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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Application in Mechanics and Rate Measurer

  • question_answer
    The speed \[v\] of a particle moving along a straight line is given by \[a+b{{v}^{2}}={{x}^{2}}\] (where x is its distance from the origin). The acceleration of the particle is [MP PET 2002]

    A)            \[bx\]

    B)            \[x/a\]

    C)            \[x/b\]

    D)            \[x/ab\]

    Correct Answer: C

    Solution :

               \[a+b{{v}^{2}}={{x}^{2}}\]  Þ  \[0+b\left( 2v.\,\frac{dv}{dt} \right)=2x\,.\,\frac{dx}{dt}\]            Þ  \[v.b\frac{dv}{dt}=x\,.\,\frac{dx}{dt}\] Þ \[\frac{dv}{dt}=\frac{x}{b}\] ,  \[\left( \because \frac{dx}{dt}=v \right)\].


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