The big idea: When the acceleration is constant, four equations link the five motion quantities.
These five letters are nicknamed suvat — s displacement, u start speed, v end speed, a acceleration, t time.
You're usually given three, and you pick the equation that finds a fourth.
| Quantity | What it means | |
|---|---|---|
| s | displacement | how far it goes (m) |
| u | initial velocity | speed at the start (m s⁻¹) |
| v | final velocity | speed at the end (m s⁻¹) |
| a | acceleration | how fast the velocity changes (m s⁻²) |
| t | time | how long it takes (s) |
Spot it: Suvat only works while the acceleration stays constant (a straight v–t line). If the acceleration changes, split the motion into constant-a stages and do each one separately.
All four are given in the data booklet — you don't memorise them, you choose one. Tap any to see its booklet badge:
- displacement (m)
- initial velocity (m s⁻¹)
- final velocity (m s⁻¹)
- acceleration (m s⁻²)
- time (s)
How to choose the right one: List your three knowns and the one unknown you want.
Then pick the equation that contains those four letters and leaves out the fifth.
Example: you know u, a, t and want s → the equation missing v is s = ut + ½at².
IB-style question — find the final speed
A sled starts at 4.0 m s⁻¹ and accelerates uniformly at 2.0 m s⁻² for 5.0 s. Find its final speed.
Solution
- Knowns: u = 4.0, a = 2.0, t = 5.0; want v. No s, so use the given formula:
- Put in the numbers:
- Work it out — keep the unit:
Final answer
final speed v = 14 m s⁻¹.
IB-style question — find the displacement
The same sled (u = 4.0 m s⁻¹, a = 2.0 m s⁻², t = 5.0 s): how far does it travel in those 5.0 s?
Solution
- Knowns: u, a, t; want s. No v, so use the given formula:
- Put in the numbers:
- Work it out — keep the unit:
Final answer
displacement s = 45 m.
See how examiners mark answers
Access past paper questions with model answers. Learn exactly what earns marks and what doesn't.
How this is tested: Suvat is the workhorse of motion questions.
- Paper 1A: a one-mark calculation — often a stopping distance (a car decelerating to rest). - Paper 2: longer problems, e.g. the time for a particle to cross a gap from rest.
Classic trap: time isn't given for a stopping distance — that's the signal to use v² = u² + 2as (the equation with no t), not v = u + at.
Stopping = ends at rest: 'Comes to rest' / 'stops' means the final velocity v = 0. 'Deceleration of 5 m s⁻²' means the acceleration is negative: a = −5 m s⁻².
IB-style question — stopping distance
A car travelling at 20 m s⁻¹ brakes with a constant deceleration of 5.0 m s⁻². Find the distance it travels before stopping.
Solution
- Knowns: u = 20, v = 0 (it stops), a = −5.0 (decelerating); want s. No t, so use the given formula:
- Put in the numbers:
- Tidy up:
- Solve for s — keep the unit:
Final answer
stopping distance s = 40 m.