JIPMER Jipmer Medical Solved Paper-2008

  • 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., \[v=\frac{dx}{dt}\]    \[=\frac{d}{dt}\left( {{a}_{0}}+\frac{{{a}_{1}}t}{2}-\frac{{{a}_{2}}{{t}^{2}}}{3} \right)\]    \[=0+\frac{{{a}_{1}}}{2}-\frac{2{{a}_{2}}t}{3}\]    \[=\frac{{{a}_{1}}}{2}-\frac{2{{a}_{2}}t}{3}\] 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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