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

(A) $1 \mathrm{sec}$

(B) $2 \mathrm{sec}$

(C) $1,5 \mathrm{sec}$

(D) not possible

Solution —

$\vec{r}=\left(t_{2}-4 t+6\right) \hat{i}+t_{2} \hat{j} ;$ $\vec{v}=\frac{d \dot{r}}{d t}=(2 t-4) \hat{i}+2 t \hat{j}$, $\vec{a}=\frac{d \vec{v}}{d t}=2 \hat{i}+2 \hat{j}$

if $\vec{a}$ and $\vec{v}$ are perpendicular

$\vec{a} \cdot \vec{v}=0$ $(2 \hat{i}+2 \hat{j}) \cdot((2 t-4) \hat{i}+2 t \hat{j})=0$

$8 t-8=0$

$t=1$ sec. $\quad$

Ans. $t=1$ sec.

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