The Equations of Uniformly Accelerated Motion

Displacement is vAve * `dt; since vAve = (vf + v0) / 2 and `dt = tf - t0 we have

• `ds = (vf + v0) / 2  * `dt  or
• `ds = (vf + v0) /2 * (tf - t0).

We have obtained the two equation `ds = (vf + v0) / 2 * `dt and a = (vf - v0) / `dt.  Rearranging the second and keeping the first as it is we get the first two equations of uniformly accelerated motion:

• `ds = (vf + v0) / 2 * `dt and
• vf = v0 + a `dt.

These equations involve the five quantities `ds, `dt, v0, vf and a.  Given any three of these five quantities the two equations give us a set of two simultaneous equations for two unknowns, so we could solve to find the remaining two of the five quantities.

If we eliminate vf from the two equations (substitue v0 + a `dt into the first equation in place of vf) we get the equation

• `ds = v0 `dt + .5 a `dt^2.

If we eliminate `dt from the two equations we get the equation

• vf^2 = v0^2 + 2 a `ds.

This gives us a set of four equations in the five quantites `ds, `dt, v0, vf and a:

• `ds = (vf + v0) / 2 * `dt
• vf = v0 + a `dt
• `ds = v0 `dt + .5 a `dt^2
• vf^2 = v0^2 + 2 a `ds.

Given any three of the quantities `ds, `dt, v0, vf and a we can find an equation containing these three quantities and can therefore solve for the fourth.  Then we will have four quantities, three of which can be substituted into one of the other equations to find the fifth.