Question

The position of a particle is given by $\overrightarrow{r}=3t\hat{i}+\sqrt{3}{t}^2\hat{j}-4\hat{k}$ where $t$ is in seconds and $\overrightarrow{r}$ is meters. Find out magnitude of velocity $\overrightarrow{v}$ and angle made by velocity $\overrightarrow{v}$ with X-axis at $t=\sqrt{3}\text{s}.$

• A. $3\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\left(2\right)$
• B. $3\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$
• C. $3\sqrt{2}\text{m/s},\theta ={\tan }^{-1}\left(3\right)$
• D. $3\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

The correct option is A $3\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\left(2\right)$

It is given that the position vector $\overrightarrow{r}=3t\hat{i}+\sqrt{3}{t}^2\hat{j}-4\hat{k}$.

We know that velocity is derivative of position with respect to time,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=3\hat{i}+2\sqrt{3}t\hat{j}{\overrightarrow{v}}_{(t=\sqrt{3}s)}=3\hat{i}+6\hat{j}|\overrightarrow{v}|=\sqrt{3^2+6^2}=3\sqrt{5}\text{m/s}\theta ={\tan }^{-1}\left(\frac{6}{3}\right)={\tan }^{-1}\left(2\right)$

Hence option A is the correct answer.