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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    A particle moves along a straight line\[OX\]. At a time \[t\] (in second) the distance \[x\] of the particle from \[O\] is given by\[x=40+12t-{{t}^{3}}\]. How long would the particle travel before coming to rest?

    A) \[24\,\,m\]                        

    B) \[40\,\,m\]

    C) \[56\,\,m\]                        

    D) \[16\,\,m\]

    Correct Answer: C

    Solution :

    Distance travelled by the particle is                 \[x=40+12t-{{t}^{3}}\]                 \[v=\frac{dx}{dt}=2-3{{t}^{2}}\] But final velocity\[v=0\]                 \[12-3{{t}^{2}}=0\]                 \[{{t}^{2}}=\frac{12}{3}=4\]                 \[t=2s\]                 \[x=40+12(2)-8=56\,\,m\]


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