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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9709Paper 4 · MechanicsKinematics of motion in a straight lineConstant acceleration (suvat)

Kinematics of motion in a straight line · Subtopic 1 of 4

Constant Acceleration: the suvat Formulae

Five quantities, five formulae. Choose a positive direction, list what you know, and pick the formula that leaves out the one quantity you neither know nor need.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Know what $s$ , $u$ , $v$ , $a$ and $t$ stand for, and which are vectors.
Choose the right suvat formula and handle signs with a chosen positive direction.
Solve problems that give a quadratic in $t$ , and interpret both roots.
Break a journey into stages, and compare two moving objects.

Pause and think

A car speeds up steadily from $10$ to $30\ \text{m s}^{-1}$ in $8$ s. Without any formula, what was its average speed, and how far did it go?

Section 2 of 14

Core idea

Start hereBraking distance

A car at $25$ m/s brakes and stops in $50$ m. What was its deceleration, and how long did it take? With constant acceleration, five quantities ( $s,u,v,a,t$ ) are tied by five equations. Know any three, find the other two.

The big idea

For constant acceleration: $v=u+at$ , $s=\frac12(u+v)t$ , $s=ut+\frac12at^2$ , $s=vt-\frac12at^2$ , $v^2=u^2+2as$ . Each comes from the velocity–time graph.

1. Deriving the equations

A velocity–time graph: the area is the displacement

t = 0.5

u=4

a=2

. The gradient is the acceleration. The shaded trapezium has area

\tfrac12(u+v)t

: that is where

s=\tfrac12(u+v)t

comes from.

s,\ u,\ v,\ a,\ t

$s$: displacement (m).
$u$: initial velocity.
$v$: final velocity.
$a$: acceleration (constant).
$t$: time.

From the v–t graph

$a=\frac{v-u}t\Rightarrow v=u+at$
Gradient is acceleration.
$s=\tfrac12(u+v)t$
Area of the trapezium.
$s=\tfrac12(u+u+at)t=ut+\tfrac12at^2$
Substitute $v$ .
$s=\tfrac12(v-at+v)t=vt-\tfrac12at^2$
Substitute $u=v-at$ .
$s=\tfrac12(u+v)\cdot\frac{v-u}a\Rightarrow v^2=u^2+2as$
Eliminate $t$ ; difference of squares.

2. Using them

Worked exampleThe braking car

$u=25$ , $v=0$ , $s=50$ . Find $a$ and $t$ .

Goal: $a$ , $t$ .
Strategy: Pick the equation without $t$ first.

$0=625+2a(50)\Rightarrow a=-6.25$
$v^2=u^2+2as$ .
$0=25-6.25t\Rightarrow t=4$
$v=u+at$ .

Answer

a=-6.25

m s

^{-2}

;

t=4

✓ Check

s=\frac12(25+0)(4)=50

✓.

Worked exampleCatching up

Car A passes a point at a steady 20 m/s. At that instant car B starts from rest there with $a=4$ . When and where does B catch A?

Goal: Time, place.
Strategy: Equal displacements.

$s_A=20t,\quad s_B=2t^2$
Constant speed; from rest.
$2t^2=20t\Rightarrow t=10$
$t\neq0$ .

Answer

t=10

200

m on.

✓ CheckB's speed then

40

m/s, twice A's: average speed equal ✓.

Common misconceptionUnits and signs Convert km/h ÷ 3.6. Decide a positive direction; deceleration means negative

a

Edge caseOnly constant acceleration If

a

varies with time, use calculus instead.

Pause and think

From rest with $a=3$ , distance in the 4th second?

Summary

Five equations from the graph.
Choose by knowns.

Displacement: Distance in a direction.

u = 4 m s⁻¹, a = 2 m s⁻², for 5 s

The gradient is the acceleration and the area is the displacement:

s=\tfrac12(u+v)t=\tfrac12(4+14)\times5=45

HistoryGalileo rolls balls down a ramp. Free fall was too quick to time in the early 1600s, so Galileo Galilei slowed it down by rolling balls down gentle ramps, timing them with a water clock. In Two New Sciences (1638) he showed that the distance travelled from rest grows with the square of the time, which is

s=\tfrac12at^2

Fun factThe odd-number rule. From rest with constant acceleration, the distances covered in the 1st, 2nd, 3rd, 4th second are in the ratio

1:3:5:7

. Add them up and you get the square numbers

1, 4, 9, 16

: that is exactly why

s\propto t^2

. Galileo spotted this pattern on his ramps.

Section 3 of 14

Key formulae

The five suvat formulae (given in the formula list)

\begin{gathered}v=u+at \quad(\text{no } s) \\[3pt] s=ut+\tfrac12at^2 \quad(\text{no } v) \\[3pt] s=\tfrac12(u+v)t \quad(\text{no } a) \\[3pt] v^2=u^2+2as \quad(\text{no } t) \\[3pt] s=vt-\tfrac12at^2 \quad(\text{no } u)\end{gathered}

Unit conversion

1\ \text{km h}^{-1}=\frac{1000}{3600}\ \text{m s}^{-1}=\frac{1}{3.6}\ \text{m s}^{-1}

TrickFind the missing letter. Write

s=\ ,\ u=\ ,\ v=\ ,\ a=\ ,\ t=\

down the margin. Fill in the three you know and put "?" by the one you want. The fifth letter is the one you do not care about, and exactly one formula leaves it out. That is your formula.

Section 4 of 14

Worked examples

Example 1 · Choosing the formula0 / 3 steps

A cyclist accelerates uniformly from

4\ \text{m s}^{-1}

10\ \text{m s}^{-1}

over a distance of

42

m. Find the acceleration and the time taken.

Example 2 · Two roots0 / 2 steps

A particle passes

O

with velocity

12\ \text{m s}^{-1}

and has constant acceleration

-2\ \text{m s}^{-2}

. Find the times when it is

27

m from

O

on the positive side.

Example 3 · Catching up0 / 2 steps

Car

A

passes a point

O

at a constant

15\ \text{m s}^{-1}

. At that moment car

B

starts from rest at

O

with acceleration

2\ \text{m s}^{-2}

in the same direction. How long does

B

take to catch

A

Section 5 of 14

Explore the graph

Displacement and velocity with constant acceleration

Loading the interactive graph…

Try this

The horizontal axis is time $t$ . The straight line is $v=u+at$ ; the curve is $s=ut+\tfrac12at^2$ .
Set $a=0$ . Why does $s$ become a straight line?
Set $u=12$ and $a=-2$ . Where does the velocity line cross the axis? What is the curve doing at that moment?

The velocity line is the gradient of the displacement curve: where $v=0$ , the curve has a turning point.

In the real worldWhy speeding is so dangerous. With

v=0

, the formula

v^2=u^2+2as

gives a braking distance

s=\dfrac{u^2}{2|a|}

. It is proportional to the square of the speed: drive

50\%

faster and you need

2.25

times the distance to stop. That is why the braking distances in driving-test charts grow so quickly.

Section 6 of 14

Common mistakes

IncorrectDeceleration

3

used as

a=+3

Correct

a=-3

Acceleration against the direction of motion is negative.

Incorrect

72

km/h used as

u=72

Correct

u=20\ \text{m s}^{-1}

Divide km/h by

3.6

before using suvat.

IncorrectRejecting the second root of a quadratic in

t

without thinking

CorrectInterpret both roots

A particle can pass the same point twice, once each way.

IncorrectUsing suvat when

a

depends on

t

CorrectUse calculus

suvat is only valid for constant acceleration.

Expectation

Pick a suvat equation at random

Reality

List $s,u,v,a,t$ , cross out the one you don't need, pick the equation without it

Section 7 of 14

Exam tip

Exam tipDraw the journey. For two stages or two vehicles, sketch a line with the start, the change-over point and the end marked, and write the suvat values for each stage underneath. The final velocity of stage 1 is the initial velocity of stage 2, which is where most marks are lost.

Section 8 of 14

Guided practice

SA-9709-P4-KM-G013 marksL1 FoundationCalculatorsuvat

A particle has initial velocity

4\ \text{m s}^{-1}

and constant acceleration

2\ \text{m s}^{-2}

. Find its velocity after

5

s and its displacement in that time.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P4-KM1-0013 marksL1 FoundationCalculatorsuvat

A particle starts from rest and moves with constant acceleration

3\ \text{m s}^{-2}

. Find its velocity and its displacement after

4

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM1-0023 marksL1 FoundationCalculatorsuvat

A car accelerates uniformly from

5\ \text{m s}^{-1}

17\ \text{m s}^{-1}

4

s. Find its acceleration and the distance travelled.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM-0013 marksL1 FoundationCalculatorsuvat

A car travelling at

20\ \text{m s}^{-1}

brakes uniformly and stops in

50

m. Find the deceleration and the time taken.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM-0103 marksL2 StandardCalculatorsuvat

A car travelling at

72

km/h brakes with deceleration

5\ \text{m s}^{-2}

. Find the stopping distance.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM1-0034 marksL2 StandardCalculatorsuvat

A particle passes

O

with velocity

12\ \text{m s}^{-1}

and moves with constant acceleration

-2\ \text{m s}^{-2}

. Find both times at which its displacement from

O

27

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM1-0064 marksL2 StandardCalculatorMulti-stage

A particle starts from rest and accelerates at

1.5\ \text{m s}^{-2}

for

8

s. It then moves at constant speed for

20

s. Find its greatest speed and the total distance travelled.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM-0093 marksL2 StandardCalculatorMulti-stage

A particle starts from rest and accelerates at

2\ \text{m s}^{-2}

for

5

s. It then decelerates uniformly to rest in

4

s. Find the total distance travelled.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-KM1-0075 marksL3 ExamCalculatorMulti-stage

Particle

P

passes point

O

with velocity

3\ \text{m s}^{-1}

and constant acceleration

2\ \text{m s}^{-2}

. At the same instant particle

Q

starts from rest at

O

with constant acceleration

3\ \text{m s}^{-2}

in the same direction. Find the time when

Q

catches

P

and the distance from

O

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P4-KM-0124 marksL3 ExamCalculatorMulti-stage

Car A starts from rest at a junction with acceleration

2\ \text{m s}^{-2}

just as car B passes it at a constant

10\ \text{m s}^{-1}

. Find the time when A catches B, and the distance from the junction.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

suvat only works when acceleration is constant.
Pick a positive direction; $s$ , $u$ , $v$ and $a$ take signs from it.
Choose the formula that leaves out the quantity you neither know nor need.
A quadratic in $t$ can have two meaningful roots.
In multi-stage problems, the end of one stage is the start of the next.

Section 12 of 14

Quick quiz

Q1 · SA-9709-P4-KM1-0091 markL1 FoundationCalculatorsuvat

You know

u

a

and

s

and want

v

. The time is not needed. Which formula should you use?

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P4-KM1-0052 marksL2 StandardCalculatorsuvat

A particle with constant acceleration

3\ \text{m s}^{-2}

has velocity

20\ \text{m s}^{-1}

at the end of a

4

s interval. Find the displacement during that interval.

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P4-KM1-0083 marksL2 StandardCalculatorsuvat

A train slows uniformly from

30\ \text{m s}^{-1}

10\ \text{m s}^{-1}

over a distance of

200

m. Find the deceleration and the time taken.

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 9709 AS & A Level Mathematics
Where: Paper 4, Mechanics · 4.2 Kinematics of motion in a straight line
Expected: Use the formulae for motion in a straight line with constant acceleration, understanding displacement, velocity and acceleration as vectors with a sign.
Calculator: Permitted. Use g = 10 m s⁻² unless told otherwise.
Typical style: Usually a 4–6 mark question with two stages, or one vehicle catching another.
Open in: 9709 · Paper 4

Section 14 of 14

Past paper questions

5 questions (35 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

9709/42 · Oct/Nov 2025Question 13 marks
Distance and acceleration from a displacement–time graph.
Question paper, page 3 Mark scheme
9709/42 · May/June 2025Question 411 marks
Sketch a velocity–time graph and use constant acceleration to find a time from a total distance.
Question paper, page 6 Mark scheme
9709/42 · Feb/March 2025Question 25 marks
Motion in stages with constant acceleration and deceleration.
Question paper, page 4 Mark scheme
9709/42 · Oct/Nov 2023Question 47 marks
Coefficient of friction, then the distance covered in one particular second.
Question paper, page 5 Mark scheme
9709/42 · May/June 2022Question 49 marks
Two particles with constant acceleration on parallel tracks, and displacement–time sketches.
Question paper, page 6 Mark scheme

All 9709 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9709, 2026–2027 (official specification).