A particle of mass m is moving in the $$xy$$-plane such that its velocity at a point $$(x, y)$$ is given as $$\vec{v} = \alpha(y\hat{x} + 2x\hat{y})$$, where $$\alpha$$ is a non-zero constant. What is the force $$\vec{F}$$ acting on the particle?

The velocity field is $$\vec v = \alpha\bigl(y\,\hat x + 2x\,\hat y\bigr)$$.
Hence

$$v_x = \alpha y , \qquad v_y = 2\alpha x$$

Acceleration is the material (total) derivative of velocity:

$$\vec a = \frac{d\vec v}{dt} = v_x\frac{\partial\vec v}{\partial x} + v_y\frac{\partial\vec v}{\partial y}$$

Component $$a_x$$:

$$a_x = v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y}$$

Since $$v_x = \alpha y$$, we have $$\frac{\partial v_x}{\partial x}=0$$ and $$\frac{\partial v_x}{\partial y}=\alpha$$.

Therefore $$a_x = v_x(0) + v_y(\alpha) = (2\alpha x)(\alpha) = 2\alpha^{2}x$$.

Component $$a_y$$:

$$a_y = v_x\frac{\partial v_y}{\partial x} + v_y\frac{\partial v_y}{\partial y}$$

Because $$v_y = 2\alpha x$$, we have $$\frac{\partial v_y}{\partial x}=2\alpha$$ and $$\frac{\partial v_y}{\partial y}=0$$.

Thus $$a_y = v_x(2\alpha) + v_y(0) = (\alpha y)(2\alpha) = 2\alpha^{2}y$$.

Combining the two components:

$$\vec a = 2\alpha^{2}\bigl(x\,\hat x + y\,\hat y\bigr)$$

Using Newton’s second law $$\vec F = m\vec a$$:

$$\vec F = 2m\alpha^{2}\bigl(x\,\hat x + y\,\hat y\bigr)$$

Therefore, the correct option is:
Option A which is: $$\vec{F} = 2m\alpha^{2}(x\hat{x} + y\hat{y})$$.