The position of a particle is given by$\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)=4\mathrm{t}\hat{\mathrm{i}}+2{\mathrm{t}}^2\hat{\mathrm{j}}+5\hat{\mathrm{k}}$ where t is in seconds and r in metre. Find the magnitude and direction of velocity v(t), at t = 1s, with respect to x-axis

Position vector, 

$\begin{array}{rl} & r=4\widehat{ti}+2t^2\hat{j}+5\hat{k}\\ & \text{Velocity}=\frac{d\overrightarrow{r}}{dt}=4\hat{i}+4t\hat{j}\\ & \text{at }t=1\mathrm{s},4\hat{i}+4\hat{j}=\overrightarrow{v}\\ & \left|\overrightarrow{v}\right|=\sqrt{4^2+4^2}=4\sqrt{2},\end{array}$

direction   $\tan \theta =\frac{4}{4}=1$

$\theta ={45}^{\circ }$

So, it has a magnitude of  $4\sqrt{2}{\mathrm{ms}}^{-1}$ at an angle 45° with positive x-axis at t = 1 s.