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.