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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • 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)  24m                                     

    B)  40m

    C)  56m                                     

    D)  16m

    Correct Answer: C

    Solution :

                     Distance travelled by the particle is \[x=40+12t-{{t}^{3}}\] We know that, speed is rate of change of distance, \[v=\frac{dx}{dt}\] \[\therefore \]  \[v=\frac{d}{dt}(40+12t+{{t}^{3}})\]                 \[=12+3{{t}^{3}}\] But final velocity,\[v=0\] \[\therefore \]  \[12-3{{t}^{2}}=0\]                 \[{{t}^{2}}=\frac{12}{3}=4\] Or           \[t=2s\] \[\therefore \]  \[x=40+12(2)-8\]                 \[=56\,m\]         


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