# Kinematics

> CIE A-Level Mathematics · 9709 Unit 3 Mechanics
> Source: https://www.owlsprep.com/study/cie-9709-u3-kinematics/

This subtopic covers one-dimensional kinematics for motion with variable acceleration. You will use differentiation and integration to relate displacement, velocity and acceleration, and solve common applied motion problems for CIE A-Level Mechanics.

**Prerequisites:** [Basic differentiation and integration](https://www.owlsprep.com/study/cie-9709-u1-basic-calculus/); Constant acceleration kinematics (SUVAT)

## Learning objectives

- Relate displacement, velocity and acceleration using calculus for variable acceleration motion
- Solve problems involving one-dimensional motion with non-constant acceleration
- Find maximum and minimum values of velocity/displacement using differentiation
- Apply the chain rule identity for acceleration as a function of displacement

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 M1 past papers (2016–2025): **Kinematics** appears **221 times** in the last 10 years — **39.7%** of all M1 questions (221 of 443).

Most-tested forms: SUVAT equations (constant acceleration) (136), Variable acceleration (calculus kinematics) (62), Velocity-time graphs (41), Distance-time graphs (5), Velocity and acceleration (2).

## Relationships Between Motion Quantities

**Variable Acceleration Kinematics** — When acceleration is not constant, SUVAT formulas are invalid, so we use calculus to relate displacement, velocity and acceleration as functions of time $t$.

*Example:* Acceleration that changes with time, $a = 6t - 2$, requires calculus to solve.

Velocity is the first derivative of displacement with respect to time, and acceleration is the first derivative of velocity (second derivative of displacement):

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

Working backwards, we integrate to get velocity from acceleration, and displacement from velocity, where $C_1$ and $C_2$ are constants of integration found from initial conditions:

$$v(t) = \int a(t) dt + C_1, \quad s(t) = \int v(t) dt + C_2$$

**Worked example:** A particle moves in a straight line with acceleration $a = 6t - 2 \text{ ms}^{-2}$ at time $t$. Given that $v = 4 \text{ ms}^{-1}$ and $s = 0 \text{ m}$ at $t=0$, find $s$ in terms of $t$.

1. Integrate acceleration to get velocity:
2. $$v = \int (6t - 2) dt = 3t^2 - 2t + C$$
3. Substitute $v=4, t=0$ to find $C$:
4. $$4 = 0 - 0 + C \implies C = 4$$
5. Integrate velocity to get displacement:
6. $$s = \int (3t^2 - 2t + 4) dt = t^3 - t^2 + 4t + D$$
7. Substitute $s=0, t=0$ to find $D$:
8. $$0 = 0 - 0 + 0 + D \implies D = 0$$
9. Final expression for displacement:
10. $$s = t^3 - t^2 + 4t$$

> **tip**
>
> Examiners always award at least one mark for correctly finding the constant of integration, so never skip this step.

*Calculator:* allowed

## Maxima and Minima of Motion Quantities

We use differentiation to find maximum or minimum values of velocity or displacement, just like we do for any other function. A maximum or minimum occurs when the derivative of the quantity equals zero:

- Maximum/minimum displacement occurs when $v = \frac{ds}{dt} = 0$
- Maximum/minimum velocity occurs when $a = \frac{dv}{dt} = 0$

> **info**
>
> You must confirm the nature of the stationary point (maximum/minimum) using the second derivative test if the question asks you to justify your answer.

**Worked example:** A particle's velocity is given by $v = t^3 - 6t^2 + 9t \text{ ms}^{-1}$ for $t \geq 0$. Find the values of $t$ where velocity is maximum or minimum, and state which is which.

1. Differentiate $v$ to get acceleration:
2. $$a = \frac{dv}{dt} = 3t^2 - 12t + 9$$
3. Set $a=0$ to find stationary points:
4. $$3(t^2 - 4t + 3) = 0 \implies (t-1)(t-3) = 0 \implies t = 1, t=3$$
5. Find the second derivative of $v$ for the test:
6. $$\frac{d^2v}{dt^2} = 6t - 12$$
7. Test at $t=1$:
8. $$6(1) - 12 = -6 < 0 \implies \text{Maximum at } t=1$$
9. Test at $t=3$:
10. $$6(3) - 12 = 6 > 0 \implies \text{Minimum at } t=3$$

*Calculator:* allowed

## Acceleration as a Function of Displacement

**Chain Rule Identity for Acceleration** — When acceleration is given as a function of displacement $s$, we use the chain rule to rewrite acceleration in an integrable form.

*Notation:* a = v \frac{dv}{ds}

Separating variables and integrating both sides gives the relationship between $v$ and $s$:

$$\int a ds = \int v dv$$

**Worked example:** A particle starts from rest at $s=0$, with acceleration $a = 2s + 1 \text{ ms}^{-2}$. Find $v$ when $s = 2 \text{ m}$. Assume the particle moves in the positive direction.

1. Substitute into the identity $a = v \frac{dv}{ds}$:
2. $$v \frac{dv}{ds} = 2s + 1$$
3. Separate variables and integrate with initial bounds:
4. $$\int_{0}^{v} v dv = \int_{0}^{s} (2s + 1) ds$$
5. Evaluate both integrals:
6. $$\left[ \frac{1}{2}v^2 \right]_0^v = \left[ s^2 + s \right]_0^s \implies \frac{1}{2}v^2 = s^2 + s$$
7. Substitute $s=2$:
8. $$\frac{1}{2}v^2 = 4 + 2 = 6 \implies v^2 = 12 \implies v = 2\sqrt{3} \text{ ms}^{-1}$$

> **exam_tip**
>
> This identity is one of the most frequently examined concepts in this topic, so memorise it!

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using SUVAT formulas for problems with variable acceleration
  - Why it fails: SUVAT only works for constant acceleration, it will always give the wrong answer for non-constant acceleration
  - Correct: Always use calculus (differentiation/integration) if acceleration is not stated to be constant
- **Wrong:** Forgetting to calculate the constant of integration
  - Why it fails: Most problems have non-zero initial conditions, so omitting the constant gives an incorrect final expression
  - Correct: Find the constant of integration immediately after integrating, using given initial conditions
- **Wrong:** Using $a = \frac{d^2s}{dt^2}$ when acceleration is a function of $s$
  - Why it fails: This results in a second-order differential equation that is difficult to solve at A-Level
  - Correct: Always use the chain rule identity $a = v \frac{dv}{ds}$ when acceleration depends on displacement
- **Wrong:** Skipping the test to confirm if a stationary point is maximum/minimum
  - Why it fails: Examiners require you to justify the nature of the turning point, so you will lose a mark for skipping the test
  - Correct: Always use the second derivative test to confirm whether a stationary point is a maximum or minimum

## Cheatsheet

| Relationship | Formula |
| --- | --- |
| Velocity from displacement | $v = \frac{ds}{dt}$ |
| Acceleration from velocity | $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ |
| Velocity from acceleration | $v = \int a \; dt + C$ |
| Displacement from velocity | $s = \int v \; dt + C$ |
| $a$ as function of $s$ | $a = v \frac{dv}{ds}$ |
| Max/min of $v$ occurs at | $a = 0$ |
| Max/min of $s$ occurs at | $v = 0$ |

## What's next

Kinematics is the foundation for all mechanics topics in CIE A-Level 9709, as it describes how objects move, which you will extend to connect to forces, energy and momentum. Mastery of calculus-based kinematics is essential for almost all applied mechanics questions on the exam, so practice integrating and differentiating motion functions until you are confident.

- [Momentum](https://www.owlsprep.com/study/cie-9709-u3-momentum/)
- [Forces and Equilibrium](https://www.owlsprep.com/study/cie-9709-u3-forces-and-equilibrium/)
- [Newton's laws of motion](https://www.owlsprep.com/study/cie-9709-u3-newton-s-laws-of-motion/)

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