Answer :

Correct answer is : 4.24

ψ = kx³y, k = 1 m^-2s^-1

at x = 1 m, y = 1 m

$u=-\frac{d\psi }{dy},v=\frac{d\psi }{dx}$

u = - kx³ , v = 3kx²y

u = - x³ , v = 3x²y

Since the flow is steady and two-dimensional:

acceleration in x-direction $a_x=u\frac{du}{dx}+v\frac{du}{dy}$

a_x = x³ (-3x²) - 0 = - 3 m/s²

acceleration in y-direction $a_y=u\frac{dv}{dx}+v\frac{dv}{dy}$

a_y = - x³ (6xy) + (3x²y) (3x²)

a_y = 1 (-6) + (3)(3) = -6 + 9

a_y = 3 m/s²

The magnitude of acceleration :

$|a|=\sqrt{a_x^2+a_y^2}=\sqrt{(-3{)}^2+3^2}$

|a| = 3√2 = 4.24