# Find the velocity vector and its magnitude at a point $$p\left ( 3,2,4 \right )$$ at $$t=5$$ seconds.

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.

Asked on 6th May 2021 in
Velocity components are $$u=\left ( 6x^{2}y+t \right )$$; $$v=\left ( 3yx+t^{2}+3 \right )$$; $$w=\left ( 3+3t^{2}y \right)$$
$\vec{V}=u\hat{i}+v\hat{j}+w\hat{k}$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$
$\vec{V}=113\hat{i}+46\hat{j}+153\hat{k}$Velocity magnitude is, $V=\sqrt{u^{2}+v^{2}+w^{2}}$$=\sqrt{113^{2}+46^{2}+153^{2}}=195.69\,m/s$