Question

The position vector of a particle is $\overrightarrow{r}=\left(a\cos \omega t\right)\hat{i}+\left(a\sin \omega t\right)\hat{j}$. the velocity vector of the particle is:

• A. parallel to the position vector
• B. perpendicular to the position vector
• C. directed towards the origin
• D. directed away from the origin

Answer 1

The correct option is B perpendicular to the position vector

Position vector $\overrightarrow{r}=\left(acoswt\right)\hat{i}+\left(asinwt\right)\hat{j}$

Thus velocity vector $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=(-aw$ $sinwt)\hat{i}+(aw$ $coswt)\hat{j}$

Calculating $\overrightarrow{v}.\overrightarrow{r}=$ $\left[\right(-aw$ $sinwt)\hat{i}+(aw$ $coswt\left)\hat{j}\right]$ $.\left[\right(acoswt)\hat{i}+(asinwt\left)\hat{j}\right]$

$\therefore$ $\overrightarrow{v}.\overrightarrow{r}=-a{w}^2$ $sinwtcoswt+$ $a{w}^2$ $sinwtcoswt=0$

Hence velocity vector is perpendicular to the position vector.