Question

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

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

Answer 1

Answer: (B)

Velocity is the rate of change of position vector,
$\vec{r}=a(\cos \omega t)\hat{i}+(a\sin \omega t)\hat{j}$
Velocity vector, $\vec{v}=\frac{d\vec{r}}{dt}$
$=\frac{d}{dt}[(a\cos \omega t)]\hat{i}+(a\sin \omega t)\hat{j}]$
$=(-a\sin \omega t)\hat{i}+(a\cos \omega t)\hat{j}$
Now, $\vec{v}\cdot \vec{r}=[(-a\sin \omega t)\hat{i}+(a\cos \omega t)\hat{j}][(a\cos \omega t)\hat{i}+(a\sin \omega t)\hat{j}]$
$=(-a\sin \omega t)(a\cos \omega t)+(a\cos \omega t)(a\sin \omega t)$
$=-a^2\sin \omega t\cos \omega t+a^2\cos \omega t\sin \omega t$
$=0$
As dot product of velocity vector with position vector is zero, so velocity vector of the particle is perpendicular to position vector.