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

(See also my post on parabolic trajectories.)


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