Kinematics Question 142
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 [CBSE PMT 1995]
Options:
A) Parallel to the position vector
B) Perpendicular to the position vector
C) Directed towards the origin
D) Directed away from the origin
Correct Answer: B $ \vec{r}=(a\cos \omega t)\hat{i}+(a\sin \omega t)\hat{j} $ $ \vec{v}=\frac{d\vec{r}}{dt}=-a\omega \sin \omega t\hat{i}+a\omega \cos \omega t\hat{j} $ As $ \vec{r}.\vec{v}=0 $ therefore velocity of the particle is perpendicular to the position vector.
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