Eliminate t from v = u + at and s = ut + ½at². Step-by-step derivation.

Derive v² = u² + 2as — Third Equation of Motion

Question

Derive the third equation of motion: v² = u² + 2as

Given that you know:

First equation: v = u + at
Second equation: s = ut + ½at²

Derive the third equation by eliminating time t.

Solution — Step by Step

From v = u + at, we can isolate t directly:

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

We’re eliminating t because the third equation relates velocity and displacement without time — useful when the problem doesn’t tell you how long something took.

Put this expression for t into s = ut + ½at²:

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

Don’t rush this step — expand each term separately to avoid sign errors.

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

Simplify the second term by cancelling one a:

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

Multiply through by 2a on both sides:

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

Expand using the identity (v-u)² = v² - 2uv + u²:

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

The 2uv terms cancel. We’re left with:

2as = v^2 - u^2

Rearrange:

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

Why This Works

We have three kinematic variables beyond displacement: initial velocity u, final velocity v, and time t. The first two equations each involve all three. By solving one for t and substituting into the other, we eliminate that variable entirely.

This is standard simultaneous equation technique — the same thing you do in Class 10 algebra, just applied to physics formulas. The result is an equation that only cares about velocities and displacement, not how long the motion took.

This is why the third equation is so powerful for problems like “a car brakes from 72 km/h to rest over 40 m — find deceleration.” No time is given, no time is needed.

Alternative Method — Using Calculus (Class 11)

For students who have started differentiation, there’s a cleaner route.

We know v = ds/dt and a = dv/dt. Write acceleration as:

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

This is the chain rule. Now separate variables and integrate:

a \, ds = v \, dv

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

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

Which gives v² = u² + 2as directly. For JEE, this derivation is more elegant and takes fewer steps.

Common Mistake

Squaring before substituting. Many students try to square the first equation to get v² = (u + at)² and then substitute at somehow. This leads to a mess because you still have a t term inside the expansion. Always substitute t = (v-u)/a into the second equation — that’s the move that eliminates t cleanly.

Memory hook: The three equations cover three “missing variable” cases — missing s, missing v, and missing t. The third equation v² = u² + 2as is your go-to whenever the problem has no time data. In board exams, roughly 40% of kinematics numericals are solved fastest with this equation.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Calculus (Class 11)

Common Mistake

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