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JEE Main & Advanced Physics Two Dimensional Motion Question Bank Self Evaluation Test - Motion in a Plane

  • question_answer
    The position of particle is given by \[\vec{r}=2\,{{t}^{2}}\widehat{i}+3\,t\widehat{j}+4\widehat{k},\] where \[t\] is in second and the coefficients have proper units for \[\vec{r}\] to be in meter. The \[\vec{a}\,(t)\] of the particle at \[t=1s\,\] is

    A) \[\text{4}\,\text{m }{{\text{s}}^{-2}}\] along y-direction

    B) \[\text{3}\,\text{m }{{\text{s}}^{-2}}\] along x-direction

    C) \[\text{4 m }{{\text{s}}^{-2}}\] along x-direction

    D) \[\text{2 m }{{\text{s}}^{-2}}\] along z-direction

    Correct Answer: C

    Solution :

    [c] \[\text{\vec{r}}=2{{\text{t}}^{2}}\hat{i}+3t\hat{j}+4\hat{k}\] \[\therefore \text{\vec{v}=}\frac{d\vec{r}}{dt}=\frac{d}{dt}=\left( 2{{\text{t}}^{2}}\hat{i}+3t\hat{j}+4\hat{k} \right)=4\text{t}\hat{i}+3\hat{j}\] \[\text{\vec{a}=}\frac{d\text{\vec{v}}}{dt}=\frac{d}{dt}\left( 4\text{t}\hat{i}+3\hat{j} \right)=4\hat{i}\] \[\therefore \text{\vec{a}}=4m{{s}^{-2}}\text{ along x-direction}\]


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