Once an object has constant acceleration, four neat formulas — the equations of motion, often called SUVAT — let you solve almost any motion problem. They tie together five quantities, and the trick is simply choosing the right equation for what you know and what you want.
📘 What you need to know
The equations of motion work only for constant (uniform) acceleration in a straight line.
They link five variables: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time).
The nickname SUVAT comes from those five letters.
Each equation uses four of the five — so you pick the one that’s missing the variable you don’t care about.
“Starts from rest” → u = 0. “Falling under gravity” → a = g = 9.81 m s⁻².
Always be consistent with direction: if you call up positive, then downward quantities are negative.
The five SUVAT variables
Before the formulas, get comfortable with the five letters. Every kinematics question is really just hunting for one of these, given some of the others.
The five quantities every kinematics problem is built from.
The four equations
Here are the four equations of motion. Notice each one is missing exactly one of the five variables — that’s your clue for which to use.
v = u + atno s
s = ut + ½at²no v
v² = u² + 2asno t
s = (u + v)2tno a
Think of it as a “which one’s missing?” game. If a question never mentions time and never asks for it, reach for v² = u² + 2as — the equation with no t. Match the equation to the one variable you neither know nor want.
Where v = u + at comes from
You don’t need to memorise the derivations, but seeing one makes the equations feel less like magic. Acceleration is the change in velocity over time: a = (v − u) ÷ t. Multiply both sides by t to get at = v − u, then rearrange to v = u + at. That’s the first equation, straight from the definition of acceleration.
Key phrases to watch for
Exam questions hide useful information inside ordinary words. Train yourself to translate these:
“Starts from rest” → u = 0. • “Comes to a stop” → v = 0. • “Falling under gravity” → a = g = 9.81 m s⁻². • “Constant acceleration in a straight line” → SUVAT is the tool to use.
🧭 The 3-step SUVAT method
List your variables. Write s, u, v, a, t down the side and fill in what you know. Use clues (“from rest” → u = 0) to fill gaps, and mark the one you want.
Pick the equation that contains your knowns and your unknown — the one missing the variable you don’t have and don’t need.
Convert to SI units, substitute the numbers in, and rearrange to get the answer.
Worked examples
WE 1
A cyclist accelerating — distance and final velocity
A cyclist rides east through a flat village at 6 m s⁻¹, then accelerates constantly at 2 m s⁻² for 4 s. (a) How far do they travel in those 4 s? (b) What’s their final velocity?
List: u = 6, a = 2, t = 4(a) Want s, don’t have v → use s = ut + ½at²s = (6 × 4) + (0.5 × 2 × 4²)s = 24 + 16s = 40 m(b) Want v, don’t have s → use v = u + atv = 6 + (2 × 4)v = 14 m s⁻¹
WE 2
A braking train — how far apart should the markers be?
A train approaches at 50 m s⁻¹. The driver brakes at marker 1 so the train decelerates uniformly and passes marker 2 at no more than 10 m s⁻¹, taking 20 s between the markers. How far apart should the markers be?
List: u = 50, v = 10, t = 20, want sDon’t have a → use s = ½(u + v)ts = ½ × (50 + 10) × 20s = ½ × 60 × 20s = 600 mmarker 1 should be 600 m before marker 2.
WE 3
A stone dropped down a well
A stone is dropped from rest down a well and hits the water after falling 20 m. Taking g = 9.81 m s⁻², find the velocity with which it hits the water.
“Dropped from rest” → u = 0; falling → a = 9.81; s = 20No time given or wanted → use v² = u² + 2asv² = 0² + (2 × 9.81 × 20)v² = 392.4v = √392.4v ≈ 19.8 m s⁻¹
WE 4
One cyclist catching another
Cyclist A rides at a constant 18 m s⁻¹. As A passes friend B, B starts from rest and accelerates at 1.5 m s⁻² in the same direction. How long until B catches A?
Both cover the same displacement when B catches AA (no acceleration): s = 18tB (from rest): s = ½ × 1.5 × t² = 0.75t²Set equal: 18t = 0.75t²0.75t² − 18t = 0t(0.75t − 18) = 0t = 0 (the start) or t = 18 ÷ 0.75t = 24 st = 0 is just the moment they first pass; the catch-up is at 24 s.
💡 Top tips
Always list s, u, v, a, t first. Half the marks come from setting the problem up cleanly.
Pick the equation by what’s missing — match it to the variable you neither have nor want.
Be consistent with signs. Choose a positive direction and stick to it for every vector in the problem.
For vertical motion, a = ±9.81 m s⁻² depending on whether you called up or down positive.
Check the equation only applies for constant acceleration — if a changes, SUVAT won’t work.
⚠ Common mistakes
Using SUVAT when acceleration isn’t constant — the equations simply don’t apply then.
Mixing sign conventions mid-problem, so velocity and acceleration end up pointing the “wrong” way.
Forgetting “from rest” means u = 0 (and “stops” means v = 0).
Squaring slips in v² = u² + 2as — remember to square-root at the end to get v.
Not converting units (km h⁻¹ to m s⁻¹, etc.) before substituting.
Up next: Motion Graphs — how displacement–time, velocity–time, and acceleration–time graphs connect, and how gradients and areas under the line reveal the motion.
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