Question

The position of a particle is given by $\vec{r}=3t\hat{i}+2t^2\hat{j}+5\hat{k}$, where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in metres. The direction of velocity of the particle at $t=1s$ is

• A. $53^{°}$ with $x$-axis
• B. $37^{°}$ with $x$-axis
• C. $30^{°}$ with $y$-axis
• D. $60^{°}$ with $y$-axis

Answer 1

Answer: (A)

Given : $\vec{r}=3t\hat{i}+2t^2\hat{j}+5\hat{k}$
Velocity,
$\vec{v}=\frac{d\vec{r}}{dt}=\frac{d}{dt}\left(3t\hat{i}+2t^2\hat{j}+5\hat{k}\right)$
$=3\hat{i}+4t\hat{j}m{s}^{-1}$
Let $\theta$ be angle which the direction of $\vec{v}$ makes with the $x$-axis. Then
$tan\theta =\frac{v_y}{v_x}=\frac{4t}{3}=\frac{4}{3}$
or $\theta =ta{n}^{-1}\left(\frac{4}{3}\right)=53^{\circ }$