# A particle moves such that its position vector

Question:

A particle moves such that its position vector $\vec{r}(t)$ $=\cos \omega \mathrm{t} \hat{i}+\sin \omega \mathrm{t} \hat{j}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ of the particle:

1. (1) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed away from the origin

2. (2) $\vec{v}$ and $\vec{a}$ both are perpendicular to $\vec{r}$

3. (3) $\vec{v}$ and $\vec{a}$ both are parallel to $\vec{r}$

4. (4) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin

Correct Option: 4,

Solution:

(4) Given, Position vector,

$\vec{r}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$

Velocity, $\vec{v}=\frac{d r}{d t}=\omega(-\sin \omega t \hat{i}+\cos \omega t \hat{j})$

Acceleration,

$\vec{a}=\frac{d \vec{v}}{d t}=-\omega^{2}(\cos \omega t \hat{i}+\sin \omega t \hat{j})$

$\vec{a}=-\omega^{2} \vec{r}$

$\therefore \vec{a}$ is antiparallel to $\vec{r}$

Also $\vec{v} \cdot \vec{r}=0$

$\therefore \vec{v} \perp \vec{r}$

Thus, the particle is performing uniform circular motion.