Question

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

• A. 2 s
• B. 1 s
• C. 1 .5 s
• D. 5 s

Answer 1

Answer: (B)

$\vec{r} = (t^2 - 4t + 6) \hat{i} + t^2 \hat{j} : \vec{v} = \frac{d\vec{r}}{dt}$
$ = (2t - 4) \hat{i} + 2t\,\hat{j}, \vec{a} = \frac{d\vec{v}}{dt} $
$ = 2\hat{i} + 2\hat{j}$
If $\vec{a}$ and $\vec{v}$ are perpendicular, $\vec{a} \cdot \vec{v} = 0$
$( 2\hat{i} + 2\hat{j}) \cdot (( 2t - 4) \hat{i} + 2t\,\hat{j}) = 0$
$ 8t - 8 = 0, t = 1\, \sec$