In a flow field, the velocity components are, $u=\left ( 6x^{2}y+t \right ),v=\left ( 3yx+t^{2}+3 \right )$ and $w=\left ( 3+3t^2y \right )$. Find the velocity vector and its magnitude at a point $p\left ( 3,2,4 \right )$ at $t=5$ . Units of $x,y,z$ are in meters and time $t$ is in seconds.

Velocity components are $u=\left ( 6x^{2}y+t \right )$; $v=\left ( 3yx+t^{2}+3 \right )$; $w=\left ( 3+3t^{2}y \right)$
Velocity vector is given as,
$$\vec{V}=u\hat{i}+v\hat{j}+w\hat{k}$$ At point ($3,2,4)$ and at $t=5$, velocity components will be,
$$u=\left ( 6x^{2}y+t \right )=\left ( 6\left ( 3 \right )^{2}2+5 \right )=113$$ $$v=\left ( 3yx+t^{2}+3 \right )=\left ( \left ( 3\times 2\times 3 \right )+25+3 \right )=\left ( 18+25+3 \right )=46$$ $$w=\left ( 3+3t^{2}y \right )=\left ( 3+\left ( 3\left ( 5 \right )^{2}2 \right ) \right )=\left ( 3+150 \right )=153$$
Therefore, velocity vector is,
$$\vec{V}=113\hat{i}+46\hat{j}+153\hat{k}$$ Velocity magnitude is, $$V=\sqrt{u^{2}+v^{2}+w^{2}}$$ $$=\sqrt{113^{2}+46^{2}+153^{2}}=195.69\,m/s$$