# Find the acceleration of a fluid particle in the $$x$$ direction.

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.

Asked on 5th May 2021 in
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 $$=12x= 12\times 3= 36\,m/s^{2}$$