AFMC AFMC Solved Paper-2006

  • question_answer
    A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) 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 :

    Key Idea: Speed is rate of change of distance. Distance travelled by the particle is\[x=40+12t-{{t}^{3}}\] We know that, speed is rate of change of distance i.e, \[v=\frac{dx}{dt}\] \[\therefore \]     \[v=\frac{d}{dt}(40+12t-{{t}^{3}})\] \[=0+12-3{{t}^{2}}\] 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      \[x=40+12(2)-{{(2)}^{3}}\]      \[=40+24-8=64-8\]       \[\text{= 56}\,\text{m}\]                       


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