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

  • question_answer
     The distance in seconds, described by a particle in t seconds is given by \[s=a{{e}^{t}}+\frac{b}{{{e}^{t}}}\]. Then acceleration of the particle at time t is

    A)            Proportional to t

    B)            Proportional to s

    C)            s

    D)            Constant

    Correct Answer: C

    Solution :

               Given that \[s=a{{e}^{t}}+\frac{b}{{{e}^{t}}}\]            Differentiating w.r.t. time t, we get            \[\frac{ds}{dt}=\]velocity \[=a{{e}^{t}}-\frac{b}{{{e}^{t}}}\]            Again \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\] acceleration \[=a{{e}^{t}}+\frac{b}{{{e}^{t}}}=s\].


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