Second Equation of Motion

What this formula says

A body starts with initial velocity $u$ and moves under constant acceleration $a$ for time $t$. The total displacement covered is $s$.

$s = ut + \frac{1}{2}at^2$

There are two parts to this:

• $ut$ — the displacement the body would have covered if there were no acceleration (just constant velocity $u$ for time $t$)
• $\frac{1}{2}at^2$ — the extra displacement gained because acceleration kept increasing the velocity throughout

Derivation using the average velocity method

When acceleration is constant, the velocity increases uniformly from $u$ to $v$. A quantity that increases uniformly has its average exactly at the midpoint:

$\text{average velocity} = \frac{u + v}{2}$

Displacement is average velocity multiplied by time:

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

Now substitute $v = u + at$ (the first equation of motion):

$s = \frac{u + (u + at)}{2} \times t$

$s = \frac{2u + at}{2} \times t$

$s = \left(u + \frac{at}{2}\right) \times t$

$\boxed{s = ut + \frac{1}{2}at^2}$

Why average velocity is $\frac{u+v}{2}$ only for constant acceleration

This is worth understanding carefully. Average velocity is always defined as:

$\text{average velocity} = \frac{\text{total displacement}}{\text{total time}}$

For uniform acceleration, velocity increases linearly with time — it forms a straight line on a velocity-time graph. The average of a linearly changing quantity is simply the average of its starting and ending values. That is why $\frac{u+v}{2}$ works here.

If acceleration were not constant — if velocity changed in a curve — then $\frac{u+v}{2}$ would give the wrong average, and this derivation would break down.

The calculus derivation (for advanced learners)

Velocity is the derivative of displacement with respect to time:

$v = \frac{ds}{dt}$

We know $v = u + at$, so:

$\frac{ds}{dt} = u + at$

Integrate both sides with respect to time, from $t = 0$ (when $s = 0$) to time $t$:

$\int_0^s ds = \int_0^t (u + at) \, dt$

Left side:

$s$

Right side — integrate term by term:

$\int_0^t u \, dt + \int_0^t at \, dt = ut + a \cdot \frac{t^2}{2}$

Therefore:

$\boxed{s = ut + \frac{1}{2}at^2}$

The calculus derivation makes no use of the average velocity shortcut — it works directly from the definition of velocity as rate of change of displacement.

Understanding the two terms geometrically

On a velocity-time graph, the body starts at velocity $u$ at $t = 0$ and rises linearly to $v = u + at$ at time $t$. The displacement is the area under this graph.

The area under the graph is a trapezium, which can be split into:

• A rectangle of height $u$ and width $t$ → area = $ut$
• A triangle of base $t$ and height $at$ → area = $\frac{1}{2} \times t \times at = \frac{1}{2}at^2$

Total area = $ut + \frac{1}{2}at^2$ = displacement. This is the geometric meaning of both terms.

Sign convention

Displacement $s$ is positive if the body moves in the chosen positive direction, negative otherwise. Since $u$, $a$, and $t$ can each be positive or negative (except $t^2$ which is always positive), the formula handles all cases automatically when signs are assigned correctly.