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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

Learn9231Paper 3 · Further MechanicsLinear motion under a variable forceVariable forces and differential equations

Linear motion under a variable force · Subtopic 1 of 1

Motion Under a Variable Force

When force changes, Newton's law becomes a differential equation. Write $a=\frac{dv}{dt}$ if you want time, or $a=v\frac{dv}{dx}$ if you want distance, separate the variables and integrate with the initial conditions.

Cambridge 9231 Paper 3Edexcel YMA01 WME03Edexcel 9FM0 Option≈ 50 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Choose $\frac{dv}{dt}$ or $v\frac{dv}{dx}$ for acceleration.
Set up and solve separable equations for $F(t)$ , $F(v)$ , $F(x)$ .
Solve problems with resistance proportional to $v$ or $v^2$ .

Pause and think

A boat cuts its engine. Water resistance is proportional to speed. Does it ever stop?

Section 2 of 14

Core idea

Start hereCoasting to a stop

A motorboat cuts its engine. Water drag is proportional to speed, so as the boat slows, the drag weakens too. Does the boat ever stop? Newton's second law, written as a differential equation, gives a surprising answer.

The big idea

With variable forces, $F=ma$ is a differential equation. Use $a=\frac{dv}{dt}$ to connect $v$ and $t$ , or $a=v\frac{dv}{dx}$ to connect $v$ and $x$ . Separate, integrate, apply initial conditions.

1. Two forms of acceleration

Where

v\frac{dv}{dx}

comes from

$a=\frac{dv}{dt}$
Definition.
$=\frac{dv}{dx}\cdot\frac{dx}{dt}$
Chain rule.
$=v\frac{dv}{dx}$
$\frac{dx}{dt}=v$ .

Use this whenever the force depends on

x

, or the question asks about distance.

Worked exampleResistance proportional to v

A $2$ kg particle has only resistance $4v$ ; $v=10$ at $t=0$ . Find $v(t)$ .

Goal: $v$ in terms of $t$ .
Strategy: $\frac{dv}{dt}$ form, separate.

$2\frac{dv}{dt}=-4v$
Newton's second law; resistance opposes motion.
$\int\frac{dv}v=\int-2\,dt$
Separate.
$\ln v=-2t+c,\ c=\ln10$
Integrate; use $v(0)=10$ .

Answer

v=10e^{-2t}

✓ CheckNever reaches 0; total distance

\int_0^\infty10e^{-2t}dt=5

m, finite.

2. Force depending on position

Worked exampleInverse square repulsion

A $1$ kg particle is repelled from $O$ by $\frac8{x^2}$ N. At rest at $x=1$ . Find $v$ at $x=4$ .

Goal: $v$ .
Strategy: $v\frac{dv}{dx}$ form.

$v\frac{dv}{dx}=\frac8{x^2}$
$F=ma$ , $m=1$ .
$\tfrac12v^2=-\frac8x+c$
Integrate both sides.
$0=-8+c\Rightarrow c=8$
$v=0$ at $x=1$ .
$\tfrac12v^2=-2+8=6$
At $x=4$ .

Answer

v=\sqrt{12}\approx3.46

m s

^{-1}

✓ CheckWork done

\int_1^48x^{-2}dx=6

=

KE gained ✓.

Choosing the form

Drawing diagram…

Let the question decide the form.

Common misconceptionSigns Resistance acts against motion:

-kv

-kv^2

when

v>0

in the positive direction.

Key takeawayTerminal velocity Falling with resistance

kv^2

: acceleration is zero when

kv^2=mg

, giving

v_T=\sqrt{\frac{mg}k}

Pause and think

With resistance $kv^2$ (no other force), does the particle travel a finite distance?

Summary

$a=\frac{dv}{dt}=v\frac{dv}{dx}$ .
Separate and integrate.
Initial conditions fix constants.

Variable force: Force that changes with time, speed or position.
Terminal velocity: Constant speed when forces balance.

HistoryNewton again. Book II of Newton's Principia (1687) is devoted to motion in resisting media, with resistance proportional to

v

and to

v^2

Fun factTerminal velocity. A skydiver reaches about

55

m s

^{-1}

belly-down, when air resistance

kv^2

equals weight.

Section 3 of 14

Key formulae

Acceleration

a=\frac{dv}{dt}=v\frac{dv}{dx}=\frac{d^2x}{dt^2}

Newton

m\frac{dv}{dt}=F(t,v)\quad\text{or}\quad mv\frac{dv}{dx}=F(x,v)

TrickWhich form? Question mentions time →

\frac{dv}{dt}

. Mentions distance →

v\frac{dv}{dx}

. Force depends on

x

→ always

v\frac{dv}{dx}

Section 4 of 14

Worked examples

Example 1 · Resistance kv0 / 3 steps

2

kg particle has resistance

4v

and no other force;

v=10

t=0

. Find

v

at time

t

Example 2 · Force of x0 / 3 steps

1

kg particle feels force

\frac{8}{x^2}

away from

O

. At

x=1

v=0

. Find

v

x=4

Section 5 of 14

Explore the graph

Speed under resistance kv

Loading the interactive graph…

Try this

$x$ is time. The curve never reaches zero.
Increase $m$ : heavier bodies slow down more gradually.

$v=ue^{-kt/m}$ .

In the real worldRocket launches. Rocket thrust is roughly constant, but mass falls as fuel burns and drag depends on speed and height: engineers solve these equations numerically.

Section 6 of 14

Common mistakes

IncorrectUsing suvat with variable force

CorrectSolve the differential equation

suvat needs constant acceleration.

IncorrectResistance with the wrong sign

CorrectOpposes motion:

-kv

It acts against the velocity.

IncorrectForgetting the constant

CorrectUse initial conditions

Each integration needs one.

Force depends on x

v dv/dx enters the chat.

Section 7 of 14

Exam tip

Exam tipSeparate early. Get all

v

terms with

dv

and all

t

x

terms with

dt

dx

before integrating.

Section 8 of 14

Guided practice

SA-FM-VF-G013 marksL2 StandardNon-calculatorF(t), F(v)

2

kg particle starts from rest under force

F=6t

N. Find its speed at

t=2

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FM-VF-0013 marksL3 ExamCalculatorResistance

2

kg particle moves with resistance

4v

N only, starting at

10

m s

^{-1}

. Find its speed after

0.5

s (3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FM-VF-0023 marksL3 ExamCalculatorv dv/dx

1

kg particle is repelled from

O

by a force

\frac8{x^2}

N. It starts at rest at

x=1

. Find its speed at

x=4

(3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FM-VF-0033 marksL3 ExamNon-calculatorResistance

2

kg particle moves with resistance

4v

N only, starting at

10

m s

^{-1}

. Find the total distance it travels.

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-FM-VF-0042 marksL4 ChallengingCalculatorResistance

70

kg skydiver falls with air resistance

0.25v^2

N. Find the terminal speed in m s

^{-1}

(

g=9.8

, 3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$a=\frac{dv}{dt}=v\frac{dv}{dx}$ .
Separate and integrate.
Resistance opposes motion.

Section 12 of 14

Quick quiz

Q1 · SA-FM-VF-Q011 markL1 FoundationNon-calculatorv dv/dx

If the force depends on displacement

x

, write acceleration as

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 9231 AS & A Level Further Mathematics
Where: Paper 3 · Linear motion under a variable force
Expected: Use F = m dv/dt and F = mv dv/dx; set up and solve separable differential equations for forces depending on t, v or x.
Calculator: Permitted on all papers; exact answers are expected where the question asks.
Typical style: Typical: a particle moves against resistance kv²; find the distance travelled while its speed halves.
Open in: 9231 · Paper 3

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).