Discussion
For a steady flow, the velocity field is \(\vec{V} = (-x^{2} + 3y)\hat{i} + (2xy)\hat{j}\). The magnitude of the acceleration of a particle at \((1, -1)\) is
Answer: C
The acceleration magnitude is \(2\sqrt{5}\).
For steady flow the acceleration is purely convective. With \(u = -x^{2} + 3y\) and \(v = 2xy\), at (1, -1): u = -4 and v = -2.
\[ \begin{aligned} a_{x} &= u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = (-4)(-2) + (-2)(3) = 2 \\ a_{y} &= u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = (-4)(-2) + (-2)(2) = 4 \end{aligned} \]
So \(|a| = \sqrt{4 + 16} = 2\sqrt{5}\). Zero would be right only if one forgot that steady flow still has convective acceleration; 2 is the x component alone.
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