Kinematics with calculus (displacement, velocity, acceleration and distance)

IB Mathematics Applications and Interpretation Higher Level· AHL 5.13: Kinematics — displacement, velocity, acceleration and total distance with calculus· 12 min read

1. Core Calculus Relationships Between Motion Quantities★★★☆☆⏱ 10 min

For 1D motion of a particle along a straight line, the three core motion quantities are linked directly by differentiation and integration. These relationships hold for any motion, including non-constant acceleration that cannot be modelled with SUVAT equations.

📘 Definition

Instantaneous Rate of Motion

$\frac{d s}{d t}$ , $\frac{d v}{d t}$

The value of velocity or acceleration at a single exact point in time, rather than an average over a time interval.

🔬 Derivation

Goal:

Derive the link between displacement, velocity and acceleration

1
Velocity is defined as the rate of change of displacement with respect to time
2
$v(t) = \frac{ds}{dt}$
3
Acceleration is defined as the rate of change of velocity with respect to time
4
$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Result:

Differentiate displacement twice to get acceleration, or integrate acceleration twice to get displacement.

📐 Worked Example

A particle's displacement is given by $s (t) = 2 t^{3} - 9 t^{2} + 12 t + 5$ for $t \geq 0$ , where s is in metres and t in seconds. Find the velocity and acceleration at t=2 seconds.

1
Step 1: Differentiate the displacement function to get the velocity function
2
$v(t) = \frac{ds}{dt} = 6t^2 - 18t + 12$
3
Step 2: Substitute t=2 into the velocity function
4
$v(2) = 6(2)^2 - 18(2) + 12 = 24 - 36 + 12 = 0 \text{ m/s}$
5
Step 3: Differentiate the velocity function to get the acceleration function
6
$a(t) = \frac{dv}{dt} = 12t - 18$
7
Step 4: Substitute t=2 into the acceleration function
8
$a(2) = 12(2) - 18 = 24 - 18 = 6 \text{ m/s}^2$

2. Calculating Total Distance Travelled★★★★☆⏱ 12 min

Net displacement only measures the difference between the final and initial position of a particle, but total distance counts every metre travelled even if the particle moves backwards and reverses direction. This is the most frequently tested trick on IB AI HL kinematics questions.

📐 Worked Example

Using the same displacement function $s (t) = 2 t^{3} - 9 t^{2} + 12 t + 5$ , calculate the total distance travelled by the particle from t=0 to t=3 seconds.

1
Step 1: Find all points in the interval where velocity equals zero (direction change points)
2
$v(t) = 6t^2 - 18t + 12 = 0 \implies t^2 - 3t + 2 = 0 \implies t=1, t=2$
3
Step 2: Split the interval [0, 3] into sub-intervals separated by the zero velocity points: [0,1], [1,2], [2,3]
4
Step 3: Calculate displacement at each boundary point
5
$s(0)=5, s(1)=10, s(2)=9, s(3)=14$
6
Step 4: Sum the absolute value of displacement across each sub-interval
7
$d = |10-5| + |9-10| + |14-9| = 5 + 1 + 5 = 11 \text{ m}$

✓ Quick check

Test your understanding before moving on:

A particle moves with velocity v(t) = 2t - 4 from t=0 to t=5. What is the total distance travelled?
- 5 m
- 13 m
- 9 m
- 25 m
Reveal answer
13 m —
Velocity crosses zero at t=2, split interval to sum absolute displacements to get 4 + 9 = 13 m total.

3. Deriving Motion Functions from Boundary Conditions★★★★☆⏱ 10 min

Most IB AI HL kinematics problems give you the acceleration function first, and ask you to work backwards to find velocity and displacement. You will always be given initial conditions (usually velocity and displacement at t=0) to solve for the constants of integration.

📐 Worked Example

A particle has acceleration $a (t) = 6 t - 4$ m/s². At t=0, its velocity is -3 m/s and its displacement is 2 m. Find the full displacement function s(t).

1
Step 1: Integrate the acceleration function to get the general velocity function
2
$v(t) = \int (6t - 4) dt = 3t^2 - 4t + C_1$
3
Step 2: Substitute the initial velocity condition v(0) = -3 to solve for C₁
4
$-3 = 3(0)^2 -4(0) + C_1 \implies C_1 = -3$
5
Step 3: Integrate the velocity function to get the general displacement function
6
$s(t) = \int (3t^2 - 4t - 3) dt = t^3 - 2t^2 - 3t + C_2$
7
Step 4: Substitute the initial displacement condition s(0) = 2 to solve for C₂
8
$2 = 0 - 0 - 0 + C_2 \implies C_2 = 2$
9
Final displacement function
10
$s(t) = t^3 - 2t^2 - 3t + 2$

4. Interpreting Kinematics Graphs★★★☆☆⏱ 8 min

IB AI HL exam questions often ask you to interpret graphs of velocity or acceleration against time, without giving you the explicit algebraic function. You can use your GDC to calculate definite integrals directly from the graph to find total distance or displacement.

Graph Type	Gradient Represents	Area Under Curve Represents
Displacement vs Time	Instantaneous velocity	No standard physical meaning
Velocity vs Time	Instantaneous acceleration	Net displacement (if signed) or total distance (if absolute)
Acceleration vs Time	Rate of change of acceleration	Change in velocity over the interval

5. Common Pitfalls

Wrong move:

Integrating velocity directly over a full interval to find total distance

Why:

Negative velocity regions cancel positive areas, returning net displacement instead of total path length

Correct move:

Find all points where v(t)=0, split the interval at those points, and sum the absolute value of displacement across each segment

Wrong move:

Forgetting to add the constant of integration when integrating acceleration or velocity

Why:

The constant represents initial velocity or initial displacement, which is required to get the exact motion function

Correct move:

Always write +C after every indefinite integral, then substitute the given boundary condition to solve for C explicitly

Wrong move:

Stating that a particle with zero velocity is stationary for all time

Why:

Zero velocity at a single instant only means the particle is not moving at that exact moment, and may reverse direction immediately after

Correct move:

Check the sign of velocity just before and after the zero point to confirm a direction change

Wrong move:

Using SUVAT equations for non-constant acceleration problems

Why:

SUVAT only applies to motion with fixed acceleration, which is almost never the case for AI HL kinematics questions

Correct move:

Use differentiation and integration for all motion problems unless explicitly told acceleration is constant

Wrong move:

Writing acceleration units as m/s instead of m/s²

Why:

IB mark schemes explicitly penalise incorrect units for motion quantities, costing you 1 mark per error

Correct move:

Write the correct unit next to every motion quantity you calculate in your working

6. Quick Reference Cheatsheet

Relationship	Calculus Operation	Formula
Displacement → Velocity	Differentiate once	$v (t) = s^{'} (t) = \frac{d s}{d t}$
Velocity → Acceleration	Differentiate once	$a (t) = v^{'} (t) = \frac{d v}{d t}$
Acceleration → Velocity	Indefinite Integrate	$v (t) = \int a (t) d t + v_{0}$
Velocity → Displacement	Indefinite Integrate	$s (t) = \int v (t) d t + s_{0}$
Total Distance over [a,b]	Integrate absolute value of v(t)	$d = \int_{a}^{b} ∣ v (t) ∣ d t$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 2
Particle motion integration problem
2022 · Paper 3
Kinematics with boundary conditions
2021 · Paper 2
Total distance from velocity function

What's Next

Now that you have mastered the core calculus relationships for 1D kinematics, you are ready to tackle more complex IB AI HL problems that combine kinematics with other topics, such as modelling motion with trigonometric or exponential functions, and 2D relative velocity problems. This skill is also a foundational requirement for Paper 3 questions that frequently combine kinematics with differential equations and numerical integration. Practise a range of exam-style questions to ensure you can reliably distinguish between displacement and total distance, which is the single most common source of lost marks on this topic.