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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Application in Mechanics and Rate Measurer

  • question_answer
    The motion of stone thrown up vertically is given by \[s=13.8t-4.9{{t}^{2}}\], where s is in metre and t is in seconds. Then its velocity at \[t=1\] second is

    A)            3 m/s

    B)            5 m/s

    C)            4 m/s

    D)            None of these

    Correct Answer: C

    Solution :

               \[s=13.8t-4.9{{t}^{2}}\]                    Its velocity \[v=\frac{ds}{dt}=13.8-9.8t\]                    Hence \[{{\left( \frac{ds}{dt} \right)}_{t=1}}=13.8-9.8\times 1=4.0=4m/\sec .\]


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