Question

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

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

Answer 1

Answer: (D)

Position vector of the particle
$(r)=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}$
velocity vector
$\vec{v}= \frac{d \vec{r}}{d t}=(-a \omega \sin \omega t) \hat{i}+(a \omega \cos \omega t) \hat{j} $
$=\omega[(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j}] $
$\vec{v} \cdot \vec{r} =\omega[(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j})] \cdot[(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j})]$
$=\omega\left[-a^{2} \sin \omega t \cos \omega t+a^{2} \cos \omega t \sin \omega t\right]=0$
Therefore velocity vector is perpendicular to the displacement vector.