Kinematics

CIE A-Level Mathematics· 45 min read

1. Relationships Between Motion Quantities★★☆☆☆⏱ 15 min

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📘 Definition

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, $a = 6 t - 2$ , 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 $a = 6 t - 2 ms^{- 2}$ at time $t$ . Given that $v = 4 ms^{- 1}$ and $s = 0 m$ 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
$s = t^3 - t^2 + 4t$

2. Maxima and Minima of Motion Quantities★★★☆☆⏱ 20 min

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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 $v = \frac{d s}{d t} = 0$
Maximum/minimum velocity occurs when $a = \frac{d v}{d t} = 0$

📐 Worked Example

A particle's velocity is given by $v = t^{3} - 6 t^{2} + 9 t ms^{- 1}$ 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
$6(3) - 12 = 6 > 0 \implies \text{Minimum at } t=3$

3. Acceleration as a Function of Displacement★★★★☆⏱ 25 min

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📘 Definition

Chain Rule Identity for Acceleration

a = v \frac{dv}{ds}

When acceleration is given as a function of displacement $s$ , we use the chain rule to rewrite acceleration in an integrable form.

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 $a = 2 s + 1 ms^{- 2}$ . Find $v$ when $s = 2 m$ . Assume the particle moves in the positive direction.

1
Substitute into the identity $a = v \frac{d v}{d s}$ :
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 $s = 2$ :
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$\frac{1}{2}v^2 = 4 + 2 = 6 \implies v^2 = 12 \implies v = 2\sqrt{3} \text{ ms}^{-1}$

4. Common Pitfalls

Wrong move:

Using SUVAT formulas for problems with variable acceleration

Why:

SUVAT only works for constant acceleration, it will always give the wrong answer for non-constant acceleration

Correct move:

Always use calculus (differentiation/integration) if acceleration is not stated to be constant

Wrong move:

Forgetting to calculate the constant of integration

Why:

Most problems have non-zero initial conditions, so omitting the constant gives an incorrect final expression

Correct move:

Find the constant of integration immediately after integrating, using given initial conditions

Wrong move:

Using $a = \frac{d ^{2} s}{d t ^{2}}$ when acceleration is a function of $s$

Why:

This results in a second-order differential equation that is difficult to solve at A-Level

Correct move:

Always use the chain rule identity $a = v \frac{d v}{d s}$ when acceleration depends on displacement

Wrong move:

Skipping the test to confirm if a stationary point is maximum/minimum

Why:

Examiners require you to justify the nature of the turning point, so you will lose a mark for skipping the test

Correct move:

Always use the second derivative test to confirm whether a stationary point is a maximum or minimum

5. Quick Reference Cheatsheet

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

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 · 33
Variable acceleration problem
2022 · 31
Max/min velocity problem
2021 · 32
a as function of s problem

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.