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J & K CET Medical J & K - CET Medical Solved Paper-2002

  • question_answer
    Position of a particle moving along x-axis is given by \[x=3t-4{{t}^{2}}+{{t}^{3}},\] where x is in metre and t in seconds. Find the average velocity of the particle in the time interval from t = 2 s to 4 s.

    A)  7 m/s                                   

    B)  9 m/s

    C)  13 m/s                                

    D)  none of these

    Correct Answer: B

    Solution :

                    Rate of change of displacement is velocity, i.e.,                 \[v=\frac{dx}{dt}\]                 \[v=\frac{dx}{dt}=3-8t+3{{t}^{2}}\] Given, \[x=3t-4{{t}^{2}}+{{t}^{3}}\]                 \[v=\frac{dx}{dt}=3-8t+3{{t}^{2}}\] At \[t=2,\,{{v}_{1}}=3-16+12=-1\] At \[t=4,\,{{v}_{2}}=3-32+48=19\] Average velocity\[=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]                                 \[=\frac{-1+19}{2}=9\,m/s\]


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