s = ut + ½at² — Displacement under Constant Acceleration

The second SUVAT equation: displacement from initial velocity, acceleration and time — with a worked example.

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

Displacement equals initial velocity times time plus one-half acceleration times time squared.

What it means

Displacement has two contributions: coasting at the initial velocity ( $ut$ ) plus the extra distance from speeding up ( $\tfrac{1}{2}at^2$ ). Because time is squared, doubling the acceleration time more than doubles the distance gained from it.

Variables

Symbol	Name	Unit	Meaning
$s$	Displacement	m	Distance from the start position
$u$	Initial velocity	m/s	Velocity at time zero
$a$	Acceleration	m/s²	Rate of change of velocity
$t$	Time	s	Duration of motion

Assumptions

Acceleration is constant over the time interval
Motion is along a straight line
s measured from the starting position

Derivation

Displacement is the area under the velocity–time graph. With $v = u + at$ the graph is a straight line, so the area is a rectangle ( $ut$ ) plus a triangle ( $\tfrac{1}{2}at^2$ ) — hence $s = ut + \tfrac{1}{2}at^2$ .

Worked example

Problem

A runner moving at 5 m/s accelerates at 2 m/s² for 4 seconds. How far does she travel?

Given:

u = 5 m/s
a = 2 m/s²
t = 4 s

Write the equation: $s = ut + \tfrac{1}{2}at^2$
Substitute: $s = 5 \times 4 + 0.5 \times 2 \times 4^2$
Calculate: $s = 20 + 16 = 36\ \mathrm{m}$

36 m

The runner covers 36 m in 4 s — 20 m from the initial speed plus 16 m gained by accelerating. The squared term means extra time pays off quadratically.