Given :

$ \vec{v} = \alpha(y \hat{x} + 2x \hat{y}) $

For the x-component, $ v_x = \alpha y $ :

$ \frac{dv_x}{dt} = \alpha \frac{dy}{dt} = \alpha v_y $

Given $ v_y = 2\alpha x $ :

$ \frac{dv_x}{dt} = \alpha (2\alpha x) = 2\alpha^2 x $

For the y-component, $ v_y = 2\alpha x $ :

$ \frac{dv_y}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v_x $

Given $ v_x = \alpha y $ :

$ \frac{dv_y}{dt} = 2\alpha (\alpha y) = 2\alpha^2 y $

Thus, combining the components :

$ \vec{a} = 2\alpha^2 x \hat{x} + 2\alpha^2 y \hat{y} $

Now, the force acting on the particle is given by :

$ \vec{F} = m\vec{a} $

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

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

This is equivalent to :

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

The correct answer is Option A :

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