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

> IB Mathematics Applications and Interpretation Higher Level · IB Math AI HL
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-kinematics-with-calculus/

This guide covers core calculus relationships between motion quantities, how to calculate total distance travelled, and common IB AI HL exam problem structures for kinematics.

**Prerequisites:** [Basic polynomial, trigonometric and exponential differentiation rules](https://www.owlsprep.com/study/ib-math-ai-hl-u4-differentiation-basics/); [Definite and indefinite integration fundamentals](https://www.owlsprep.com/study/ib-math-ai-hl-u4-integration-fundamentals/)

## Learning objectives

- Define the calculus relationship between displacement, velocity and acceleration for 1D motion
- Distinguish between net displacement and total distance travelled for particles that reverse direction
- Derive motion functions from acceleration and given initial boundary conditions
- Interpret velocity-time and acceleration-time graphs using integral and derivative rules
- Solve full IB AI HL exam-style kinematics problems efficiently

## Core Calculus Relationships Between Motion Quantities

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.

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

*Notation:* $\frac{ds}{dt}$, $\frac{dv}{dt}$

**Derivation:** 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. $$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$

*Conclusion:* Differentiate displacement twice to get acceleration, or integrate acceleration twice to get displacement.

**Worked example:** A particle's displacement is given by $s(t) = 2t^3 - 9t^2 + 12t + 5$ 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. $$a(2) = 12(2) - 18 = 24 - 18 = 6 \text{ m/s}^2$$

**Exam command terms**

IB exam questions use specific command terms for kinematics problems:

- **Show that the particle is at rest at t=k** — Prove v(k) = 0 by substituting t=k into your derived velocity function

- **Find when the particle changes direction** — Solve v(t)=0 and confirm the sign of v(t) flips across that point

## Calculating Total Distance Travelled

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.

> **Common Exam Trap**
>
> If you integrate v(t) directly over an interval where velocity is negative for part of the range, positive and negative areas will cancel, giving you net displacement instead of total distance. This will cost you full method marks in almost all cases.

**Worked example:** Using the same displacement function $s(t) = 2t^3 - 9t^2 + 12t + 5$, 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. $$d = |10-5| + |9-10| + |14-9| = 5 + 1 + 5 = 11 \text{ m}$$

**Check your understanding**

Test your understanding before moving on:

1. 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

   *Why:* Velocity crosses zero at t=2, split interval to sum absolute displacements to get 4 + 9 = 13 m total.

## Deriving Motion Functions from Boundary Conditions

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 $a(t) = 6t - 4$ 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. $$s(t) = t^3 - 2t^2 - 3t + 2$$

## Interpreting Kinematics Graphs

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 |

> **GDC Shortcut**
>
> You do not need to find the algebraic equation of a plotted velocity function to calculate the area under the curve: use your GDC's integral tool directly on the graph to save time in Paper 2.

## Common pitfalls

- **Wrong:** Integrating velocity directly over a full interval to find total distance
  - Why it fails: Negative velocity regions cancel positive areas, returning net displacement instead of total path length
  - Correct: Find all points where v(t)=0, split the interval at those points, and sum the absolute value of displacement across each segment
- **Wrong:** Forgetting to add the constant of integration when integrating acceleration or velocity
  - Why it fails: The constant represents initial velocity or initial displacement, which is required to get the exact motion function
  - Correct: Always write +C after every indefinite integral, then substitute the given boundary condition to solve for C explicitly
- **Wrong:** Stating that a particle with zero velocity is stationary for all time
  - Why it fails: Zero velocity at a single instant only means the particle is not moving at that exact moment, and may reverse direction immediately after
  - Correct: Check the sign of velocity just before and after the zero point to confirm a direction change
- **Wrong:** Using SUVAT equations for non-constant acceleration problems
  - Why it fails: SUVAT only applies to motion with fixed acceleration, which is almost never the case for AI HL kinematics questions
  - Correct: Use differentiation and integration for all motion problems unless explicitly told acceleration is constant
- **Wrong:** Writing acceleration units as m/s instead of m/s²
  - Why it fails: IB mark schemes explicitly penalise incorrect units for motion quantities, costing you 1 mark per error
  - Correct: Write the correct unit next to every motion quantity you calculate in your working

## Cheatsheet

| Relationship | Calculus Operation | Formula |
| --- | --- | --- |
| Displacement → Velocity | Differentiate once | $v(t) = s'(t) = \frac{ds}{dt}$ |
| Velocity → Acceleration | Differentiate once | $a(t) = v'(t) = \frac{dv}{dt}$ |
| Acceleration → Velocity | Indefinite Integrate | $v(t) = \int a(t) dt + v_0$ |
| Velocity → Displacement | Indefinite Integrate | $s(t) = \int v(t) dt + s_0$ |
| Total Distance over [a,b] | Integrate absolute value of v(t) | $d = \int_a^b \|v(t)\| dt$ |

## 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.

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