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JCECE Engineering JCECE Engineering Solved Paper-2002

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

    A) \[zero\]                                               

    B) \[2\,\,m\]

    C)  \[3\,\,m\]                                          

    D)  \[-2\,\,m\]

    Correct Answer: A

    Solution :

    Key Idea: Acceleration is equal to rate of change of velocity. Given,                                 \[s={{t}^{3}}-3{{t}^{2}}+2\] Velocity                                \[v=\frac{ds}{dt}=3{{t}^{2}}-6t\] Acceleration       \[a=\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{dv}{dt}=6t-6\]     At\[a=0\], we have                                 \[6t-6=0\] \[\Rightarrow \]                               \[t=1s\] Hence,                  \[s={{(1)}^{3}}-3{{(1)}^{2}}+2\]                 \[=1-3+2=0\]


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