The position of a particle is given by $\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k} \; m$,where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in metres.
$(a)$ Find the velocity $\vec{v}$ and acceleration $\vec{a}$ of the particle.
$(b)$ What is the magnitude and direction of velocity of the particle at $t = 2.0 \; s$?

(N/A) The position of the particle is given by:
$\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}$
Velocity $\vec{v}$ is the time derivative of position:
$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}) = 3.0 \hat{i} - 4.0 t \hat{j} \; m/s$
Acceleration $\vec{a}$ is the time derivative of velocity:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3.0 \hat{i} - 4.0 t \hat{j}) = -4.0 \hat{j} \; m/s^{2}$
$(b)$ At $t = 2.0 \; s$,the velocity vector is:
$\vec{v} = 3.0 \hat{i} - 4.0(2.0) \hat{j} = 3.0 \hat{i} - 8.0 \hat{j} \; m/s$
The magnitude is $|\vec{v}| = \sqrt{(3.0)^{2} + (-8.0)^{2}} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54 \; m/s$
The direction $\theta$ with respect to the $x$-axis is:
$\theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{-8.0}{3.0}\right) \approx -69.45^{\circ}$
The negative sign indicates the direction is $69.45^{\circ}$ below the positive $x$-axis.