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

Show Answer

Answer:

Correct Answer: B

Solution:

$ \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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