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 in meters. Find out magnitude and direction of velocity $\overrightarrow{v}$ with horizontal 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. $3\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$

Answer 1

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

Given that , $\overrightarrow{r}=3t\hat{i}+\sqrt{3}{t}^2\hat{j}-4\hat{k}$.
We know that,velocity is derivative of position with respect to time,
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=3\hat{i}+2\sqrt{3}t\hat{j}$

Now to find the magnitude and direction of velocity at $t=\sqrt{3}\text{s}$ we put the value in the above equation

$\overrightarrow{v}(t=\sqrt{3}\text{s})=3\hat{i}+6\hat{j}$,
magnitude of the vector $a\hat{i}+b\hat{j}$ is given by $\sqrt{a^2+b^2}$ and direction is given by $\tan \theta =\frac{b}{a}$.
Hence,
$|\overrightarrow{v}|=\sqrt{3^2+6^2}=3\sqrt{5}\text{m/s}$

$\theta ={\tan }^{-1}\left(\frac{6}{3}\right)={\tan }^{-1}\left(2\right)$