# Kinematics: motion with calculus

> IB Mathematics Analysis and Approaches SL · IB Math AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-kinematics-motion-with-calculus/

This module connects core differential and integral calculus to 1D kinematics, covering relationships between displacement, velocity and acceleration, variable acceleration problem solving, and graphical interpretations.

**Prerequisites:** [Basic differentiation of polynomial, trigonometric and exponential functions](https://www.owlsprep.com/study/ib-math-aa-sl-u5-basic-differentiation/); [Definite and indefinite integration rules for standard functions](https://www.owlsprep.com/study/ib-math-aa-sl-u5-definite-integrals/)

## Learning objectives

- Define displacement, velocity and acceleration as time-dependent 1D motion functions
- Relate the three kinematic quantities via differentiation and indefinite/definite integration
- Solve exam-style problems for motion with non-constant acceleration
- Distinguish between net displacement and total distance travelled for changing direction motion

## Core Kinematic Relationships via Differentiation

For a particle moving along a straight line, the three core kinematic quantities are linked by rates of change. No constant acceleration kinematic equations (SUVAT) apply here, as acceleration can vary with time.

**Kinematic Derivative Relationships** — Velocity is the first time derivative of displacement, and acceleration is the first time derivative of velocity, or the second time derivative of displacement.

$$v(t) = \frac{ds}{dt}, \quad a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$

**Worked example:** A particle moves along a straight line with displacement function $s(t) = 2t^3 - 9t^2 + 12t$ for $t \geq 0$, where $s$ is in metres and $t$ in seconds. Find the velocity and acceleration of the particle at $t=2$ seconds.

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

**Check your understanding**

Confirm you understand the derivative relationships:

1. If $s(t) = 5\sin(t)$, what is $a(t)$?

   - $5\cos(t)$
   - $-5\sin(t)$
   - $-5\cos(t)$
   - $5\sin(t)$

   *Why:* Differentiate twice: first derivative is $5\cos(t)$, second is $-5\sin(t)$.

*Calculator:* allowed

## Finding Displacement from Velocity via Integration

Integration reverses the differentiation process: integrating acceleration gives velocity, and integrating velocity gives displacement. For indefinite integrals, you must use a given initial condition to solve for the constant of integration.

$$v(t) = \int a(t) dt + c_1, \quad s(t) = \int v(t) dt + c_2$$

**Worked example:** A particle has velocity $v(t) = 6t^2 - 18t + 12$ for $t \geq 0$. At $t=0$, the particle is at displacement $s=0$. Find the displacement function $s(t)$ and the net displacement between $t=0$ and $t=3$.

1. Integrate $v(t)$ to find general displacement function:
2. $$s(t) = \int (6t^2 - 18t + 12) dt = 2t^3 - 9t^2 + 12t + c$$
3. Apply initial condition $s(0)=0$: $0 = 0 - 0 + 0 + c$, so $c=0$.
4. Evaluate definite integral from $t=0$ to $t=3$ for net displacement:
5. $$\int_0^3 (6t^2 - 18t + 12) dt = \left[2t^3 -9t^2 +12t\right]_0^3 = (54 - 81 + 36) - 0 = 9 \text{ m}$$

> **tip**
>
> Net displacement can be negative, which means the particle ends up to the left of its starting position, not that it travelled a negative distance.

*Calculator:* allowed

## Calculating Total Distance Travelled

> **warning**
>
> Total distance travelled is NOT equal to net displacement if the particle changes direction during the time interval.

To calculate total distance, first find all times in the interval where $v(t)=0$ (these are the points where the particle reverses direction). Split the interval at these roots, and integrate the absolute value of $v(t)$ over each sub-interval, then sum the results.

**Worked example:** Using the same velocity function $v(t) = 6t^2 - 18t +12$, calculate the total distance travelled by the particle between $t=0$ and $t=3$.

1. Find roots of $v(t) = 0$: factor to get $6(t-1)(t-2)=0$, so roots at $t=1$ and $t=2$.
2. Test sign of $v(t)$ on each interval: positive 0<t<1, negative 1<t<2, positive 2<t<3.
3. Integrate absolute value of $v(t)$ over each interval:
4. $$\int_0^1 v(t) dt = 5, \quad \int_1^2 -v(t) dt = 1, \quad \int_2^3 v(t) dt = 4$$
5. Sum the three results to get total distance: $5 + 1 + 4 = 10$ m.

**Exam command terms**

IB exam questions use specific command terms for kinematics:

- **Find displacement** — Calculate net signed change in position, no need to split intervals

- **Find total distance** — You must split at v(t)=0 points and sum absolute values of integrals

*Calculator:* allowed

## Graphical Interpretation of Kinematic Functions

| Graph | Slope represents | Area under curve represents |
| --- | --- | --- |
| Displacement vs time | Instantaneous velocity | No standard kinematic meaning |
| Velocity vs time | Instantaneous acceleration | Net displacement / total distance |
| Acceleration vs time | Rate of change of acceleration | Change in velocity |

**Worked example:** A velocity vs time graph is a straight line from $v=4$ m/s at $t=0$ down to $v=-2$ m/s at $t=3$. Use area calculation to find total distance travelled.

1. Find root of v(t) at t=2, where velocity crosses zero.
2. Calculate area of positive triangle from t=0 to t=2: $\frac{1}{2} \times 2 \times 4 = 4$ m.
3. Calculate area of negative triangle from t=2 to t=3: $\frac{1}{2} \times 1 \times 2 = 1$ m.
4. Sum areas for total distance: $4 + 1 = 5$ m.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using a single definite integral of $v(t)$ to calculate total distance directly
  - Why it fails: Negative velocity regions cancel positive displacement values, returning net displacement not total path length
  - Correct: Find all roots of $v(t)$ in the interval, split the integral at these points, and sum absolute values of each sub-integral
- **Wrong:** Treating acceleration as the first derivative of displacement
  - Why it fails: Mixing up the order of differentiation skips the velocity step, leading to wrong units and values
  - Correct: Follow the chain: differentiate displacement once for velocity, differentiate velocity once for acceleration
- **Wrong:** Forgetting the constant of integration when finding $s(t)$ from $v(t)$
  - Why it fails: Indefinite integration produces an arbitrary constant that corresponds to the initial position of the particle
  - Correct: Use the given initial condition (e.g. $s(0)=3$) to solve for the constant before finalizing your displacement function
- **Wrong:** Assuming velocity is zero when acceleration is zero
  - Why it fails: Zero acceleration means velocity is constant, not that it equals zero
  - Correct: Set $v(t) = 0$ explicitly to find points where the particle is stationary or changes direction
- **Wrong:** Stating acceleration units as m/s
  - Why it fails: Acceleration is the rate of change of velocity, so units are metres per second squared
  - Correct: Always confirm units match the quantity you are calculating to avoid losing method marks

## Cheatsheet

| Quantity | Mathematical Relationship | Standard Units |
| --- | --- | --- |
| Displacement $s(t)$ | $s(t) = \int v(t) dt + c$ | m |
| Velocity $v(t)$ | $v(t) = \frac{ds}{dt} = \int a(t) dt + c$ | m/s |
| Acceleration $a(t)$ | $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ | m/s² |
| Net Displacement $t_1$ to $t_2$ | $\int_{t_1}^{t_2} v(t) dt$ | m |
| Total Distance $t_1$ to $t_2$ | $\int_{t_1}^{t_2} \|v(t)\| dt$ | m |

## What's next

Mastering kinematics with calculus gives you a high-yield skill that appears in almost every IB Math AA SL Paper 2, often worth 6-8 marks. This topic builds directly on your prior differentiation and integration fluency, and prepares you for applied calculus problems in related rates, motion modelling, and optimization contexts. You will frequently combine these kinematic rules with your knowledge of function roots, graph sketching, and definite integral evaluation to score full marks on extended response questions. This is one of the most commonly tested applied calculus topics on the syllabus, so consistent practice will directly boost your exam score.

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