Motion in a Plane - Result Question 27
30. 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
[1995]
(a) directed towards the origin
(b) directed away from the origin
(c) parallel to the position vector
(d) perpendicular to the position vector
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Answer:
Correct Answer: 30. (d)
Solution:
- (d) Position vector,
$\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}$
Velocity vector,
$\vec{v}=\frac{d(\vec{r})}{d t}=\frac{d}{d t}{(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}}$
$=(-a \omega \sin \omega t) \hat{i}+(a \omega \cos \omega t) \hat{j}$
$=\omega[(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j}]$
Slope of position vector $=\frac{a \sin \omega t}{a \cos \omega t}=\tan \omega t$
Slope of velocity vector, $=\frac{-a \cos \omega t}{a \sin \omega t}=\frac{-1}{\tan \omega t}$
$\therefore$ velocity is perpendicular to the displacement.
