The position of a particle is given by $r = 3.0 t \hat{i} + 2.0 t^{2} \hat{j} + 5.0 \hat{k}$,where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres. $(a)$ Find $v(t)$ and $a(t)$ of the particle. $(b)$ Find the magnitude and direction of $v(t)$ at $t = 1.0 \ s$.

(N/A) $v(t) = \frac{dr}{dt} = \frac{d}{dt}(3.0 t \hat{i} + 2.0 t^{2} \hat{j} + 5.0 \hat{k}) = 3.0 \hat{i} + 4.0 t \hat{j} \ m/s$.
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(3.0 \hat{i} + 4.0 t \hat{j}) = 4.0 \hat{j} \ m/s^{2}$.
At $t = 1.0 \ s$,the velocity vector is $v = 3.0 \hat{i} + 4.0 \hat{j} \ m/s$.
The magnitude is $|v| = \sqrt{3.0^{2} + 4.0^{2}} = \sqrt{9 + 16} = 5.0 \ m/s$.
The direction $\theta$ with the $x$-axis is given by $\theta = \tan^{-1}(\frac{v_{y}}{v_{x}}) = \tan^{-1}(\frac{4.0}{3.0}) \approx 53^{\circ}$.