OK I redid the problem from scratch:
From second (translation) Newton's law:
$$a_o= \frac{f}{M}$$
From second (rotation) Newton's law:
$$\tau = I \alpha = fa$$
$$k^2 \alpha= a_o a$$
$$\alpha = \frac{a_o a}{k^2}$$
Where ##a_o## is the acceleration of the cylinder measured from the ground.
Assuming that the acceleration of the belt is constant (##a_b##):
$$ \frac{V}{t} = a_b$$
The acceleration of the cylinder with respect to the ground accounts for the acceleration of the belt and the angular acceleration of the cylinder (which has a negative sign because I considered the cylinder spinning clockwise i.e. the belt accelerating to the left and regarding that direction as the positive one). Therefore:
$$a_o = a_b - a\alpha$$
$$a_o = \frac{V}{t} -a\frac{a_o a}{k^2}$$
$$a_o = \frac{Vk^2}{(k^2 + a^2)t}$$
We know by kinematics that:
$$v_{cm} = \frac{Vk^2}{k^2 + a^2}$$
The issue here is that
I had to assume the acceleration of the belt being constant so as to solve the problem. The answer makes sense to me but I would like to know how to solve it without assuming that ##a_b = const##. (Recall that @kuruman suggested on #2 comment that we should not assume it).