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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

LearnYMA01WME03 · Mechanics 3Further kinematicsKinematics with calculus and vectors

Further kinematics · Subtopic 1 of 1

Further Kinematics

Differentiate position to get velocity and acceleration; integrate the other way, using initial conditions. With vectors, do it component by component. When acceleration is given in terms of $x$ or $v$ , use $a=v\frac{dv}{dx}$ or $a=\frac{dv}{dt}$ and separate variables.

Edexcel YMA01 WME03Edexcel 9FM0 Option≈ 40 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Use calculus with vector position, velocity and acceleration.
Solve straight-line motion with $a(v)$ or $a(x)$ .
Interpret direction and speed from velocity vectors.

Pause and think

A drone's position is $\mathbf r=t^2\mathbf i+(4t-t^2)\mathbf j$ . When is it moving horizontally?

Section 2 of 14

Core idea

Start hereTracking a drone

A drone's position is logged as a vector at every instant. From that one formula you can find its velocity, speed, acceleration and when it flies level. Calculus works on each component separately.

The big idea

$\mathbf v=\frac{d\mathbf r}{dt}$ and $\mathbf a=\frac{d\mathbf v}{dt}$ , component by component; integrate the other way with a constant vector. In one dimension with $a(x)$ , use $v\frac{dv}{dx}$ .

1. Vector kinematics

Worked exampleLevel flight

$\mathbf r=t^2\mathbf i+(4t-t^2)\mathbf j$ . Find when the drone moves parallel to $\mathbf i$ and its speed at $t=1$ .

Goal: $t$ and speed.
Strategy: Differentiate; set the $\mathbf j$ -component to 0.

$\mathbf v=2t\mathbf i+(4-2t)\mathbf j$
Differentiate each component.
$4-2t=0\Rightarrow t=2$
No vertical velocity.
$\mathbf v(1)=2\mathbf i+2\mathbf j$
Substitute.
$|\mathbf v|=\sqrt{8}=2\sqrt2$
Magnitude.

Answer

t=2

; speed

2\sqrt2

t=1

✓ CheckAt

t=2

y=4

is the highest point of

y=4t-t^2

✓.

Integrating with a constant vector

$\mathbf a=6t\mathbf i-2\mathbf j$
Suppose this is given, with $\mathbf v(0)=\mathbf i+3\mathbf j$ .
$\mathbf v=3t^2\mathbf i-2t\mathbf j+\mathbf c$
Integrate each component.
$\mathbf c=\mathbf i+3\mathbf j$
At $t=0$ .
$\mathbf v=(3t^2+1)\mathbf i+(3-2t)\mathbf j$
Combine.

2. Acceleration depending on x

Worked exampleOscillating particle

$a=-4x$ ; $v=6$ at $x=0$ . Find $v$ at $x=1$ and the greatest distance from $O$ .

Goal: $v(1)$ , $x_{\max}$ .
Strategy: $v\frac{dv}{dx}=-4x$ .

$\tfrac12v^2=-2x^2+c$
Integrate.
$c=18$
$v=6$ at $x=0$ .
$v^2=36-4x^2$
Rearrange.
$v(1)=\sqrt{32};\ v=0\Rightarrow x=3$
Substitute.

Answer

v=4\sqrt2

x=1

; greatest distance

3

✓ CheckThis is SHM with

\omega=2

, amplitude

\frac{6}{2}=3

✓.

Kinematics chain

Drawing diagram…

Differentiate down, integrate up.

Common misconceptionSpeed is not a sum

|2\mathbf i+2\mathbf j|=2\sqrt2

, not

4

TrickDirections "Moving north-east" means equal positive

\mathbf i

and

\mathbf j

components of velocity.

Pause and think

When is the drone's velocity perpendicular to its acceleration?

Summary

Vectors: componentwise calculus.
Constant vectors from initial conditions.
$a(x)$ : $v\frac{dv}{dx}$ .

Velocity vector: Rate of change of position vector.
Speed: Magnitude of velocity.

HistoryGibbs and Heaviside. Vector notation as we use it (

\mathbf i,\mathbf j,\mathbf k

, dot and cross products) was developed by Gibbs and Heaviside in the 1880s to simplify physics.

Fun factGPS. Your phone estimates velocity by differentiating position fixes from satellites, then smooths the noise.

Section 3 of 14

Key formulae

Vectors

\mathbf v=\frac{d\mathbf r}{dt},\quad\mathbf a=\frac{d\mathbf v}{dt}

Integrating

\mathbf r=\int\mathbf v\,dt+\mathbf c

Acceleration in x

a=v\frac{dv}{dx}=\frac{d}{dx}\left(\tfrac12v^2\right)

TrickSpeed vs velocity. Speed is

|\mathbf v|

. For "parallel to

\mathbf i+\mathbf j

", set the components equal; for "due north", set the

\mathbf i

-component to zero.

Section 4 of 14

Worked examples

Example 1 · Vectors0 / 2 steps

\mathbf r=t^2\mathbf i+(4t-t^2)\mathbf j

. Find the speed at

t=1

Example 2 · a(x)0 / 2 steps

a=-4x

v=6

x=0

. Find

v

x=1

Section 5 of 14

Explore the graph

Path of the drone

Loading the interactive graph…

Try this

At $s=2$ the path is momentarily horizontal: highest point.
The points spread out: speed increases with time.

$\mathbf r=t^2\mathbf i+(4t-t^2)\mathbf j$ .

In the real worldAir traffic control. Controllers predict aircraft positions from velocity vectors to keep planes separated.

Section 6 of 14

Common mistakes

IncorrectForgetting the vector constant

Correct

\mathbf c=c_1\mathbf i+c_2\mathbf j

Each component has its own constant.

IncorrectSpeed as sum of components

Correct

\sqrt{v_x^2+v_y^2}

Magnitude, not sum.

Incorrect

a=\frac{dv}{dx}

Correct

v\frac{dv}{dx}

Chain rule:

\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}

Differentiate r

Each component, one at a time, no shortcuts.

Section 7 of 14

Exam tip

Exam tipCheck units. Velocity in m s

^{-1}

, acceleration in m s

^{-2}

: a quick unit check spots a missed derivative.

Section 8 of 14

Guided practice

SA-FM-FK-G012 marksL1 FoundationNon-calculatorVectors

\mathbf r=t^2\mathbf i+(4t-t^2)\mathbf j

. Find the time when the particle moves parallel to

\mathbf i

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FM-FK-0012 marksL2 StandardCalculatorVectors

For the same particle, find the speed at

t=1

(3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FM-FK-0021 markL2 StandardNon-calculatorVectors

For the same particle, find the

\mathbf j

-component of acceleration.

Salam Academy Original Practice · Proposed mark scheme

SA-FM-FK-0033 marksL3 ExamCalculatora(v), a(x)

A particle has

a=-4x

and

v=6

x=0

. Find

v

x=1

(3 s.f.).

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-FK-0042 marksL4 ChallengingNon-calculatora(v), a(x)

For the same particle, find the greatest distance from

O

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Differentiate down, integrate up.
Vectors: componentwise.
$a(x)$ : use $v\frac{dv}{dx}$ .

Section 12 of 14

Quick quiz

Q1 · SA-FM-FK-Q011 markL1 FoundationNon-calculatorVectors

To find position from velocity vector

\mathbf v(t)

you

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: A level Further Mathematics 9FM0
Where: Further Mechanics 2 · Further kinematics
Expected: Motion in a straight line with acceleration depending on time, velocity or displacement; vector kinematics in two and three dimensions using calculus.
Calculator: Permitted; show enough working to earn the method marks.
Typical style: Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
Open in: YMA01 · WME03 9FM0 · Option

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).