2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Applications of Derivatives

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

  • question_answer
    A particle moves along a straight line so that its distance s in time t sec is \[s=t+6{{t}^{2}}-{{t}^{3}}\]. After what time is the acceleration zero [AMU 1999]

    A)            2 sec

    B)            3 sec

    C)            4 sec

    D)            6 sec

    Correct Answer: A

    Solution :

               \[s=t+6{{t}^{2}}-{{t}^{3}}\]                    Þ  \[\frac{ds}{dt}=1+12t-3{{t}^{2}}\]Þ\[v=5\,cm/\sec \]                    Þ  \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=0\] Þ \[12-6t=0\Rightarrow t=2\].


You need to login to perform this action.
You will be redirected in 3 sec spinner