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Q. 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 $t = 1\, s$ is

Motion in a Plane

Solution:

Given : $\vec{r}=3t\,\hat{i}+2t^{2}\,\hat{j}+5\hat{k}$
Velocity,
$\vec{v}=\frac{d\vec{r}}{dt}=\frac{d}{dt}\left(3t\,\hat{i}+2t^{2}\,\hat{j}+5\hat{k}\right)$
$=3\hat{i}+4t\,\hat{j}\,ms^{-1}$
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 $\theta=tan^{-1}\left(\frac{4}{3}\right)=53^{\circ}$