BCECE Medical BCECE Medical Solved Papers-2002

  • question_answer
    The displacement x of a particle moving along a straight line at time t is given by \[x={{a}_{0}}+{{a}_{1}}\,t+{{a}_{2}}{{t}^{2}}\] The acceleration of the particle is :

    A)  \[4\,{{a}_{2}}\]

    B)  \[2\,{{a}_{2}}\]

    C)  \[2\,{{a}_{1}}\]

    D)  \[{{a}_{2}}\]

    Correct Answer: B

    Solution :

    Key Idea: Acceleration is known as rate of change of velocity. The given equation is                 \[x={{a}_{0}}+{{a}_{1}}\,t+{{a}_{2}}\,{{t}^{2}}\] First differentiate the above equation with respect to r, to obtain velocity that is, \[v=\frac{dx}{dt}\] (velocity is rate of change of displacement)                 \[\therefore \] \[v=\frac{dx}{dt}={{a}_{1}}+2\,{{a}_{2}}\,t\]                 \[\left( using\frac{d}{dx}{{x}^{n}}=n\,{{x}^{n-1}} \right)\] Then second differentiation of above equation gives acceleration i.e., \[a=\frac{dv}{dt}=2\,{{a}_{2}}\]


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