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JEE Main & Advanced AIEEE Solved Paper-2011

  • question_answer
    An object, moving with a speed of \[6.25\,m/s\], is decelerated at a rate given by \[\frac{dv}{dt}=-2.5\sqrt{v}\]where v is the instantaneous speed. The time taken by the object, to come to rest, would be   AIEEE  Solved  Paper-2011

    A)              8 s                                             

    B)              1 s

    C) 2 s                                             

    D) 4 s

    Correct Answer: C

    Solution :

                 \[\int\limits_{v}^{0}{\frac{dv}{\sqrt{v}}=-\int\limits_{0}^{t}{2.5\,dt}}\] \[2\sqrt{v}=2.5\,t\] \[2\sqrt{6.25}=2.5\,t\] \[t=2\,s\]


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