Question

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

• A. $1\text{sec}$
• B. $1.5\text{sec}$
• C. $2\text{sec}$
• D. $\text{Not Possible}$

Answer 1

The correct option is A $1\text{sec}$

Given, $\overrightarrow{r}=(t^2-4t+6)\hat{i}+t^2\hat{j}$
$\Rightarrow \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=(2t-4)\hat{i}+2t\hat{j}$
Also, $\Rightarrow \overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=2\hat{i}+2\hat{j}$
As the two vectors i.e. $\overrightarrow{r}$ and $\overrightarrow{a}$ will be perpendicular to each other
$\Rightarrow \overrightarrow{v}\cdot \overrightarrow{a}=0$
$\Rightarrow \left[\right(2t-4)\hat{i}+2t\hat{j}]\cdot [2\hat{i}+2\hat{j}]=0$
$\Rightarrow 2(2t-4)+4t=0$
$\Rightarrow t=1\text{sec}$