2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    The position of a particle as a function of time t, is given by \[x(t)=at\,+b{{t}^{2}}-c{{t}^{3}}\] where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:                         [JEE Main 9-4-2019 Afternoon]

    A) \[a+\frac{{{b}^{2}}}{4c}\]                       

    B) \[a+\frac{{{b}^{2}}}{c}\]

    C) \[a+\frac{{{b}^{2}}}{2c}\]                       

    D) \[a+\frac{{{b}^{2}}}{3c}\]

    Correct Answer: D

    Solution :

    \[x=at+b{{t}^{2}}-c{{t}^{3}}\]             \[\text{v}=\frac{dx}{dt}=a+2bt-3c{{t}^{2}}\]             \[a=\frac{dv}{dt}=2b-6ct=0\Rightarrow t=\frac{b}{3c}\]             \[{{^{\text{v}}}_{\left( at\,t=\frac{b}{3c} \right)}}=a+2b\left( \frac{b}{3c} \right)-3c\left( \frac{b}{3c} \right)\]\[=a+\frac{{{b}^{2}}}{3c}.\]            


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