Position at constant acceleration: no calculus allowed

Remember our conventions: position (x), velocity (v), acceleration (a) and time (t).

We already know that, at constant v
x = vt

We also know that, at constant a
v = v0 + at

What happens to position with constant acceleration? Well...

If we begin at a point x0, and travel at an average velocity vavg, we travel a distance given by
x = x0 + vavg t

The definition of vavg is
v(avg) = (v0 + vf)/2

But, under constant acceleration a, the final velocity vf is simply
vf = v0 + at
so that
v(avg) = v0 + (1/2)at

Substituting this into our equation for position x = x0 + vavg t, we obtain x = x0 + (v0 + ½at) t

x = x0 + v0 t + ½at2