Answer :

Solution :

$\overrightarrow{V}=(4xy)\hat{i}+2(x^2-y^2)\hat{j}$

U = 4xy , V = 2(x² - y²)

δU/δx = 4y , δU/δy = 4x

δV/δx = 4x, δV/δy = -4y

For Incompressible flow : $\overrightarrow{▽}.\overrightarrow{V}=0$

$\overrightarrow{▽}.\overrightarrow{V}=\frac{\delta U}{\delta x}+\frac{\delta V}{\delta y}$ = 4y - 4y = 0

Hence the flow is incompressible.

For irrotational flow ω_z = 0

ω_z = $\frac{1}{2}(\frac{\delta V}{\delta x}-\frac{\delta U}{\delta y})$  = (4x - 4x) /2 = 0

Hence flow is irrotational.

Acceleration at x = 0, y = 0

a_x = $\frac{DU}{Dt}=(U\frac{\delta U}{\delta x}+V\frac{\delta U}{\delta y})$  = 0

a_y = $\frac{DV}{Dt}=(U\frac{\delta V}{\delta x}+V\frac{\delta V}{\delta y})$  = 0

Hence acceleration in x and y particle is zero.