Calculus for Kinematics

Revise Calculus for Kinematics for Additional Mathematics 0606 (O Level) — revision notes and instant AI marking.

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Cambridge IGCSE Additional Maths

Calculus for Kinematics

Big idea: if you know an object's position over time, differentiating tells you how fast it's moving and how its speed is changing — and integrating does the reverse, rebuilding position and velocity from acceleration.

Displacement · Velocity · Acceleration Differentiation & Integration Travel Graphs

Summary — the whole chapter in one glance

Kinematics models motion in a straight line only — either horizontal (positive = right) or vertical (positive = up).
Displacement (s) is position relative to a fixed point and can be negative. Distance (d) is always positive — it's how far something has actually travelled.
Velocity (v) is the rate of change of displacement, and can be negative. Speed |v| is just velocity's size — always positive.
Acceleration (a) is the rate of change of velocity. Same sign as velocity → speeding up. Opposite sign → slowing down.
Differentiating goes s → v → a. Integrating goes the opposite way: a → v → s (and each integration step needs a constant, found from a given condition).
On a velocity-time graph: gradient = acceleration, area = displacement (net, with sign) or distance (total area, ignoring sign).
Displacement-time, velocity-time, and acceleration-time graphs are all linked — you can sketch one from another using gradients and areas.

1 · The Kinematics Toolkit — knowing your vocabulary cold

Before any calculus happens, you need to be fluent in five words: displacement, distance, velocity, speed, and acceleration. In everyday English these words get used loosely — "speed" and "velocity" basically mean the same thing to most people. In maths, they don't. Getting this distinction wrong is the single most common way students lose marks in this topic, so let's build it properly.

aDisplacement vs. Distance — the "can it go negative?" test

Picture a number line. You're standing at 0, and you can only move left or right along it (that's the "straight line" restriction — kinematics never deals with curved paths in this course).

Displacement is your position relative to that starting point, right now, in a straight line. If you're 5 m to the right, your displacement is +5. If you're 5 m to the left, it's −5. Crucially, displacement doesn't care about the wiggly path you took to get there — only where you ended up relative to the fixed point.

Distance is the total length of the path you actually walked, and it only ever adds up — it can never be negative and it can never "cancel out." Walk 3 m right then 3 m left, and your displacement is back to 0, but you've covered a distance of 6 m.

Analogy

Think of a bus route that starts and ends at the same depot. When the bus gets back, its displacement is zero — it's exactly where it started. But the odometer doesn't reset — the distance travelled is the full length of the route. Same journey, two totally different numbers, because they're answering two different questions ("where are you?" vs. "how far did you go?").

Practice Question 1

A particle starts at the origin, moves 8 m to the right, then moves 3 m back to the left. State (a) its final displacement, and (b) the total distance it has travelled.

bVelocity vs. Speed — the same relationship, one level up

Velocity Speed

v = 4 |v| = 4 v = −6 |v| = 6

at rest stationary

c

Velocity sign

negative

d

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Also in the full note

2 · Differentiation for Kinematics — going "up" the chain
3 · Integration for Kinematics — going back "down" the chain
4 · Sketching Travel Graphs — seeing the whole story at once
What to Memorise
Concepts Checklist
Exam Tips — where marks are won and lost
cAcceleration — the trickiest one, because sign alone isn't enough
dReading a velocity-time graph like a pro

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Calculus for Kinematics