$d = v_0 t + \tfrac{1}{2}at^2$ — read d equals v-nought t plus one-half a t-squared. The letters are the same cast as before: $d$ is the displacement in metres, $v_0$ the velocity at the start in m/s, $a$ the acceleration in $\mathrm{m/s^2}$, $t$ the elapsed time in seconds. Notice which letter is absent: the final velocity. This is the relation for when you know how the motion BEGAN and how hard it was pushed, and want to know how far it got.

The $\tfrac{1}{2}$ is not a fudge factor, and here is where it comes from. Draw the velocity–time line again. Under it sits a rectangle of height $v_0$ and width $t$ — that is the $v_0 t$ term, the distance the object would have covered had it never accelerated. Sitting on top of the rectangle is a triangle of base $t$ and height $at$, the speed the acceleration added. A triangle is half its box, so its area is $\tfrac{1}{2}(t)(at) = \tfrac{1}{2}at^2$. Add the two and you have the formula, built rather than memorised.

In calculus language, position is the antiderivative of velocity: integrate $v_0 + at$ with respect to t and out comes $v_0 t + \tfrac{1}{2}at^2 + C$, where C is where you started measuring from. The half is the integral's own bookkeeping. Run the units as a check: $(\mathrm{m/s})(\mathrm{s}) = \mathrm{m}$ and $(\mathrm{m/s^2})(\mathrm{s^2}) = \mathrm{m}$ — both terms land in metres, which is the only way they were ever allowed to be added.