Historical Context & Motivation
Humans have always wanted to predict how things move — whether tracking planets, launching cannonballs, or designing engines. For centuries, mathematicians struggled with questions like: if you know where an object is at every moment, how do you find its speed? And if you know its speed, how do you find how far it has traveled? These questions sit at the heart of kinematics, the study of motion using mathematics rather than forces. The development of calculus in the 17th century finally gave us the tools to answer them precisely.
The central question this topic addresses is elegant in its simplicity: given one piece of information about a changing quantity — its position, its rate of change, or its accumulated total — how can calculus help us find the others? In SL 5.7, you will learn to move fluently between displacement, velocity, and acceleration using differentiation and integration, and then apply the same reasoning to any real-world rate or accumulation problem.
Core Principles & Definitions
Before diving into calculations, you need a solid grasp of the key concepts and how they connect. In kinematics, we describe the motion of an object along a straight line using three related functions of time. Each one can be obtained from the next by either differentiating or integrating. This chain of relationships is the backbone of SL 5.7.
Displacement s(t)
Velocity v(t)
Acceleration a(t)
Differentiation Links "Down"
Integration Links "Up"
The Kinematics Chain — A Visual Explanation
The diagram below shows the fundamental relationship chain in kinematics. Notice how differentiation moves you from left to right (position → velocity → acceleration), while integration moves you from right to left. This visual will be your roadmap for every kinematics problem you encounter in SL 5.7.
This diagram is the single most important visual in SL 5.7. Whenever you face a kinematics problem, ask yourself: "Where am I on the chain, and which direction do I need to go?" If the question gives you displacement and asks for velocity, you differentiate. If it gives you acceleration and asks for displacement, you integrate twice, using initial conditions to find each constant of integration.
Mathematical Framework
Now let's formalise the relationships from the kinematics chain into equations you can apply. In SL 5.7 you are expected to work with polynomial, exponential, and trigonometric functions. The key operations are straightforward: differentiation gives rates, and definite integration gives accumulated change.
Beyond kinematics, the same framework applies to any rate/accumulation context. If R(t) is the rate at which water flows into a tank (in litres per minute), then ∫R(t) dt over an interval gives the total volume of water added. If C(x) represents a marginal cost function (cost per additional unit), then ∫C(x) dx from a to b gives the total cost of producing units a through b. The calculus is identical — only the physical interpretation changes.
Reading Motion from Graphs
One of the most powerful skills in SL 5.7 is reading information from displacement–time and velocity–time graphs. The slope of a displacement graph gives velocity, and the area under a velocity graph gives displacement. Let's see what this looks like for a particle that speeds up, slows down, reverses direction, and then stops.
From the velocity–time graph above, you can read several things. The slope of the line gives acceleration: from t = 0 to t = 2, the slope is (4 − 0)/(2 − 0) = 2 m/s², so the particle accelerates at 2 m/s². From t = 2 to t = 6, the velocity is constant at 4 m/s, so acceleration is zero. From t = 6 to t = 10, velocity drops from 4 to −4, so acceleration is (−4 − 4)/(10 − 6) = −2 m/s².
The area between the velocity curve and the time axis gives displacement. Area above the axis counts as positive; area below counts as negative. The total displacement from t = 0 to t = 12 is the sum of all the signed areas. If the question asks for total distance, you take the absolute value of each area piece and add them together.
| Graph Feature | What It Tells You | Mathematical Operation |
|---|---|---|
| Slope of s–t graph | Instantaneous velocity | v(t) = ds/dt |
| Slope of v–t graph | Instantaneous acceleration | a(t) = dv/dt |
| Area under v–t graph | Displacement (signed) | ∫v(t) dt |
| Area under |v–t| graph | Total distance | ∫|v(t)| dt |
| v–t graph crosses t-axis | Particle changes direction | v(t) = 0 |
Worked Example — Particle Motion
A particle moves along a straight line. Its displacement from the origin, in metres, is given by s(t) = t³ − 6t² + 9t + 2 for t ≥ 0, where t is in seconds. Find: (a) the velocity function v(t), (b) the acceleration function a(t), (c) when the particle is at rest, (d) the total distance travelled in the first 4 seconds.
Beyond Kinematics — Other Rate and Accumulation Applications
While kinematics is the classic example, SL 5.7 also expects you to interpret derivatives and integrals in non-motion contexts. The table below shows how the same mathematical framework applies to completely different scenarios. In every case, the structure is the same: a quantity changes at some rate, and integration accumulates that change over time.
| Context | Quantity (like s) | Rate (like v = ds/dt) | What Integration Gives |
|---|---|---|---|
| Kinematics | Displacement s (m) | Velocity v (m/s) | Change in position |
| Water flow | Volume V (litres) | Flow rate dV/dt (L/min) | Total volume added/removed |
| Population growth | Population P | Growth rate dP/dt (people/year) | Change in population |
| Economics | Total cost C ($) | Marginal cost dC/dx ($/unit) | Total cost of producing x₁ to x₂ units |
| Temperature | Temperature T (°C) | Cooling rate dT/dt (°C/min) | Total temperature change |
Connection to Advanced Theory
SL 5.7 gives you the introductory toolkit for kinematics and rate problems, but these ideas extend much further. At the HL level and in university mathematics and physics, you will encounter more complex versions of the same principles. The table below shows how the SL content connects to what comes next.
| SL 5.7 (This Course) | HL / University Extension |
|---|---|
| Motion in one dimension (straight line) | Motion in 2D/3D using vector calculus — velocity and acceleration become vectors |
| Finding when v(t) = 0 to determine direction changes | Analysing critical points and phase portraits of differential equations |
| Polynomial and simple displacement functions | Differential equations: solving a = f(v) or a = f(s) to find motion |
| Rate of change interpreted in context | Related rates: how two or more changing quantities depend on each other |
| Definite integral for total change | Fundamental Theorem of Calculus applied rigorously, improper integrals |
Understanding the relationship between a function and its derivative is one of the most transferable skills in mathematics. Every topic you study after SL 5.7 — from optimisation to differential equations to probability distributions — builds on the idea that the derivative measures instantaneous rate of change, and the integral accumulates that change. Mastering this foundation now will pay dividends throughout your mathematical journey.
Practice Problems
Lesson Summary
In SL 5.7, you learned to navigate the kinematics chain: differentiate displacement s(t) to get velocity v(t), and differentiate again to get acceleration a(t). Going the other way, integrate acceleration to recover velocity, and integrate velocity to recover displacement, using initial conditions to determine constants of integration. A particle is at rest when v(t) = 0 and changes direction when the velocity changes sign.
The crucial distinction between displacement (net position change, found by ∫v(t) dt) and total distance (total path length, found by ∫|v(t)| dt) is an IB exam favourite. Beyond kinematics, the same rate-and-accumulation framework applies to any context: water flow, population growth, marginal cost, and more. In every case, the derivative gives the instantaneous rate of change and the definite integral gives the total accumulated change over an interval.