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AMU Medical AMU Solved Paper-2014

  • question_answer
    The distance x of a particle moving in one I dimension under the action of constant force is related to the time t by the relation,\[t=\sqrt{x}+3\]Find the displacement of the particle when its velocity is 6.0 m/s.

    A)  9.0m                    

    B)  6.0m  

    C)  4.0 m                   

    D)  0.0 m

    Correct Answer: A

    Solution :

                    The given equation can be written as                 \[\sqrt{x}=t-3\] Squaring on both sides, we get, \[x={{t}^{2}}-6t+9\] Differentiating                 \[\frac{dx}{dt}=2\,t-6\] For 6 m/s velocity, we have                 \[6=2\,t-6\]                 \[t=6\,s\] At t = 6 s, the displacement is                 \[x={{t}^{2}}-6\,t+9\]                 \[x={{6}^{2}}-6\times 6+9\]                 \[x=9.0\,m\]


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