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

Answer 1

The correct option is A $\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$ and $\overrightarrow{a}$ is directed towards the origin


The position vector,
$\overrightarrow{r}\left(t\right)=\cos \omega t\hat{i}+\sin \omega t\hat{j}\cdots \left(i\right)$
On differentiating equation $\left(i\right)$ we get
$\overrightarrow{v}=-\omega \sin \omega t\hat{i}+\omega \cos \omega t\hat{j}\cdots \left(ii\right)$
Again, differentiating equation $\left(ii\right)$
$\overrightarrow{v}=-{\omega }^2\overrightarrow{r}$
$\overrightarrow{v}\cdot \overrightarrow{r}=0$
Final Answer: $\left(a\right)$