First Equation of Motion

What this formula says

A body is moving. At some moment, we start observing it. At that moment, its velocity is $u$ — we call this the initial velocity.

A constant force acts on it, giving it a constant acceleration $a$. After a time $t$ has passed, the body now has a new velocity $v$ — we call this the final velocity.

The formula tells us exactly what $v$ is :

$v = u + at$

The velocity changed by an amount $at$. That is the acceleration multiplied by the time. This is the entire content of the formula.

Starting from the definition of acceleration

Acceleration is defined as the rate of change of velocity. In plain words: how much does the velocity change per second?

$a = \frac{\text{change in velocity}}{\text{time taken}}$

The change in velocity is the final velocity minus the initial velocity: $v - u$. The time taken is $t$.

So:

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

This is not a derived result. This is the definition of acceleration. We are simply writing in symbols what acceleration means in words.

Rearranging to get $v$

We have:

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

Multiply both sides by $t$:

$at = v - u$

Add $u$ to both sides:

$u + at = v$

Which we write as:

$\boxed{v = u + at}$

That is the complete derivation. No calculus needed. The formula follows directly from the definition of acceleration.

Why this only works for constant acceleration

The definition $a = \frac{v-u}{t}$ gives the average acceleration over the time interval. When acceleration is constant, the average acceleration equals the instantaneous acceleration at every moment — so the formula holds exactly.

If acceleration is changing — for example, a car that keeps pressing harder on the accelerator — then $a$ is not a single fixed number, and this formula cannot be used directly.

The calculus derivation (for advanced learners)

For those who know calculus, the same result follows more rigorously.

Acceleration is the derivative of velocity with respect to time:

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

Since $a$ is constant, we can separate variables and integrate both sides:

$\int_u^v dv = \int_0^t a \, dt$

Left side:

$\int_u^v dv = v - u$

Right side ($a$ is a constant, comes out of the integral):

$\int_0^t a \, dt = a \cdot t$

Putting it together:

$v - u = at$

$\boxed{v = u + at}$

The same result. The calculus derivation is more powerful because it makes the assumption of constant $a$ explicit — it is the step where $a$ comes outside the integral.

Sign convention

Velocity and acceleration are vectors — they have direction. In one-dimensional motion, we assign a positive direction (usually the direction of initial motion) and treat quantities in the opposite direction as negative.