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NEET AIPMT SOLVED PAPER SCREENING 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: Velocity is 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 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 \[\therefore 12-3{{t}^{2}}=0\] \[or{{t}^{2}}=\frac{12}{3}=4\] \[ort=2\,s\]                 Hence, distance travelled by the particle before coming to rest is given by                 \[x=40+12(2)-{{(2)}^{3}}\]                  = 40 + 24 - 8 = 64 - 8                  = 56 m


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