Calculus in Kinematics — SL 5.7 Kinematics using calculus (v, a; displacement/distance) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

For centuries, people observed objects in motion—falling stones, orbiting planets, rolling carts—and tried to describe their behaviour with mathematics. The ancient Greeks studied motion qualitatively, but they lacked the algebraic tools to express how velocity and acceleration change from one instant to the next. It was not until the seventeenth century that two brilliant minds independently invented the mathematical language needed to unlock the secrets of motion: calculus.

1638 • Galileo's Two New Sciences

Galileo published experiments on falling bodies and inclined planes, establishing that distance fallen is proportional to the square of time—a relationship that begged for a rate-of-change framework.

1665 • Newton's 'Method of Fluxions'

Isaac Newton developed his calculus ('fluxions') partly to solve problems of planetary motion. He treated velocity as the instantaneous rate of change of position—the concept we now call a derivative.

1684 • Leibniz Publishes His Calculus

Gottfried Wilhelm Leibniz independently published his version of calculus with the dx/dt notation still used today, making it easier to express differentiation and integration in kinematics.

1687 • Newton's Principia Mathematica

Newton's masterwork unified his laws of motion with calculus, showing that the same mathematics that describes a thrown ball also governs the orbit of the Moon.

The central question that drove these developments was deceptively simple: if you know where an object is at every moment in time, how do you figure out how fast it is going, and how quickly that speed is changing? Conversely, if you know its acceleration, can you work backwards to find its velocity and position? These are exactly the questions that kinematics using calculus answers, and they form the heart of IB SL 5.7.

SECTION 2

Core Principles & Definitions

Kinematics is the branch of mechanics that describes how objects move without worrying about the forces causing the motion. In this topic, you work with three fundamental quantities—displacement, velocity, and acceleration—and you connect them through differentiation and integration. The direction of the connection is straightforward: differentiating takes you 'down' from displacement to velocity to acceleration, while integrating takes you 'up' in the opposite direction.

Displacement s(t)

Displacement is the signed (positive or negative) distance from a reference point. It is a function of time, s(t). Positive and negative values indicate direction along a line.

Velocity v(t) = s′(t)

Velocity is the rate of change of displacement with respect to time. It tells you how fast and in which direction the object moves. Found by differentiating s(t).

Acceleration a(t) = v′(t)

Acceleration is the rate of change of velocity. It tells you how quickly the object speeds up, slows down, or changes direction. Found by differentiating v(t).

Displacement vs. Distance

Displacement can be negative (the object moved backward). Total distance travelled is always positive and is found by integrating the absolute value of velocity, |v(t)|.

Initial Conditions

When you integrate, you get a constant of integration (+C). You determine C using an initial condition such as s(0) = 0 or v(0) = 5.

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KEY TAKEAWAY

Think of displacement, velocity, and acceleration as three floors of a building. Differentiating is like taking the stairs down one floor (s → v → a), and integrating is like taking the stairs back up (a → v → s). You always need a 'starting position' (initial condition) when you go upstairs, because integration introduces an unknown constant.

SECTION 3

Visual Explanation — The Differentiation–Integration Chain

The top row shows the three kinematic quantities connected by differentiation (left to right, shown in purple) and integration (right to left, shown in green). The bottom row summarises how the sign of velocity and acceleration tells you the direction and behaviour of the particle.

The diagram above is worth memorising. Whenever you face a kinematics problem, first identify which quantity you are given and which one you need. If you need to go 'down' (from displacement to velocity, or velocity to acceleration), you differentiate. If you need to go 'up' (from acceleration to velocity, or velocity to displacement), you integrate and use the given initial condition to find the constant C.

SECTION 4

Mathematical Framework

Let s(t) represent the displacement of a particle at time t. All three core relationships in this topic follow directly from the definition of the derivative and its reverse, the integral.

VELOCITY FROM DISPLACEMENT

v(t) = ds/dt = s′(t)

v(t) is the instantaneous velocity at time t. It equals the derivative of displacement with respect to time. A positive value means the particle moves in the positive direction; a negative value means it moves in the negative direction.

ACCELERATION FROM VELOCITY

a(t) = dv/dt = v′(t) = s″(t)

a(t) is the instantaneous acceleration. It is the derivative of velocity, which is the same as the second derivative of displacement. Positive acceleration does not always mean 'speeding up'—you must compare the signs of v and a.

DISPLACEMENT FROM VELOCITY (INTEGRATION)

s(t) = ∫ v(t) dt + C

Integrating velocity recovers displacement. The constant C is determined from an initial condition, such as s(0) = 2. Note: the definite integral ∫ from t₁ to t₂ of v(t) dt gives the net displacement (change in position) over that interval.

TOTAL DISTANCE TRAVELLED

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

Because distance is always positive, you integrate the absolute value of velocity. In practice, find where v(t) = 0, split the interval at those points, evaluate each integral, and add the absolute values.

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IB Exam Tip

The IB frequently asks for total distance in one part and net displacement in another. For displacement, integrate v(t) directly. For distance, integrate |v(t)| by splitting the interval wherever v(t) changes sign. Mixing these up is one of the most common errors on the exam.

SECTION 5

Displacement vs. Distance — A Detailed Look

One of the trickiest parts of this topic is the distinction between displacement and distance. Consider a particle that moves forward 5 metres, then reverses and comes back 3 metres. Its displacement is 5 − 3 = 2 metres (net change in position), but the total distance travelled is 5 + 3 = 8 metres. The velocity function captures this: whenever v(t) is negative, the particle is moving backward, and that backward movement subtracts from the displacement integral but adds to the distance integral.

The curve represents v(t). The cyan shaded area A₁ above the t-axis is positive displacement (forward motion). The pink shaded area A₂ below the t-axis represents negative displacement (backward motion). Displacement = A₁ + A₂ (signed sum), while total distance = |A₁| + |A₂|.

In the graph above, the particle changes direction at the instant when v(t) = 0 (the yellow dot at t = 4). To find total distance, you evaluate the integral of v(t) from 0 to 4 separately from the integral of v(t) from 4 to 8, take the absolute value of each, and add them together. This is the key procedure that the IB expects you to perform confidently.

Quantity	Calculation	Can be negative?
Displacement	∫ from t₁ to t₂ of v(t) dt	Yes — negative means net movement in the negative direction
Total distance	∫ from t₁ to t₂ of \|v(t)\| dt (split at sign changes)	No — distance is always ≥ 0
Average velocity	Displacement ÷ total time	Yes — can be negative or zero

Summary of displacement, distance, and average velocity calculations

SECTION 6

Worked Example

A particle moves along a straight line. Its displacement (in metres) at time t seconds is given by s(t) = t³ − 6t² + 9t + 2, for t ≥ 0. Find (a) the velocity and acceleration functions, (b) the times when the particle is at rest, (c) the total distance travelled in the first 4 seconds.

Full Solution

1
Step 1 — Find v(t) by differentiating s(t)

Differentiate s(t) = t³ − 6t² + 9t + 2 with respect to t. v(t) = ds/dt = 3t² − 12t + 9

v(t) = 3t² − 12t + 9
2
Step 2 — Find a(t) by differentiating v(t)

Differentiate v(t) = 3t² − 12t + 9 with respect to t. a(t) = dv/dt = 6t − 12

a(t) = 6t − 12
3
Step 3 — Find when the particle is at rest (v = 0)

Set v(t) = 0: 3t² − 12t + 9 = 0. Divide through by 3: t² − 4t + 3 = 0. Factorise: (t − 1)(t − 3) = 0.

t = 1 and t = 3
4
Step 4 — Determine direction of motion in each interval

Test the sign of v(t) in each sub-interval: • 0 < t < 1: v(0.5) = 3(0.25) − 12(0.5) + 9 = 0.75 − 6 + 9 = 3.75 > 0 → moving forward • 1 < t < 3: v(2) = 3(4) − 12(2) + 9 = 12 − 24 + 9 = −3 < 0 → moving backward • 3 < t < 4: v(3.5) = 3(12.25) − 12(3.5) + 9 = 36.75 − 42 + 9 = 3.75 > 0 → moving forward
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Step 5 — Calculate displacement in each sub-interval

Evaluate s(t) at the critical times: 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 Displacement from t = 0 to t = 1: s(1) − s(0) = 6 − 2 = 4 m (forward) Displacement from t = 1 to t = 3: s(3) − s(1) = 2 − 6 = −4 m (backward) Displacement from t = 3 to t = 4: s(4) − s(3) = 6 − 2 = 4 m (forward)
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Step 6 — Total distance = sum of absolute values

Total distance = |4| + |−4| + |4| = 4 + 4 + 4 = 12 metres. Note: The net displacement from t = 0 to t = 4 is s(4) − s(0) = 6 − 2 = 4 metres (much less than the distance, because part of the journey was in reverse).

Total distance = 12 m

SECTION 7

Strengths, Limitations & Common Pitfalls

Calculus-based kinematics is a powerful framework, but students often lose marks on the IB exam through a handful of predictable errors. The table below compares correct techniques with common mistakes.

Situation	Correct Approach ✓	Common Mistake ✗
Finding total distance	Split the integral at every zero of v(t) and sum absolute values of each part	Integrating v(t) straight from t₁ to t₂ without splitting — this gives displacement, not distance
Determining if particle speeds up	Check if v(t) and a(t) have the same sign at that instant	Assuming positive acceleration always means speeding up
Integrating to find s(t) from v(t)	Include +C and use the initial condition to solve for it	Forgetting the +C or ignoring the given initial condition
Particle 'changes direction'	v(t) = 0 AND v(t) changes sign at that point	Saying the particle changes direction just because v(t) = 0 (it might touch zero and bounce back)

Correct techniques vs common mistakes in calculus kinematics

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KEY TAKEAWAY

Think of velocity like a car's speedometer that can go negative. Displacement is like your GPS telling you how far you are from your starting point—it doesn't care about detours. Distance is like your car's odometer, which only ever counts up no matter which direction you drive. The odometer reading is always ≥ the GPS displacement.

SECTION 8

Connection to Advanced Theory

The SL 5.7 treatment deals with motion in one dimension (along a straight line). At the HL level and in university physics, these ideas extend in exciting directions. Understanding the SL foundations makes the transition to these advanced topics much smoother.

SL 5.7 (This Lesson)	HL / University Extension
Motion along a line: s(t), v(t), a(t) are scalar functions	Motion in 2D/3D: position, velocity, acceleration become vector functions r(t), v(t), a(t) with components
Polynomial and simple displacement functions	Parametric equations and differential equations model more complex trajectories (projectiles, circular motion)
Distance = ∫\|v(t)\| dt (split at zeros)	Arc length = ∫√(x′(t)² + y′(t)²) dt for curves in the plane
Kinematics only (description of motion)	Dynamics: Newton's second law F = ma connects force to the calculus of motion

How SL kinematics ideas extend at higher levels

If you continue with HL Mathematics or university-level physics, you will see that the differentiate-down / integrate-up chain you learned here is the backbone of classical mechanics. Mastering it at the SL level gives you a significant head start.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A particle has velocity v(t) = t² − 4. Explain in words what happens to the particle at t = 2 and how you know.

Show answer

At t = 2, v(2) = 4 − 4 = 0, so the particle is momentarily at rest. For t < 2, v(t) is negative (e.g., v(1) = 1 − 4 = −3), meaning the particle moves in the negative direction. For t > 2, v(t) is positive (e.g., v(3) = 9 − 4 = 5), meaning it moves in the positive direction. Since v(t) changes sign at t = 2, the particle changes direction at that instant.

Question 2: BASIC CALCULATION

The displacement of a particle is s(t) = 2t³ − 9t² + 12t, where s is in metres and t in seconds. Find v(t) and a(t).

Show answer

Differentiate s(t): v(t) = ds/dt = 6t² − 18t + 12. Differentiate again: a(t) = dv/dt = 12t − 18. You can verify by checking units and by noting that v(0) = 12 m/s (the particle starts with a positive velocity) and a(0) = −18 m/s² (the particle is initially decelerating).

Question 3: INTERMEDIATE

A particle moves along a line with velocity v(t) = 6t − t², for 0 ≤ t ≤ 8. Find (a) the displacement from t = 0 to t = 8, and (b) the total distance travelled.

Show answer

(a) Displacement = ∫ from 0 to 8 of (6t − t²) dt = [3t² − t³/3] from 0 to 8 = (192 − 512/3) − 0 = (576/3 − 512/3) = 64/3 ≈ 21.3 m. (b) First find where v(t) = 0: t(6 − t) = 0 gives t = 0 and t = 6. Since v(t) > 0 on (0, 6) and v(t) < 0 on (6, 8): ∫ from 0 to 6 of (6t − t²) dt = [3t² − t³/3] from 0 to 6 = 108 − 72 = 36. ∫ from 6 to 8 of (6t − t²) dt = [3t² − t³/3] from 6 to 8 = (192 − 512/3) − (108 − 72) = 64/3 − 36 = −44/3. Total distance = 36 + 44/3 = 108/3 + 44/3 = 152/3 ≈ 50.7 m.

Question 4: APPLIED

A drone's acceleration at time t seconds is a(t) = 4 − 2t m/s². At t = 0, the drone's velocity is 3 m/s and its displacement from the launch pad is 0 m. Find the displacement function s(t), and determine when the drone is furthest from the launch pad for t ≥ 0.

Show answer

Integrate a(t) to get v(t): v(t) = ∫(4 − 2t) dt = 4t − t² + C₁. Using v(0) = 3: C₁ = 3, so v(t) = 4t − t² + 3 = −t² + 4t + 3. Integrate v(t) to get s(t): s(t) = ∫(−t² + 4t + 3) dt = −t³/3 + 2t² + 3t + C₂. Using s(0) = 0: C₂ = 0, so s(t) = −t³/3 + 2t² + 3t. The drone is furthest when v(t) = 0 and the particle changes from positive to negative velocity. Solve −t² + 4t + 3 = 0, i.e., t² − 4t − 3 = 0. Using the quadratic formula: t = (4 ± √28)/2 = 2 ± √7. Since t ≥ 0, t = 2 + √7 ≈ 4.65 seconds. This is the time of maximum displacement from the launch pad.

Question 5: CRITICAL THINKING

A particle moves so that v(t) = sin(πt/3) m/s for 0 ≤ t ≤ 6. Without evaluating any integrals numerically, explain why the displacement over the full 6 seconds is zero, but the total distance is not zero. Then set up (but do not evaluate) the integral expression for the total distance.

Show answer

The function sin(πt/3) has a period of 6 seconds (since 2π/(π/3) = 6). Over one full period, the sine function spends exactly as much time above the t-axis as below it, and the positive and negative areas cancel perfectly, giving a net displacement of zero. However, the total distance counts all movement as positive, so it does not cancel. To find total distance, identify the zeros of v(t): sin(πt/3) = 0 when πt/3 = 0, π, 2π, giving t = 0, 3, 6. On [0, 3], v(t) > 0; on [3, 6], v(t) < 0. Total distance = ∫ from 0 to 3 of sin(πt/3) dt + |∫ from 3 to 6 of sin(πt/3) dt|. By symmetry, this equals 2 × ∫ from 0 to 3 of sin(πt/3) dt. This question illustrates that symmetric velocity functions always produce zero net displacement over a full period but nonzero total distance.

SUMMARY

Lesson Summary

Summary

In SL 5.7, three quantities form a connected chain: displacement s(t), velocity v(t) = s′(t), and acceleration a(t) = v′(t) = s″(t). Moving down the chain requires differentiation; moving back up requires integration plus an initial condition to determine the constant of integration.

The sign of v(t) tells you the direction of motion; the particle changes direction when v(t) = 0 and v changes sign. Displacement is the signed integral of v(t), while total distance is found by integrating |v(t)|—in practice, split the interval at each zero of v(t) and sum the absolute values of each piece. Confusing displacement with distance is the single most common error on the IB exam for this topic.