AFMC AFMC Solved Paper-2006

  • question_answer
    The displacement of particle is given by \[x={{a}_{0}}+\frac{{{a}_{1}}t}{2}-\frac{{{a}_{2}}{{t}^{2}}}{3}\]What is its acceleration?

    A) \[\frac{2{{a}_{2}}}{3}\]                                 

    B) \[-\frac{2{{a}_{2}}}{3}\]

    C) \[{{a}_{2}}\]                                      

    D) Zero

    Correct Answer: B

    Solution :

    Key Idea: Acceleration is the rate of change of velocity and velocity is the rate of change of displacement. The displacement equation is given by         \[x={{a}_{0}}+\frac{{{a}_{1}}t}{2}-\frac{{{a}_{2}}{{t}^{2}}}{3}\] Velocity = rate of change of displacement i.e.,                   \[v=\frac{dx}{dt}\] \[=\frac{d}{dx}\left( {{a}_{0}}+\frac{{{a}_{1}}t}{2}-\frac{{{a}_{2}}{{t}^{2}}}{3} \right)\] \[=0+\frac{{{a}_{1}}}{2}-\frac{2{{a}_{2}}t}{3}\] \[=\frac{{{a}_{1}}}{2}-\frac{2{{a}_{2}}t}{3}\] Acceleration = rate of change of velocity i.e.,        \[a=\frac{dv}{dt}\] \[=\frac{d}{dt}\left( \frac{{{a}_{1}}}{2}-\frac{2{{a}_{2}}}{3}t \right)\] \[=0-\frac{2{{a}_{2}}}{3}\] \[=-\frac{2{{a}_{2}}}{3}\]


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