Find the acceleration of a fluid particle in the \(x\) direction at \((x,y) = (3,1)\) for a velocity field, which is given by\(\vec{V} = 3y\hat{i}+4x\hat{j}\). \(x\) and \(y\) are in meters.

Velocity field is \(\vec{V}=3y\hat{i}+4x\hat{j}\)
Here, \(\vec{V}=u\hat{i}+v\hat{j}\)
Acceleration of the fluid particle in the \(x\)-direction is given as\[a_{x}=u\frac{\partial u}{\partial x}+v\frac{\partial u }{\partial y}\]\[\Rightarrow a_{x}=3y\frac{\partial }{\partial x}\left ( 3y \right )+4x\frac{\partial }{\partial y}\left ( 3y \right )\]\[=\left ( 3y\times 0 \right )+\left ( 4x\times 3 \right )=12x\]At point \((x,y)\)=\((3,1)\), acceleration of the fluid particle \(=12x= 12\times 3= 36\,m/s^{2}\)