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

  • question_answer
    The equation of motion  of a particle is given by \[s=2{{t}^{3}}-9{{t}^{2}}+12t+1\],where s and t are measured in cm and sec. The time when the particle stops momentarily is

    A)            1 sec

    B)            2 sec

    C)            1, 2 sec

    D)            None of these

    Correct Answer: C

    Solution :

               \[\frac{ds}{dt}=6{{t}^{2}}-18t+12\] = velocity = 0 (when particle stopped)            Þ \[6{{t}^{2}}-18t+12=0\Rightarrow (t-1)\,(t-2)=0\]            Hence time 1, 2 sec.


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