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.

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    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, \[V=\sqrt{u^{2}+v^{2}+w^{2}}\]\[=\sqrt{113^{2}+46^{2}+153^{2}}=195.69\,m/s\]

    Answered on 20th May 2021.
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