Whaaaaat?

If you have an object undergoing constant acceleration (in the direction opposite to the velocity, ie a deceleration) and an initial velocity v0 then the stopping distance may be calculated as the position from the starting point where the velocity goes to 0. ie.

0 = v0^2 + 2ad

d = -v0^2/(2a)

(The negative sign appears because I have tacitly assumed a positive direction as the direction of the initial velocity, and hence the acceleration, a, is negative.)

Neither of the equations:

t = as^2

d = v +(v^2/20)

come out in the correct units. I would treat them as highly suspect. (Or more likely as applications to specific problems.)

-Dan