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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    A particle moves along a straight line such that its displacement at any time t is given by \[S={{t}^{3}}-6{{t}^{2}}+3t+4.\] The velocity when the accelerations zero in m/s is:

    A)  -9

    B)                                         -12                        

    C)  42                         

    D)         3

    Correct Answer: A

    Solution :

    Given   \[s={{t}^{3}}-6{{t}^{2}}+3t+4\] \[\frac{ds}{dt}=3{{t}^{2}}-12t+3\] and        \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=6t-12\] If acceleration is zero, \[6t-12=0\] or                                            \[t=2\sec \] Hence, velocity at t = 2 sec \[v=\frac{ds}{dt}=3{{t}^{2}}-12t+3\] \[=12-24+3=-\,9\,m\text{/}s\]


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