To get the position of an object we can use the equation: $$x=x_0+v_0t+\frac{1}{2}at^2$$
But now I don't have a constant acceleration, both $x_0$ and $v_0$ start at $0$ and $v$ ends at $16 \text{ m/s}$ and $t$ ends at $12 \text{ s}$. So to calculate the position I can use the average acceleration which is about $1.33 \ \frac{m}{s^2}$.
Am I right?
Best to draw the graph. The distance travelled (hence position) is equal to the area under the curve.
This equation you are showing is one of the four kinematic motion equations. They have been derived from the assumption of constant acceleration. It thus only applies for scenarios where the acceleration is constant.
If you don't have constant acceleration then you might have to fall back to the general definitions of and calculate what you need from there:
$$v=\frac{ds}{dt} \quad, \quad a=\frac{dv}{dt}.$$
These always apply, but are mathematically harder to use and require more information about the scenario. It is possible that you can avoid these and use yet other methods - maybe geometric methods for instance that might be much easier. That depends on what information you've been given in the task.
You cannot determine the position from only the average acceleration.
For example, suppose the acceleration is given by $a(t)=j t + a_0$, where $j$ is the constant jerk and $a_0$ is the initial acceleration. Then with the given initial conditions the position is $x(t)=\frac{1}{2} a_0 t^2 + \frac{1}{6} j t^3$ and the velocity is $v(t)=a_0 t + \frac{1}{2} j t^2$.
With the additional condition that $v(12)=16$ we find that $a_0=\frac{4}{3}-6j$. Substituting back and simplifying we get $x(12)=96-144 j$. So if $j=0$ then the final position will be 96, but if $j=\frac{2}{3}$ then the final position will be 0. In all cases, the average acceleration is indeed $\frac{4}{3}$, but the way the actual acceleration varies about that average determines the distance.
