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, $\vec{a}\cdot \vec{v}=0$

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

$8t-8=0,t=1\sec$