Manipal Medical Manipal Medical Solved Paper-2007

  • question_answer
    A particle moves along a straight line OX. At a time t (in seconds) the distance\[x\] (in meters) 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)  56m           

    D)  16m

    Correct Answer: C

    Solution :

     Key Idea: Velocity is the rate of change of distance or displacement. 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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