BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    For a particle moving in a straight line, if timer be regarded as a function of velocity v, then the rate of change of the acceleration a is given by:

    A) \[{{a}^{2}}\frac{{{d}^{2}}t}{d{{v}^{2}}}\]                               

    B)                   \[{{a}^{3}}\frac{{{d}^{2}}t}{d{{v}^{2}}}\]                              

    C)                    \[-{{a}^{3}}\frac{{{d}^{2}}t}{d{{v}^{2}}}\]                            

    D)                   none of these

    Correct Answer: C

    Solution :

    Let \[t=f(v),\]then \[\frac{dt}{dv}=f(v)\Rightarrow \frac{dv}{dt}=\frac{1}{f(v)}\] \[\Rightarrow \]\[a=\frac{1}{f(v)}=1\Rightarrow af(v)\frac{dv}{dt}+\frac{da}{dt}f(v)=0\]\[\Rightarrow \]                \[{{a}^{2}}f(v)+\frac{da}{dt}\times \frac{1}{a}=0\] \[\Rightarrow \]               \[\frac{da}{dt}=-{{a}^{3}}f(v)\] \[\Rightarrow \]\[\frac{da}{dt}=-{{a}^{3}}\frac{{{d}^{2}}t}{d{{v}^{2}}}\]


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