Calculus Applications — SL 5.7 Kinematics and other rate/accumulation applications (intro) - IB Mathematics Applications and Interpretation

SECTION 1

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.

~350 BCE • Zeno's Paradoxes

Greek philosopher Zeno posed paradoxes about motion and infinity — such as Achilles never catching a tortoise — revealing the need for rigorous mathematics to handle continuous change.

1638 • Galileo's Laws of Motion

Galileo Galilei published results showing that falling objects accelerate uniformly. He described distance as proportional to the square of time, hinting at the relationship between position and its derivatives.

1687 • Newton's Principia

Isaac Newton published the Principia Mathematica, unifying his laws of motion with calculus (which he called "fluxions"). Velocity became the fluxion of position, and acceleration the fluxion of velocity.

1684 • Leibniz's Calculus Notation

Gottfried Wilhelm Leibniz independently developed calculus and introduced the notation ds/dt and the integral sign ∫ that we still use today, making kinematics calculations more accessible.

Modern Era • Applications Everywhere

Today, rate-of-change and accumulation models drive fields from epidemiology (infection rates) to economics (marginal cost) to engineering (fluid flow), far beyond the study of physical motion.

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.

SECTION 2

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)

Displacement is the position of an object relative to a reference point, measured as a function of time t. Unlike distance, displacement can be negative (meaning the object is on the opposite side of the origin).

Velocity v(t)

Velocity is the rate of change of displacement with respect to time: v(t) = ds/dt. A positive velocity means motion in the positive direction; a negative velocity means motion in the negative direction.

Acceleration a(t)

Acceleration is the rate of change of velocity: a(t) = dv/dt = d²s/dt². It tells you how quickly the velocity itself is changing. Positive acceleration doesn't always mean speeding up — it depends on the sign of velocity too.

Differentiation Links "Down"

Differentiating displacement gives velocity. Differentiating velocity gives acceleration. Each derivative tells you the instantaneous rate of change of the function above it.

Integration Links "Up"

Integrating acceleration gives velocity (plus a constant). Integrating velocity gives displacement (plus a constant). Integration accumulates the quantity over a time interval.

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KEY TAKEAWAY

Think of a car's dashboard. The odometer tracks total displacement, the speedometer shows velocity (how fast the odometer reading is changing), and the feeling of being pushed back into your seat is acceleration (how fast the speedometer reading is changing). Differentiation moves from odometer → speedometer → that push. Integration moves back the other way.

SECTION 3

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.

The kinematics chain: differentiate to move from displacement → velocity → acceleration. Integrate (with initial conditions) to move back. The cyan arrows represent differentiation (d/dt), and the emerald arrows represent integration (∫ dt).

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.

SECTION 4

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.

VELOCITY FROM DISPLACEMENT

v(t) = ds/dt

where s(t) is the displacement function (in metres) and t is time (in seconds). The derivative ds/dt gives the instantaneous velocity at time t.

ACCELERATION FROM VELOCITY

a(t) = dv/dt = d²s/dt²

Acceleration is the derivative of velocity, or equivalently the second derivative of displacement. Units are m/s².

DISPLACEMENT FROM VELOCITY (DEFINITE INTEGRAL)

s(t₂) − s(t₁) = ∫ from t₁ to t₂ of v(t) dt

The definite integral of velocity from t₁ to t₂ gives the change in displacement over that interval. Be careful: this gives displacement (net change), not total distance. If velocity changes sign, you must split the integral or take ∫|v(t)| dt for total distance.

TOTAL DISTANCE TRAVELLED

Total distance = ∫ from t₁ to t₂ of |v(t)| dt

To find the total distance (always positive), integrate the absolute value of velocity. On the IB exam, you typically split the integral at points where v(t) = 0, then add the absolute values of each piece.

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Displacement vs. Distance — Don't Mix Them Up!

Displacement is the net change in position (can be negative). Distance is the total path length (always ≥ 0). If a particle moves 5 m right then 3 m left, displacement = 2 m but distance = 8 m. IB examiners love testing this distinction.

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.

SECTION 5

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.

A velocity–time graph for a particle. The shaded cyan region above the t-axis represents positive displacement (forward movement). The shaded pink region below the t-axis represents negative displacement (backward movement). The slope at any point gives instantaneous acceleration.

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

Summary of graphical interpretations in kinematics

SECTION 6

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.

Kinematics with a Polynomial Displacement Function

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Step 1 — Find the velocity function

Velocity is the derivative of displacement with respect to time. Differentiate s(t) = t³ − 6t² + 9t + 2 term by term using the power rule: v(t) = ds/dt = 3t² − 12t + 9.

v(t) = 3t² − 12t + 9
2
Step 2 — Find the acceleration function

Acceleration is the derivative of velocity. Differentiate v(t) = 3t² − 12t + 9: a(t) = dv/dt = 6t − 12.

a(t) = 6t − 12
3
Step 3 — Find when the particle is at rest

A particle is at rest when v(t) = 0. Set 3t² − 12t + 9 = 0. Factor out 3: 3(t² − 4t + 3) = 0, so t² − 4t + 3 = 0. Factorise: (t − 1)(t − 3) = 0, giving t = 1 and t = 3.

The particle is at rest at t = 1 s and t = 3 s
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Step 4 — Determine direction of motion in each interval

We need to check the sign of v(t) in each interval. For 0 < t < 1: try t = 0.5, v(0.5) = 3(0.25) − 12(0.5) + 9 = 0.75 − 6 + 9 = 3.75 > 0 (moving forward). For 1 < t < 3: try t = 2, v(2) = 3(4) − 12(2) + 9 = 12 − 24 + 9 = −3 < 0 (moving backward). For 3 < t < 4: try t = 3.5, v(3.5) = 3(12.25) − 12(3.5) + 9 = 36.75 − 42 + 9 = 3.75 > 0 (forward again).

Forward on (0, 1), backward on (1, 3), forward on (3, 4)
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Step 5 — Calculate positions at key times

Evaluate s(t) at t = 0, 1, 3, 4. s(0) = 0 − 0 + 0 + 2 = 2. s(1) = 1 − 6 + 9 + 2 = 6. s(3) = 27 − 54 + 27 + 2 = 2. s(4) = 64 − 96 + 36 + 2 = 6.

s(0) = 2, s(1) = 6, s(3) = 2, s(4) = 6
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Step 6 — Calculate total distance

Total distance = |s(1) − s(0)| + |s(3) − s(1)| + |s(4) − s(3)| = |6 − 2| + |2 − 6| + |6 − 2| = 4 + 4 + 4 = 12 metres. Note that the displacement from t = 0 to t = 4 is only s(4) − s(0) = 6 − 2 = 4 metres, because the backward motion partially cancels the forward motion.

Total distance = 12 metres

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IB Exam Tip

On the IB exam, you can also compute total distance using your GDC (graphing calculator). Enter ∫|v(t)| dt from 0 to 4 directly using the absolute value function. This avoids having to split the integral manually. Just make sure you show your setup and state what the integral represents.

SECTION 7

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

Rate and accumulation applications share the same calculus structure

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KEY TAKEAWAY

Think of a bathtub. The derivative (rate) is like the tap — it tells you how fast water is flowing in at any instant. The integral (accumulation) tells you the total amount of water that has been added over a period of time. Whether it's water, money, people, or distance, the calculus works exactly the same way. Master kinematics, and you've mastered the pattern for all rate-accumulation problems.

SECTION 8

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

How SL 5.7 connects to higher-level mathematics and physics

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.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A particle has velocity v(t) > 0 for 0 < t < 3 and v(t) < 0 for 3 < t < 5. At t = 3, v(3) = 0. Explain in words what happens to the particle at t = 3, and explain the difference between displacement and total distance over the interval [0, 5].

Show answer

At t = 3, the particle is momentarily at rest (v = 0), and it changes direction — it was moving in the positive direction and begins moving in the negative direction. The displacement over [0, 5] is the net change in position: ∫₀⁵ v(t) dt. Since the velocity is positive then negative, the positive and negative parts partially cancel. The total distance is ∫₀⁵ |v(t)| dt, which adds the absolute values of both portions and is therefore always greater than or equal to the magnitude of the displacement. Total distance = ∫₀³ v(t) dt + |∫₃⁵ v(t) dt|.

Question 2: BASIC CALCULATION

A particle has displacement s(t) = 4t² − t³ for t ≥ 0 (metres, seconds). Find the velocity v(t) and the acceleration a(t). At what time is the velocity zero?

Show answer

Differentiate s(t) = 4t² − t³ to get v(t) = ds/dt = 8t − 3t². Differentiate again: a(t) = dv/dt = 8 − 6t. Set v(t) = 0: 8t − 3t² = 0, so t(8 − 3t) = 0, giving t = 0 or t = 8/3 ≈ 2.67 seconds. The velocity is zero at t = 0 (the particle starts from rest) and at t = 8/3 seconds (the particle momentarily stops before reversing direction).

Question 3: INTERMEDIATE

A particle moves with velocity v(t) = 6t − 2t² m/s for t ≥ 0. Find the total distance travelled in the first 4 seconds.

Show answer

First find when v(t) = 0: 6t − 2t² = 0 → 2t(3 − t) = 0 → t = 0 or t = 3. So the particle changes direction at t = 3. For 0 < t < 3, v(t) > 0 (check: v(1) = 4 > 0). For 3 < t < 4, v(t) < 0 (check: v(3.5) = 21 − 24.5 = −3.5 < 0). Total distance = |∫₀³ (6t − 2t²) dt| + |∫₃⁴ (6t − 2t²) dt|. First integral: [3t² − (2/3)t³]₀³ = 27 − 18 = 9. Second integral: [3t² − (2/3)t³]₃⁴ = (48 − 128/3) − (27 − 18) = (48 − 42.67) − 9 = 5.33 − 9 = −3.67. Total distance = 9 + |−3.67| = 9 + 3.67 = 12.67 metres (or exactly 38/3 metres).

Question 4: APPLIED

Water flows into a tank at a rate of R(t) = 20 − 4t litres per minute, where t is in minutes. At t = 0, the tank contains 50 litres. (a) How much water is in the tank at t = 3 minutes? (b) At what time does the water stop flowing in? (c) After the flow stops, does the tank start to drain? Explain.

Show answer

(a) Volume added = ∫₀³ (20 − 4t) dt = [20t − 2t²]₀³ = 60 − 18 = 42 litres. So the tank has 50 + 42 = 92 litres at t = 3. (b) The flow rate is zero when R(t) = 0: 20 − 4t = 0 → t = 5 minutes. (c) For t > 5, R(t) = 20 − 4t becomes negative. A negative rate means water is flowing out of the tank (or the model predicts outflow). So yes, according to this model, the tank starts to drain after t = 5 minutes. This is analogous to a particle changing direction when velocity passes through zero — the same mathematics applies to completely different contexts.

Question 5: CRITICAL THINKING

A particle has velocity v(t) = sin(πt/2) m/s for 0 ≤ t ≤ 4. (a) Sketch the velocity–time graph. (b) Find the displacement from t = 0 to t = 4 using integration. (c) Find the total distance travelled. (d) Explain why the displacement is zero even though the particle has been moving.

Show answer

(a) The velocity v(t) = sin(πt/2) completes one full cycle from t = 0 to t = 4: it goes from 0, up to 1 at t = 1, back to 0 at t = 2, down to −1 at t = 3, and back to 0 at t = 4. (b) Displacement = ∫₀⁴ sin(πt/2) dt = [−(2/π)cos(πt/2)]₀⁴ = −(2/π)cos(2π) + (2/π)cos(0) = −(2/π)(1) + (2/π)(1) = 0. (c) Total distance = ∫₀² sin(πt/2) dt + |∫₂⁴ sin(πt/2) dt| = [−(2/π)cos(πt/2)]₀² + |[−(2/π)cos(πt/2)]₂⁴| = (2/π + 2/π) + |−2/π − 2/π| = 4/π + 4/π = 8/π ≈ 2.55 m. (d) The displacement is zero because the particle returns exactly to its starting position — the positive displacement in [0, 2] is perfectly cancelled by the negative displacement in [2, 4]. This is just like walking 10 steps forward then 10 steps back: you've travelled 20 steps of distance but have zero net displacement.

SUMMARY

Lesson Summary

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.