Question

A particle moves such that its position vector$\overrightarrow{r}\left(t\right)=\cos \left(\omega t\right)i+\sin \left(\omega t\right)j$ 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 $\overrightarrow{r}$ and $\overrightarrow{a}$ is directed towards the origin
• B. $\overrightarrow{v}$and $\overrightarrow{a}$ both are parallel to $\overrightarrow{r}$
• C. $\overrightarrow{v}$and $\overrightarrow{a}$both are perpendicular to$\overrightarrow{r}$
• 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

Step 1: Given data.

A particle moving with a position vector $\overrightarrow{r}\left(t\right)=\cos \left(\omega t\right)i+\sin \left(\omega t\right)j$

Step 2: Finding the velocity vector $\overrightarrow{v}\left(t\right)$ and acceleration vector $\overrightarrow{a}\left(t\right)$ of the particle.

We have,

$\overrightarrow{r}\left(t\right)=\cos \left(\omega t\right)i+\sin \left(\omega t\right)j$

We know that,

$\overrightarrow{v}\left(t\right)=\frac{d\overrightarrow{r}\left(t\right)}{dt}$

Where $\overrightarrow{v}\left(t\right)$ is the velocity vector.

$\overrightarrow{v}\left(t\right)=\frac{d\left(\cos \left(\omega t\right)i+\sin \left(\omega t\right)j\right)}{dt}$

$\overrightarrow{v}\left(t\right)=\omega \left(-\sin \left(\omega t\right)i+\cos \left(\omega t\right)j\right)$ ….$\left(i\right)$

Again,

$\overrightarrow{a}\left(t\right)=\frac{d\overrightarrow{v}\left(t\right)}{dt}$

Where $a\left(t\right)$ is the acceleration vector.

$\overrightarrow{a}\left(t\right)=\frac{d\omega \left(-\sin \left(\omega t\right)i+\cos \left(\omega t\right)j\right)}{dt}$

$\overrightarrow{a}\left(t\right)=-{\omega }^2\left(\cos \left(\omega t\right)i+c\sin \left(\omega t\right)j\right)$

$\overrightarrow{a}\left(t\right)=-{\omega }^2\overrightarrow{r}\left(t\right)$ …$\left(ii\right)$

Where the negative sign shows that acceleration is antiparallel to the position of the particle.

Therefore.

$\overrightarrow{v}\left(t\right).\overrightarrow{r}\left(t\right)=\omega (-\sin \omega t\cos \omega t+\cos \omega t\sin \omega t)$

$\overrightarrow{v}\left(t\right).\overrightarrow{r}\left(t\right)=0$

So, from above it is clear that $\overrightarrow{v}\left(t\right)$ is perpendicular to $\overrightarrow{r}\left(t\right)$.

Hence, option A is correct.