91Ó°ÊÓ

Calculate the derivative of the velocity components with respect to their respective variables. Using \( \nabla \cdot \vec{V} = \frac{ \partial V_x }{ \partial x } + \frac{ \partial V_y }{ \partial y } = 0 \), where \( V_x = A x^{3}+B x y^{2} \) and \( V_y = A y^{3}+B x^{2} y \). After differentiation and simplifying, the equation will be \( 3Ax^{2} + By^{2} + 3Ay^{2} +Bx^{2} =0 \). For an incompressible flow, the total divergence must be zero for all (x, y). Therefore, \( 3A + B = 0 \), which implies \( B = -3A = -3(0.2 \, m^{-2}s^{-1}) = -0.6 \, m^{-2}s^{-1} \). This is the value and units for B in order for the velocity field to represent an incompressible flow.

The acceleration of a particle in the flow is given by the time derivative of the velocity vector. The velocity vector does not vary with time explicitly. However, since the position of the particle changes with time, we need to take the convective derivative which gives the path-dependent change, \( a = \frac {d \vec{V}} {dt} = \frac {\partial \vec{V}} {\partial t} + \vec{V} \cdot \nabla \vec{V} \). With given values of \(x = 2, y = 1\), we find \( a_x = \vec{V} \cdot \nabla (\vec{V}_x) = (A x^{3}+B x y^{2}) \cdot (3Ax^{2}+2By) = (0.2*8 - 0.6*4) = 0.8 \, m^{-1}s^{-1}\) and \( a_y = \vec{V} \cdot \nabla (\vec{V}_y) = (A y^{3}+B x^{2} y) \cdot (3Ay^{2}+2Bx) = (0.2*1 - 0.6*4) = -1.6 \, m^{-1}s^{-1}\). Hence, acceleration \( a = a_x \hat{i} + a_y \hat{j} = 0.8 \hat{i} - 1.6 \hat{j} m^{-1}s^{-1} \).

The normal component of acceleration is obtained by first calculating the magnitude (speed) of the velocity vector at the given point. Then, find the dot product of acceleration and velocity vectors and subtract it from the total acceleration. Evaluate at \( (x, y) = (2, 1) \). The speed \( V_s = \sqrt{V_x^{2} + V_y^{2}} = \sqrt{(0.2*8)^{2} + (-1.2*2)^{2}} = 2.24 m/s \). The tangential component of acceleration \( a_t = \vec{a} \cdot \frac{\vec{V}}{V_s} = (0.8 \hat{i} - 1.6 \hat{j})\cdot (\frac{1.6\hat{i} - 2.4\hat{j}}{2.24}) = 1.28 m/s^{2} \). Hence, normal component \( a_n = \sqrt{a^{2} - a_t^{2}} = \sqrt{1^{2} - 1.28^{2}} = 0 m/s^{2} \).