If you want constant acceleration (like you do now), then you can consider it as identical to the second equation of motion from basic physics:

$$s = s_0 + v_0 t + \frac{1}{2} a t^2s$$

only with s as an angle, v as your angular velocity. Solve for a. This probably won't give you what you want, i.e. your final velocity won't be 0.

Let's say, then, that our acceleration is linear, i.e.

$$a(t) = mt + b$$

Then our velocity is necessarily quadratic

$$v(t) = \frac{m}{2} t^2 + bt + c$$

with the constant term dictated by the intial velocity

$$v(0) = 360 = c$$

and position given by the definite integral

$$s(2.5) = \int_{0}^{2.5} v(t) dt = \frac{m}{6} (2.5)^3 + \frac{b}{2} (2.5)^2 + 360(2.5)$$

so with the distance traveled and other constraints we have the other equation

$$v(2.5) = \frac{m}{2} (2.5)^2 + b (2.5) + 360 = 0$$

So you have two equations and two unknowns. Just solve for m and b and plug those into a=mt+b.

Editing posts destroys equations, so I'll just sum up here instead:

$$\frac{m}{6} (2.5)^3 + \frac{b}{2} (2.5)^2 + 360(2.5) = 234$$

$$\frac{m}{2} (2.5)^2 + b (2.5) + 360 = 0$$

are your two equations, and your acceleration will be

$$ a = m t + b$$

with m and b solved from above