2}}+qx,\]where p and q are constants. The relation between velocity v and acceleration \[f\] is

    A)  \[f\propto v\]                      

    B)         \[f\propto {{v}^{4}}\]                         

    C)         \[f\propto {{v}^{2}}\]                         

    D)         \[f\propto {{v}^{3}}\]

    Correct Answer: C

    Solution :

    On differentiating gives equation, we get \[dt=(2px+q)dx\]                 \[\therefore \]      \[v=\frac{dx}{dt}=\frac{1}{2px+q}\]  ?(i)                 \[\Rightarrow \] \[\frac{dv}{dt}=f=\frac{-2p}{{{(2px+q)}^{2}}}=-2p{{v}^{2}}\] [from Eq.(i)]                 So,           \[f\propto {{v}^{2}}\]


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