Here are four little equations that will get you through most motion problems in the whole course. They only work in one situation — when the acceleration stays constant — but that covers a huge amount, from a braking car to a falling apple. The trick isn’t memorising them (they’re in your data booklet). It’s learning to spot which one to reach for. Let’s take it slowly.
📘 What you need to know
There are four equations of motion — they only work when acceleration is constant
They link five quantities: s, u, v, a, t — nicknamed SUVAT
Each equation uses four of the five, so pick the one missing the letter you don’t have
“Starts from rest” means u = 0; “falling under gravity” means a = g = 9.81 m s−2
They come straight from a velocity–time graph — you can derive them
Use a simple 3-step method: list what you know, pick the equation, plug in
Meet the five letters: SUVAT
Every constant-acceleration problem is really about five quantities. Learn these five letters and you’re halfway there.
🔤 The five SUVAT quantities
s — displacement (how far it moves), in metres
u — the starting (initial) velocity
v — the final velocity
a — the acceleration (kept constant)
t — the time interval, in seconds
The four equations
Here they are. Notice the clever part: each equation is missing one of the five letters. That’s exactly how you choose which to use — find the equation that skips the quantity you neither know nor want.
The four equations of motion — each one leaves out a different SUVAT letter, which is how you pick the right one.
Don’t panic about memorising these — they’re printed in your data booklet, so you’ll always have them in the exam. What the booklet won’t do is choose the right one for you. So the skill to practise isn’t recall, it’s matching: “which quantity is missing from my problem?” Find the equation that also leaves that letter out, and you’ve found your equation.
Where do they come from?
These equations aren’t magic — they fall straight out of a velocity–time graph for an object with constant acceleration. On such a graph the line is straight, and two facts do all the work: the gradient is the acceleration, and the area underneath is the displacement.
The area under a velocity–time graph is the displacement: a rectangle (ut) plus a triangle (½(v−u)t). Add them and you get s = ut + ½at2.
Reading straight off that graph: the slope of the line is (v − u) ÷ t, and the slope is the acceleration — rearrange and you get v = u + at. The area is a rectangle plus a triangle, which adds up to s = ut + ½at2. The other two equations come from combining these first two. So they’re all just the graph, written in symbols.
Spotting the hidden clues
Exam questions rarely hand you all the numbers on a plate. Instead they hide one or two inside ordinary phrases. Train your eye to translate them.
🔎 Key phrases — and what they secretly tell you
“Starts from rest” → the initial velocity u = 0 (and if no starting speed is mentioned, you can usually assume this).
“Falling due to gravity” → the acceleration a = g = 9.81 m s−2, pointing down.
“Comes to rest” / “stops” → the final velocity v = 0.
“Constant / uniform acceleration” → the green light to use these equations at all.
Quick recap: five letters (SUVAT), four equations, each missing one letter. Read the question for hidden values like u = 0 or a = 9.81, then pick the equation that skips the letter you don’t have.
How to solve any SUVAT problem
Same three steps every single time. Do them in order and even a scary-looking question becomes tidy.
1. List SUVAT
→ what do I know? →
2. Pick the equation
→ plug & solve →
3. Answer
In Step 1, write out the five letters and fill in every value you’re given — including the hidden ones. In Step 2, look at which letter you want and which you’re missing, then choose the equation containing your wanted letter but not the missing one. In Step 3, put the numbers in (converted to SI units) and rearrange to get your answer.
WE 1
A cyclist starts at 4 m s−1 and accelerates steadily at 3 m s−2 for 5 s. Find their final velocity.
Step 1 — list what we know
u = 4, a = 3, t = 5, v = ? (s not needed)
Step 2 — pick the equation with no s
v = u + at
Step 3 — plug inv = 4 + (3 × 5) = 4 + 15v = 19 m s⁻¹
WE 2
A car travelling at 20 m s−1 brakes at 5 m s−2 until it stops. How far does it travel while braking?
Step 1 — list what we know
u = 20, v = 0 (“stops”), a = −5 (braking), s = ?
t isn’t given and isn’t wanted — so use the equation with no t.Step 2 — pick the equation with no t
v² = u² + 2as
Step 3 — plug in and rearrange
0 = 20² + 2(−5)s
0 = 400 − 10s → s = 400 ÷ 10s = 40 m
WE 3
A stone is dropped from rest and falls for 0.9 s. Taking g = 9.81 m s−2, how far does it fall?
Step 1 — spot the hidden clues
“dropped from rest” → u = 0
“falls” → a = g = 9.81
t = 0.9, s = ?
Step 2 — pick the equation with no v
s = ut + ½at²
Step 3 — plug in (u = 0 kills the first term)s = 0 + ½ × 9.81 × 0.9² = ½ × 9.81 × 0.81s = 3.97 m
💡 Top tips
Always write your SUVAT list first — it makes the right equation almost jump out at you.
Pick one direction as positive and stick to it — if up is positive, a downward acceleration is negative all the way through.
Convert to SI units before you plug in — km h−1 or cm will wreck your answer.
You won’t have to memorise them — they’re in the data booklet, so practise choosing, not reciting.
⚠ Common mistakes
Using these equations when the acceleration isn’t constant — they simply don’t apply then
Getting signs muddled — mixing a positive velocity with a negative acceleration without a consistent direction
Forgetting the hidden values, like u = 0 for “from rest” or v = 0 for “stops”
Leaving units un-converted — plugging in km h−1 instead of m s−1
Up next: Motion Graphs — reading displacement–time, velocity–time and acceleration–time graphs, where gradients and areas tell the whole story of a journey (and quietly hand you the SUVAT quantities too).
Want this to actually click before the exam?
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