Question

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

• A. $\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}\text{and}\overrightarrow{a}$ is directed away from the origin
• B. $\overrightarrow{v}\text{and}\overrightarrow{a}$ both are perpendicular to $\overrightarrow{r}$
• C. $\overrightarrow{v}\text{and}\overrightarrow{a}$ both are parallel to $\overrightarrow{r}$
• D. $\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}\text{and}\overrightarrow{a}$ is directed towards the origin

Answer 1

Answer: (D)

Given, Position vector,
$\overrightarrow{r}=\cos \omega t\hat{i}+\sin \omega t\hat{j}$
Velocity,
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\omega (-\sin \omega t\hat{i}+\cos \omega t\hat{j})$
Acceleration,
$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=-{\omega }^2(\cos \omega t\hat{i}+\sin \omega t\hat{j})$
$\overrightarrow{a}=-{\omega }^2\overrightarrow{r}$
$\therefore \overrightarrow{a}\text{is antiparallel to}\overrightarrow{r}$
Also$\overrightarrow{v}.\overrightarrow{r}=0$

as $\overrightarrow{v}.\overrightarrow{r}=w[-sin\left(wt\right)cos\left(wt\right)+sin\left(wt\right)cos\left(wt\right)]=0$

$\overrightarrow{v}.\overrightarrow{r}=\left|v\right|\left|r\right|cos\theta =0$

$ifcos\theta =0⟶\theta ={90}^o$

$\therefore \overrightarrow{v}\perp \overrightarrow{r}$
Thus, the particle is performing uniform circular motion.