The SUVAT Equations — Constant-Acceleration Kinematics

High School⏱ 10 min readClassical Mechanics

The five SUVAT equations for motion with constant acceleration: what s, u, v, a and t mean, how to choose the right equation, and fully worked examples.

Quick answer

The SUVAT equations are five formulas linking displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t) for motion in a straight line when acceleration is constant.

$v = u + at,\qquad s = ut + \tfrac{1}{2}at^2,\qquad v^2 = u^2 + 2as$

Unit: s in meters (m), u and v in m/s, a in m/s², t in seconds (s)

Example: A car accelerates from rest at 3 m/s² for 4 s. Its final velocity is v = 0 + 3×4 = 12 m/s, and it covers s = 0 + ½×3×16 = 24 m.

What you will learn

Define each SUVAT variable and state its SI unit
State all five constant-acceleration equations from memory
Choose the correct equation based on which variable is missing
Solve one-dimensional motion problems with correct signs

What the letters stand for

SUVAT is a memory aid built from the five quantities in constant-acceleration motion. The s is displacement (change in position, in meters), u is the initial velocity, v is the final velocity, a is the acceleration, and t is the time the motion takes.

Each letter names the quantity, not a value — think of them as labeled boxes you fill in from the question. In every SUVAT problem you know three of the five, want to find a fourth, and the fifth does not appear in the equation you choose.

The five equations

There are five equations, and each one leaves out exactly one of the five variables:

$v = u + at \qquad\text{(no } s\text{)}$

$s = ut + \tfrac{1}{2}at^2 \qquad\text{(no } v\text{)}$

$s = vt - \tfrac{1}{2}at^2 \qquad\text{(no } u\text{)}$

$s = \tfrac{1}{2}(u+v)t \qquad\text{(no } a\text{)}$

$v^2 = u^2 + 2as \qquad\text{(no } t\text{)}$

Notice the pattern: each equation contains exactly four of the five letters. Spotting which letter is missing from the question's data tells you which equation to use.

v = u + at,\quad s = ut + \tfrac{1}{2}at^2,\quad v^2 = u^2 + 2as

The three most-used SUVAT equations (the other two follow from these).

Choosing the right equation

The reliable method is to list what you know before touching any equation:

Write down the five letters: s, u, v, a, t.
Mark the ones given in the question.
Mark the one asked for.
The remaining letter appears in exactly one candidate — pick the equation that does not contain it.

For example, if you know $u$ , $a$ and $t$ and want $v$ , the letter $s$ is irrelevant — so use $v = u + at$ , the equation with no $s$ . If time is never mentioned and never asked for, $v^2 = u^2 + 2as$ is almost always the right call.

Selection trick

Do not pick an equation by what you want — pick it by what is missing. The equation that omits the unused variable is always the one you need.

Worked example: finding final velocity

A cyclist pedals from $6\ \mathrm{m/s}$ with a steady acceleration of $1.5\ \mathrm{m/s^2}$ for $8\ \mathrm{s}$ . How fast is she moving at the end?

List: $u = 6\ \mathrm{m/s}$ , $a = 1.5\ \mathrm{m/s^2}$ , $t = 8\ \mathrm{s}$ , want $v$ . The unused letter is $s$ , so use $v = u + at$ :

$v = 6 + (1.5)(8) = 6 + 12 = 18\ \mathrm{m/s}$

Because acceleration is constant, the velocity grows by the same amount each second — $1.5\ \mathrm{m/s}$ every second, eight times over.

Final velocity is the vector sum of the initial velocity and the change from acceleration.

Worked example: finding displacement without time

A train moving at $25\ \mathrm{m/s}$ brakes with a constant deceleration of $2\ \mathrm{m/s^2}$ . How far does it travel before stopping?

List: $u = 25\ \mathrm{m/s}$ , $v = 0$ (it stops), $a = -2\ \mathrm{m/s^2}$ (opposing the motion), want $s$ . Time is not given and not asked for, so use $v^2 = u^2 + 2as$ :

$0 = 25^2 + 2(-2)s \implies 4s = 625 \implies s = 156.25\ \mathrm{m}$

The train travels about $156\ \mathrm{m}$ — roughly a train and a half — before coming to rest.

Watch the sign of a

Deceleration is acceleration with the opposite sign to the velocity. If forward is positive, braking means a is negative — the minus sign does the work, so never drop it.

Signs and coordinate systems

SUVAT variables are signed because they are one-dimensional vectors. Before solving, choose which direction is positive and stick to it: velocities and accelerations pointing that way are positive, those pointing the other way are negative.

A classic trap is dropping a stone from a cliff while taking upward as positive. Then $u = 0$ , $a = -9.81\ \mathrm{m/s^2}$ , and a downward displacement is negative. The equations do not care — they give correct answers as long as every sign matches your chosen direction.

When SUVAT does not apply

Every SUVAT equation assumes constant acceleration along a straight line. It breaks down in three common situations:

Air resistance matters. A falling feather or a fast car has acceleration that changes as speed changes — SUVAT overestimates how far or fast it goes.
Acceleration changes with time. A rocket burning fuel gets lighter and accelerates harder; you need calculus, not SUVAT.
Direction changes. A car turning a corner at steady speed has changing velocity even though its speed is constant — the equations only handle straight-line motion.

Free fall near Earth's surface is the classic valid case: acceleration is essentially constant at about $9.81\ \mathrm{m/s^2}$ downward.

Derived, not fundamental

SUVAT is not a new law of physics — each equation is a rearrangement of the definitions of velocity and acceleration under the assumption that a is constant.

Variables and units

Symbol	Name	Unit	Meaning
$s$	Displacement	m	Change in position from start to finish
$u$	Initial velocity	m/s	Velocity at the start of the time interval
$v$	Final velocity	m/s	Velocity at the end of the time interval
$a$	Acceleration	m/s²	Constant rate of change of velocity
$t$	Time	s	Duration of the motion

Common mistakes

✗ s stands for speed or distance.

s is displacement — the straight-line change in position, with a sign. For motion that reverses direction, displacement and distance differ.

✗ SUVAT works for any motion problem.

It only works when acceleration is constant and motion is along a straight line. Changing acceleration or curved paths need different tools.

✗ Deceleration is a separate kind of acceleration.

Deceleration is just acceleration with a sign opposite to the velocity in your coordinate system. There is only one acceleration, and it carries a sign.

Real-world applications

Traffic engineers use SUVAT to set yellow-light durations and safe stopping distances.
Sports analysts estimate sprint performance by fitting constant-acceleration models to timing splits.
Aerospace test ranges use SUVAT to predict where a projectile lands when drag is negligible.
Elevator designers choose acceleration limits so rides stay comfortable while travel time stays short.

Frequently asked questions

Why are there five equations instead of one?

Each equation omits exactly one of the five variables. Since a typical problem gives you three knowns and asks for a fourth, there is always one irrelevant variable — and exactly one equation that avoids it. The five forms are all consequences of v = u + at plus the definition of average velocity.

What does the u in SUVAT stand for?

The letter u stands for initial velocity. Historically, v was taken for velocity and u was the next available letter, so u became the initial value and v the final value. Both are measured in meters per second.

Can acceleration be zero in SUVAT problems?

Yes. Setting a = 0 reduces the equations to constant-velocity motion: v = u and s = ut. This is a good consistency check — if your solution misbehaves at zero acceleration, something is wrong.

What is the difference between s = ut + ½at² and s = ½(u+v)t?

The first computes displacement from the starting conditions, adding the extra distance gained as velocity grows. The second uses the average velocity ½(u+v) multiplied by time. They agree exactly when acceleration is constant — use whichever matches the data you have.