SUVAT equations: choosing the right equation
Constant acceleration problems always come down to the same five equations. The skill is not memorising them all — it is picking the one that uses the variables you know and the variable you want.
Work through it in three passes
Know the variables
Learn what s, u, v, a and t mean and how signs are handled.
Use the missing variable
Pick the equation that leaves out the variable you are not told and do not need.
Practise the choice
Use the interactive picker and worked examples until the right equation feels obvious.
- Identify the five SUVAT variables in a problem.
- Choose the correct SUVAT equation by spotting the missing variable.
- Apply a consistent sign convention, especially for vertical motion under gravity.
- Solve standard constant-acceleration problems in one dimension.
What each letter means
SUVAT only works for motion with constant acceleration in a straight line. If acceleration is changing, these equations do not apply.
| Variable | Name | Unit | Notes |
|---|---|---|---|
| s | Displacement | m | Straight-line distance with direction from a chosen origin. |
| u | Initial velocity | m s-1 | Velocity at the start of the time interval. |
| v | Final velocity | m s-1 | Velocity at the end of the time interval. |
| a | Acceleration | m s-2 | Constant throughout the motion. |
| t | Time | s | Duration of the interval. |
Choose a positive direction and stick to it. If up is positive, a ball thrown upwards has u > 0 and a = -g. A car braking has a < 0 if its velocity direction is positive.
The five constant acceleration equations
Every equation contains four of the five variables. That means for each problem there is one variable you do not use — and that tells you which equation to pick.
Missing s
Use when displacement is not mentioned and you do not need it.
Missing a
Use when acceleration is unknown, often with average velocity.
Missing v
Common for finding displacement from rest or known initial velocity.
Missing u
Use when you know final velocity rather than initial velocity.
Missing t
Use when time is not given and not asked for.
Do not memorise by equation number. Read the problem, list the variables you know and the one you want, then find the equation that contains exactly those four variables. The fifth variable — the one you do not have and do not need — is your guide.
The missing-variable method
The fastest way to choose is to ask: "which variable am I not given and not asked to find?" Then use the equation that leaves it out.
| If you are not using... | Use this equation | Typical use |
|---|---|---|
| s | v = u + at | Velocity after a given time. |
| a | s = ½(u + v)t | Average speed problems, motion graphs. |
| v | s = ut + ½at² | Distance travelled from rest. |
| u | s = vt - ½at² | Distance needed to brake from a known final speed. |
| t | v² = u² + 2as | Speed after a known distance. |
Pick the right equation
For each problem, identify the missing variable and select the SUVAT equation you would use. Then check your answer.
Worked examples with equation choice
Example 1: car accelerating from rest
A car starts from rest and accelerates uniformly at 2.5 m s-2 for 8.0 s. How far does it travel?
Knowns: u = 0, a = 2.5 m s-2, t = 8.0 s. Unknown: s.
The final velocity v is not mentioned, so use the equation missing v:
Example 2: ball thrown vertically upwards
A ball is thrown upwards at 12 m s-1. Calculate the maximum height it reaches. Take g = 9.81 m s-2.
Take upwards as positive. At maximum height, v = 0.
Knowns: u = 12 m s-1, v = 0, a = -9.81 m s-2. Unknown: s.
Time is not given and not needed, so use the equation missing t:
Example 3: braking distance
A car travelling at 25 m s-1 brakes with uniform deceleration of 4.0 m s-2. How far does it travel before stopping?
Take the direction of travel as positive, so a = -4.0 m s-2.
Knowns: u = 25 m s-1, v = 0, a = -4.0 m s-2. Unknown: s.
Time is not needed, so again use the equation missing t:
Example 4: velocity after a known distance
An electron accelerates from 2.0 × 106 m s-1 to 4.0 × 106 m s-1 over 0.030 m. Find the acceleration.
Knowns: u = 2.0 × 106 m s-1, v = 4.0 × 106 m s-1, s = 0.030 m. Unknown: a.
Time is not given, so use the equation missing t:
Common mistakes and quick fixes
Forgetting g is downwards
If up is positive, acceleration is -g. If down is positive, a falling object has +g.
Using the wrong equation
Students often reach for s = ut + ½at² by habit. Check which variable is missing first.
Confusing speed and velocity
Speed is a scalar. Velocity includes direction, so a ball caught at the same height as it was thrown has v = -u.
Splitting the motion badly
For vertical motion you can use the whole trip from throw to landing. You do not always need to split at the top.
Self-check questions
1. A stone is dropped from a cliff. After 3.0 s it hits the sea below. How high is the cliff?
Method: u = 0, a = g = 9.81 m s-2, t = 3.0 s. Missing v, so use s = ut + ½at².
s = 0 + ½ × 9.81 × 3.0² = 44 m
2. A cyclist accelerates from 4.0 m s-1 to 10 m s-1 over 28 m. Find the acceleration.
Method: u = 4.0, v = 10, s = 28. Missing t, so use v² = u² + 2as.
100 = 16 + 56a → a = 1.5 m s-2
3. A ball is thrown upwards at 8.0 m s-1. How long does it take to return to the thrower's hand?
Method: Take up as positive. When it returns, s = 0. Use s = ut + ½at² with a = -9.81.
0 = 8.0t - 4.905t² → t = 0 or t = 1.63 s
4. A train travelling at 30 m s-1 decelerates at 1.2 m s-2. How long until it stops?
Method: u = 30, v = 0, a = -1.2. Missing s, so use v = u + at.
0 = 30 - 1.2t → t = 25 s
Final checklist for SUVAT problems
- Check the motion has constant acceleration.
- Choose a positive direction and write it down.
- List s, u, v, a, t with correct signs.
- Identify the variable you need to find.
- Pick the equation that contains your three knowns and one unknown.
- Rearrange before substituting numbers.
- Check the answer has the right sign, unit and sensible magnitude.