The full equation
$$ Xf = X_o + V_o t + \frac{at^2}{2} $$
is integrated from the velocity function (which was integrated from constant acceleration function), right?
The problem is, I can't seem to put my mind around why the acceleration part needs to be halved.
Doesn't $at$ already measure the change in velocity over time? Why not just add that to the velocity then use the new V to measure Xf?
Looking at the graph you can also see that the displacement is equal to the average velocity $\times$ time.
yes, the 1/2 is due to the double integral.
How $at$ evolves during the interval $[0,T]$ ? It's value is 0 at the begining, and $aT$ at the end. So the cumulated value in $v$ cannot be $aT^2$, or it would mean that the value was constantly equat to $aT$ during the interval.
When you sum-up varying quantities, you really have to compute the integral, not multiply the final value by $T$.
Taylor series of the space function $s(t)$:
$$s(t) = s_0 + s'(t_0)(t - t_0) + \frac{1}{2}s''(t_0)(t - t_0)^2$$
Now, having $s'(t_0) = v_0$ and $s'' = a$ you get:
$$s(t) = S_0 + v_0t + \frac{1}{2}at^2$$
assuming $t_0 = 0$
A simple intuitive explanation is that distance equals average speed times time. The average speed is $1/2(0+aT)$, and time is $T$ (assuming starting from rest, and constant acceleration).
