Third Equation of Motion

What this formula says

A body starts with velocity $u$ and moves under constant acceleration $a$, covering a displacement $s$. The final velocity $v$ satisfies:

$v^2 = u^2 + 2as$

Notice that time does not appear in this formula. This is what makes it uniquely useful — whenever a problem gives you displacement and asks for velocity (or vice versa), without mentioning time, this is the formula to reach for.

Derivation by eliminating time

We already have two equations of motion:

$v = u + at \quad \text{...(1)}$

$s = ut + \frac{1}{2}at^2 \quad \text{...(2)}$

We want a relation between $v$, $u$, $a$, and $s$ — with no $t$. So we eliminate $t$.

From equation (1):

$t = \frac{v - u}{a}$

Substitute this into equation (2):

$s = u \cdot \frac{v-u}{a} + \frac{1}{2}a \cdot \left(\frac{v-u}{a}\right)^2$

$s = \frac{u(v-u)}{a} + \frac{1}{2}a \cdot \frac{(v-u)^2}{a^2}$

$s = \frac{u(v-u)}{a} + \frac{(v-u)^2}{2a}$

Multiply every term by $2a$ to clear the denominators:

$2as = 2u(v-u) + (v-u)^2$

Expand $2u(v-u)$:

$2u(v-u) = 2uv - 2u^2$

Expand $(v-u)^2$:

$(v-u)^2 = v^2 - 2uv + u^2$

Add them:

$2as = 2uv - 2u^2 + v^2 - 2uv + u^2$

The $2uv$ terms cancel:

$2as = v^2 - u^2$

$\boxed{v^2 = u^2 + 2as}$

Derivation using average velocity (cleaner)

There is a more elegant route. Displacement equals average velocity times time:

$s = \frac{u+v}{2} \cdot t$

From the first equation, $t = \frac{v-u}{a}$. Substitute:

$s = \frac{u+v}{2} \cdot \frac{v-u}{a}$

$s = \frac{(v+u)(v-u)}{2a}$

The numerator is a difference of squares: $(v+u)(v-u) = v^2 - u^2$

$s = \frac{v^2 - u^2}{2a}$

$2as = v^2 - u^2$

$\boxed{v^2 = u^2 + 2as}$

This route is shorter and the algebra is cleaner.

The calculus derivation (for advanced learners)

Acceleration can be written as:

$a = \frac{dv}{dt}$

Multiply numerator and denominator of $\frac{dv}{dt}$ by $ds$:

$a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v\frac{dv}{ds}$

This is a useful identity: $a = v\dfrac{dv}{ds}$.

Now separate variables:

$a \, ds = v \, dv$

Integrate both sides — displacement from $0$ to $s$, velocity from $u$ to $v$:

$\int_0^s a \, ds = \int_u^v v \, dv$

Left side ($a$ is constant):

$a \cdot s$

Right side:

$\frac{v^2}{2} - \frac{u^2}{2}$

Therefore:

$as = \frac{v^2 - u^2}{2}$

$\boxed{v^2 = u^2 + 2as}$

When to use this formula

The three equations of motion each eliminate one quantity:

When the problem does not mention time — or when finding time is not required — reach for $v^2 = u^2 + 2as$.