## The Kinematic Equations

It seems that the kinematic equations can all be derived from some simple definitions.

We’ll begin with our symbols:
$s$ = position (or distance)
$u$ = initial velocity
$v$ = final velocity
$a$ = acceleration
$t$ = time (or duration)

We have the definitions:

$a = v' = \dfrac{dv}{dt}$ so $v = \int{a}\,{dt}$

$v = s' = \dfrac{ds}{dt}$ so $s = \int{v}\,{dt}$

Now, for our purposes, we will regard $u$ and $a$ as constant, and $v$ as a function $v(t)$ of $t$.

Since $a$ is constant, we have (abusing notation a little)
$a = \dfrac{\Delta v}{\Delta t} =\dfrac{v-u}{t}$

Thus we have our first kinematic equation:
$v = u + at$ [1]

We will now integrate this (w.r.t. $t$):
$\int v\,dt = \int u+at\, dt$
so $s = u\,\int dt + a \int t\,dt = ut + \frac{1}{2}at^2 + K$.
Using the boundary condition $s=0$ if $t=0$, we have $K=0$.
Thus we have our second kinematic equation:
$s = ut + \frac{1}{2}at^2$ [2]

Now, if we substitute [1] into [2] at $u$, we obtain
$s = (v-at)t + \frac{1}{2}at^2$
and so we have our third kinematic equation (one not normally given)
$s = vt - \frac{1}{2}at^2$ [3]

If we substitute [1] into [2] differently, at $a$, we obtain
$s = ut + \frac{1}{2}\left(\dfrac{v-u}{t}\right)t^2$
so $2s = 2ut+(v-u)t$
and so we have our fourth kinematic equation
$s=\dfrac{(u+v)t}{2}$ [4]

Finally, substituting [4] into [1] at t, we have
$v = u+a \left(\dfrac{2s}{u+v}\right)$
so $(v-u)(v+u)=2as$
and so we have our fifth kinematic equation
$v^2=u^2+2as$ [5]

To review what we have:

[1] or [¬s]: $v=u+at$ ; an equation without $s$

[2] or [¬v]: $s = ut + \frac{1}{2}at^2$ ; an equation without $v$

[3] or [¬u]: $s = vt - \frac{1}{2}at^2$ ; an equation without $u$

[4] or [¬a]: $s=\dfrac{(u+v)t}{2}$ ; an equation without $a$

[5] or [¬t]: $v^2=u^2+2as$ ; an equation without $t$