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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from 0 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 \[x=40+12t-{{t}^{3}}\] We know that, velocity is rate of change of distance ie,                                                                              \[v=\frac{dx}{dt}\] \[\therefore \]  \[v=\frac{d}{dt}(40+12t-{{t}^{3}})\]                 \[=0+12-3{{t}^{2}}\] but final velocity \[v=0\] \[\therefore \]  \[12-3{{t}^{2}}=0\] or            \[{{t}^{2}}=\frac{12}{3}=4\] or            \[t=2s\] Hence, distance travelled by the particle before coming to rest is given by                 \[x=40+12(2)-{{(2)}^{3}}\]                 \[=40+24-8=64-8\]                 \[=56m\]


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