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

  • question_answer
    The position vector of a particle changes with time according to the relation \[\vec{r}(t)=15{{t}^{2}}\hat{i}+(4-{{20}^{2}})\hat{j}.\] What is the magnitude of the acceleration at t = 1 ? [JEE Main 9-4-2019 Afternoon]

    A) 40                               

    B) 100

    C) 25                               

    D) 50

    Correct Answer: D

    Solution :

    \[\vec{r}=15{{t}^{2}}\hat{i}+(4-20{{t}^{2}})\hat{j}\] \[\text{\vec{v}}=\frac{d\vec{r}}{dt}=30t\hat{i}+(-40\,t)\hat{j}\] \[\vec{a}=\frac{d\text{\vec{v}}}{dt}=30\hat{i}-40\,t\hat{j}\] \[\left| {\vec{a}} \right|=50m/{{s}^{2}}.\]


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