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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    If the distance s metres traversed by a particle in t seconds is given by \[s={{t}^{3}}-3{{t}^{2}}\], then the velocity of the particle when the acceleration is zero, in metre/sec, is :

    A)  3                                            

    B)  - 2

    C)  - 3                                         

    D)  2

    Correct Answer: C

    Solution :

    We have \[s={{t}^{3}}-3{{t}^{2}}\] On differentiating w.r. to t, we get                 \[\frac{ds}{dt}=3\,{{t}^{2}}-6t\]                                 ?. (i) Again, differentiating equation (i), we get                 \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=6t-6=0\] \[\Rightarrow \]               \[t=1\] On put the value of t = 1 in equation (i) we get, \[\frac{ds}{dt}=3\times 1-6\times 1\]                 \[=3-6\]                 \[=-3\text{ }m/sec\]


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