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BVP Medical BVP Medical Solved Paper-2006

  • question_answer
    The position x of a particle varies with time t, as\[x=a{{t}^{2}}-d{{t}^{3}}\]. The acceleration of the particle will be zero at time t equal to :

    A)  zero                                     

    B) \[\frac{a}{3b}\]

    C) \[\frac{2a}{3b}\]                                              

    D) \[\frac{a}{b}\]

    Correct Answer: B

    Solution :

                    Key Idea: Acceleration and velocity of a particle are given by \[A=\frac{dv}{dt},\]        \[v=\frac{dx}{dt}.\] The given equation is                 \[x=a{{t}^{2}}-b{{t}^{3}}\] Velocity \[v=\frac{dx}{dt}=2at-3b{{t}^{2}}\] Acceleration \[A=\frac{dv}{dt}=2a-6bt\] But             A = 0               (given)                 \[2a-6bt=0\] or           \[6bt=2a\] or            \[t=\frac{2a}{6b}=\frac{a}{3b}\]


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