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

  • question_answer
    If the distance ?s? traveled by a particle in time t is\[s=a\sin t+b\cos 2t\], then the acceleration at t = 0 is

    A)            a

    B)            ? a

    C)            4b

    D)            ? 4b

    Correct Answer: D

    Solution :

               Given \[s=a\sin t+b\cos 2t\]                    \[\therefore \]\[\frac{ds}{dt}=a\cos t-2b\sin 2t\]                      \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=-a\sin t-4b\cos 2t\]                    At \[t=0\], \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=-a\sin 0{}^\circ -4b\cos 0{}^\circ =-4b\].


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