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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 45

  • question_answer
    The position vector of a particle is given as \[\vec{r}=\left( {{t}^{2}}-4t+6 \right)\hat{i}+\left( {{t}^{2}} \right)\hat{j}\]. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to

    A) 1 sec     

    B) 2 sec                  

    C) 15 sec                 

    D)        Not possible

    Correct Answer: A

    Solution :

    [a] \[\vec{V}=\frac{d\vec{r}}{dt}=\left( 2t-4 \right)\hat{i}+2t\hat{j}\] \[\vec{a}=\frac{\overrightarrow{dv}}{dt}=\left( 2\hat{i}+2\hat{j} \right)\] Velocity and acceleration are perpendicular so,  \[\vec{a}.\vec{v}=0\] \[\left( \left( 2t-4 \right)\hat{i}+\left( 2t \right)\hat{j} \right).\left( 2\hat{i}+2\hat{j} \right)=0\] \[\left( 2t-4 \right)\times 2+2t\times 2=0\] \[8t-8=0\] \[t=1\sec \]                   


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