Mathematics · AQA 8300 §R3
Rates of Change
- Solve speed, distance, time problems including multi-step and unit conversion questions
- Calculate density and pressure in context, including composite objects
- Convert between different units of speed (mph, km/h, m/s) and area (cm², m²)
- Interpret the gradient of distance–time graphs as speed and velocity–time graphs
- Calculate the area under a velocity–time graph to find distance travelled
🔑 Core Concepts
Speed, Distance and Time
A rate compares two different quantities — speed compares distance to time. Understanding rates lets you predict how far an object travels, how long a journey takes, or how fast something moves. The three quantities are linked by a single relationship that can be rearranged for any unknown.
DEFINITION — Speed
Speed is the distance travelled per unit of time. It tells you how quickly an object covers distance. Average speed uses total distance and total time; instantaneous speed is the speed at a single moment.The SDT Triangle
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \qquad \text{Distance} = \text{Speed} \times \text{Time} \qquad \text{Time} = \frac{\text{Distance}}{\text{Speed}}$$
$v$ = speed (m/s or km/h)
$d$ = distance (m or km)
$t$ = time (s or h)
COMMON MISTAKE
Always check units are consistent before substituting into the formula. If distance is in km and time is in minutes, convert minutes to hours first, or your speed will be in km/min rather than km/h.EXAM TIP — Average vs Instantaneous Speed
Average speed = total distance ÷ total time. On a distance–time graph, average speed over an interval is the gradient of the chord; instantaneous speed at a point is the gradient of the tangent drawn at that point.Unit Conversions for Speed
The conversion between m/s and km/h comes up very frequently. Knowing the factor removes a major source of error.
Speed Unit Conversions
$$1 \text{ m/s} = 3.6 \text{ km/h} \qquad 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \approx 0.2\overline{7} \text{ m/s}$$
To convert m/s → km/h: multiply by 3.6
To convert km/h → m/s: divide by 3.6
MEMORY TRICK — The Factor 3.6
There are 1000 m in 1 km and 3600 s in 1 hour. So $1 \text{ m/s} = \frac{1 \text{ m}}{1 \text{ s}} = \frac{1 \text{ m} \times 3600}{1000 \text{ m}} \text{ km/h} = 3.6 \text{ km/h}$. Just remember 3600 ÷ 1000 = 3.6.Area Unit Conversions
Area Conversions
$$1 \text{ m}^2 = 10\,000 \text{ cm}^2 \qquad 1 \text{ cm}^2 = 0.0001 \text{ m}^2$$
$$1 \text{ km}^2 = 1\,000\,000 \text{ m}^2$$
$1 \text{ m} = 100 \text{ cm}$, so $1 \text{ m}^2 = 100^2 \text{ cm}^2 = 10{,}000 \text{ cm}^2$
EXAM TIP — Squaring and Cubing Conversions
When converting area, square the linear conversion factor. When converting volume, cube it. So 1 m = 100 cm → 1 m² = 100² = 10,000 cm² → 1 m³ = 100³ = 1,000,000 cm³.Density
Density is a compound measure that describes how much mass is packed into a given volume. A denser material has more mass in the same space. Gold is much denser than wood — the same volume of gold has far greater mass.
DEFINITION — Density
Density is mass per unit volume. It is a property of the material itself, not of the size of the object.Density Formula
$$\rho = \frac{m}{V} \qquad m = \rho V \qquad V = \frac{m}{\rho}$$
$\rho$ (rho) = density (g/cm³ or kg/m³)
$m$ = mass (g or kg)
$V$ = volume (cm³ or m³)
COMMON MISTAKE — Mixed Units
If density is given in g/cm³, then mass must be in grams and volume in cm³. Never mix g/cm³ with kg or m³ without converting first.EXAM TIP — Composite Objects
For composite objects (e.g. two materials joined), find the total mass (mass₁ + mass₂) and total volume (V₁ + V₂), then use $\rho = m/V$. The composite density is not the average of the two individual densities.Pressure
Pressure measures how concentrated a force is. A stiletto heel exerts much higher pressure than a flat shoe, even if the person weighs the same, because the force acts on a tiny area.
DEFINITION — Pressure
Pressure is the force per unit area acting perpendicular to a surface.Pressure Formula
$$P = \frac{F}{A} \qquad F = PA \qquad A = \frac{F}{P}$$
$P$ = pressure (Pa = N/m² or N/cm²)
$F$ = force (N, Newtons)
$A$ = area (m² or cm²)
COMMON MISTAKE — Area Units in Pressure
Pressure in Pa uses Newtons and square metres (N/m²). If area is given in cm², convert to m² before using Pa. Or work in N/cm² consistently throughout.EXAM TIP
In context questions, identify which quantity changes: if force stays the same but area doubles, pressure halves. Use proportional reasoning to check your answer makes sense.Gradient as a Rate of Change
On any graph, the gradient (slope) represents how one quantity changes relative to another. This is a key Grade 9 concept: gradient has real-world meaning, not just mathematical meaning.
DEFINITION — Gradient on a Distance–Time Graph
The gradient of a distance–time graph equals the speed. A steeper line = greater speed. A horizontal line = stationary object (zero speed). A negative gradient = returning towards the starting point.Gradient of a Distance–Time Graph
$$\text{Speed} = \text{gradient} = \frac{\Delta d}{\Delta t} = \frac{d_2 - d_1}{t_2 - t_1}$$
$\Delta d$ = change in distance (vertical change)
$\Delta t$ = change in time (horizontal change)
EXAM TIP — Curved Distance–Time Graphs
If the curve is steepening, the object is accelerating. If it is flattening, it is decelerating. To find instantaneous speed at a point, draw a tangent to the curve at that point and calculate its gradient.Area Under a Velocity–Time Graph
Whilst gradient gives rate of change, the area under a graph also encodes physical meaning. For a velocity–time graph, the area enclosed between the graph and the time axis equals the distance (or displacement) travelled.
DEFINITION — Area Under a v–t Graph
The area under a velocity–time graph between two points in time equals the distance travelled during that time interval. Area above the time axis = positive displacement; area below = negative displacement (returning).Area Under v–t Graph
$$\text{Distance} = \int_{t_1}^{t_2} v \, dt = \text{area between graph and } t\text{-axis}$$
Rectangle: area = base × height = $t \times v$
Triangle: area = $\frac{1}{2} \times \text{base} \times \text{height}$
Trapezium: area = $\frac{1}{2}(v_1 + v_2) \times t$
EXAM TIP — Splitting Irregular Areas
Split complex v–t graph shapes into rectangles and triangles. Add areas of each sub-shape. For curved sections, count squares or use the trapezium rule.COMMON MISTAKE — Gradient vs Area Confusion
Students often confuse what gradient and area mean on motion graphs. Remember: gradient of v–t graph = acceleration (not distance); area under v–t graph = distance. Gradient of d–t graph = speed.🗺️ Visual Notes
Rates of Change
Speed, Distance, Time
- $v = d/t$; $d = vt$; $t = d/v$
- Average speed: total $d$ ÷ total $t$
- Instantaneous: gradient of tangent
- Units: m/s, km/h, mph
Density & Pressure
- $\rho = m/V$ (g/cm³ or kg/m³)
- $P = F/A$ (Pa = N/m²)
- Composite: total $m$ ÷ total $V$
- Watch unit consistency
Unit Conversions
- 1 m/s = 3.6 km/h
- 1 m² = 10,000 cm²
- 1 m³ = 1,000,000 cm³
- Square/cube the length factor
Distance–Time Graphs
- Gradient = speed
- Horizontal = stationary
- Steep = fast; shallow = slow
- Negative gradient = returning
Velocity–Time Graphs
- Gradient = acceleration
- Area under graph = distance
- Split into rectangles/triangles
- Area below axis = return journey
Comparison: Types of Rate
| Rate | Formula | SI Units | Graph meaning |
|---|---|---|---|
| Speed | $v = d/t$ | m/s | Gradient of d–t graph |
| Density | $\rho = m/V$ | kg/m³ | Gradient of m vs V graph |
| Pressure | $P = F/A$ | Pa (N/m²) | Gradient of F vs A graph |
| Acceleration | $a = \Delta v/t$ | m/s² | Gradient of v–t graph |
Graph Interpretation Decision Tree
What type of graph is it?
→
Distance–Time
Gradient = Speed
Horizontal = Stopped
Gradient = Speed
Horizontal = Stopped
→
Velocity–Time
Gradient = Acceleration
Area = Distance
Gradient = Acceleration
Area = Distance
→
Read off values using gradient formula: $\frac{\Delta y}{\Delta x}$
Unit Conversion Summary Table
| From | To | Multiply by | Why |
|---|---|---|---|
| m/s | km/h | 3.6 | ×3600 s/h ÷ 1000 m/km |
| km/h | m/s | ÷ 3.6 | ÷3600 × 1000 |
| cm² | m² | ÷ 10,000 | ÷100² |
| m² | cm² | × 10,000 | ×100² |
| cm³ | m³ | ÷ 1,000,000 | ÷100³ |
| mph | m/s | ÷ 2.237 | 1 mph = 0.4470 m/s |
✏️ Worked Examples
Grade 4–5
A car travels 150 km in 2.5 hours. Calculate the average speed of the car in km/h.
1
IdentifyIdentify what you are given and what you need to find.
Given: Distance $d = 150$ km, Time $t = 2.5$ h. Find: Speed $v$.
Given: Distance $d = 150$ km, Time $t = 2.5$ h. Find: Speed $v$.
2
FormulaWrite the formula needed: $$v = \frac{d}{t}$$
3
SubstituteSubstitute the known values: $$v = \frac{150}{2.5} = 60 \text{ km/h}$$
4
CheckCheck units: km ÷ h = km/h. ✓ The answer is reasonable for a car on a road.
Average speed = 60 km/h
Grade 6–7
A metal block has a mass of 540 g and a volume of 200 cm³. A second block of the same metal has a volume of 350 cm³. Calculate the mass of the second block.
1
Find densityUse the first block to find the density of the metal:
$$\rho = \frac{m}{V} = \frac{540}{200} = 2.7 \text{ g/cm}^3$$
2
Same materialSince both blocks are the same metal, they share the same density: $\rho = 2.7$ g/cm³.
3
Find massRearrange $\rho = m/V$ to get $m = \rho V$:
$$m = 2.7 \times 350 = 945 \text{ g}$$
4
CheckThe second block has greater volume than the first (350 > 200), so its mass should be greater (945 > 540). ✓
Mass of second block = 945 g
Grade 9
A velocity–time graph shows a vehicle starting from rest, accelerating uniformly to 20 m/s over 8 seconds, travelling at constant speed for 12 seconds, then decelerating uniformly to rest in 5 seconds.
(a) Find the total distance travelled.
(b) Convert the maximum speed to km/h.
(c) During the constant-speed phase, a second vehicle travels the same distance in half the time. What is the second vehicle's speed in m/s?
(a) Find the total distance travelled.
(b) Convert the maximum speed to km/h.
(c) During the constant-speed phase, a second vehicle travels the same distance in half the time. What is the second vehicle's speed in m/s?
1
Identify shapesThe v–t graph consists of three sections:
• Triangle (0 to 8 s): acceleration from 0 to 20 m/s
• Rectangle (8 to 20 s): constant 20 m/s for 12 s
• Triangle (20 to 25 s): deceleration from 20 m/s to 0
• Triangle (0 to 8 s): acceleration from 0 to 20 m/s
• Rectangle (8 to 20 s): constant 20 m/s for 12 s
• Triangle (20 to 25 s): deceleration from 20 m/s to 0
2
Area — triangle 1$$d_1 = \frac{1}{2} \times 8 \times 20 = 80 \text{ m}$$
3
Area — rectangle$$d_2 = 12 \times 20 = 240 \text{ m}$$
4
Area — triangle 2$$d_3 = \frac{1}{2} \times 5 \times 20 = 50 \text{ m}$$
5
Total distance (a)$$d_{\text{total}} = 80 + 240 + 50 = 370 \text{ m}$$
6
Unit conversion (b)$$20 \text{ m/s} \times 3.6 = 72 \text{ km/h}$$
7
Second vehicle (c)During the constant-speed phase, vehicle 1 travels $d_2 = 240$ m in 12 s. Vehicle 2 travels 240 m in half the time = 6 s:
$$v_2 = \frac{d}{t} = \frac{240}{6} = 40 \text{ m/s}$$
(a) Total distance = 370 m
(b) Maximum speed = 72 km/h
(c) Speed of second vehicle = 40 m/s
(b) Maximum speed = 72 km/h
(c) Speed of second vehicle = 40 m/s
❓ Exam Questions
Q11 mark
Write down the formula for speed in terms of distance and time.
Mark scheme:
$v = \dfrac{d}{t}$ or equivalent (distance ÷ time) — [1]
$v = \dfrac{d}{t}$ or equivalent (distance ÷ time) — [1]
Q22 marks
A cyclist travels at an average speed of 15 km/h for 40 minutes. Calculate the distance travelled. Give your answer in km.
Mark scheme:
Convert time: 40 min = 40/60 = 2/3 h — [1]
$d = v \times t = 15 \times \dfrac{2}{3} = 10$ km — [1]
Convert time: 40 min = 40/60 = 2/3 h — [1]
$d = v \times t = 15 \times \dfrac{2}{3} = 10$ km — [1]
Q33 marks
A rectangular steel plate has dimensions 50 cm × 80 cm. A force of 4000 N acts uniformly across one face of the plate. Calculate the pressure on the plate in N/m².
Mark scheme:
Area = 50 cm × 80 cm = 4000 cm² — [1]
Convert to m²: 4000 ÷ 10,000 = 0.4 m² — [1]
$P = F/A = 4000/0.4 = 10\,000$ N/m² (10 kPa) — [1]
Area = 50 cm × 80 cm = 4000 cm² — [1]
Convert to m²: 4000 ÷ 10,000 = 0.4 m² — [1]
$P = F/A = 4000/0.4 = 10\,000$ N/m² (10 kPa) — [1]
Q44 marks
A composite object is made by joining a copper block (mass 890 g, volume 100 cm³) and an aluminium block (volume 150 cm³, density 2.7 g/cm³). Calculate the overall density of the composite object.
Mark scheme:
Mass of aluminium = $\rho V = 2.7 \times 150 = 405$ g — [1]
Total mass = 890 + 405 = 1295 g — [1]
Total volume = 100 + 150 = 250 cm³ — [1]
Overall density = 1295 ÷ 250 = 5.18 g/cm³ — [1]
Mass of aluminium = $\rho V = 2.7 \times 150 = 405$ g — [1]
Total mass = 890 + 405 = 1295 g — [1]
Total volume = 100 + 150 = 250 cm³ — [1]
Overall density = 1295 ÷ 250 = 5.18 g/cm³ — [1]
Q53 marks
A car's speed is 25 m/s. Convert this speed to km/h. Show your working clearly.
Mark scheme:
Method: 25 m/s × 3.6 OR equivalent chain conversion — [1]
$25 \times 3.6 = 90$ — [1]
Answer: 90 km/h with correct units — [1]
Method: 25 m/s × 3.6 OR equivalent chain conversion — [1]
$25 \times 3.6 = 90$ — [1]
Answer: 90 km/h with correct units — [1]
Q66 marks
A train journey consists of three stages: Stage 1 — uniform acceleration from rest to 30 m/s over 60 seconds. Stage 2 — constant speed of 30 m/s for 5 minutes. Stage 3 — uniform deceleration from 30 m/s to rest over 40 seconds.
(a) Calculate the total distance of the journey in metres. [4]
(b) Calculate the average speed for the entire journey in m/s. Give your answer to 3 significant figures. [2]
Mark scheme:
(a)
Stage 1 (triangle): $\frac{1}{2} \times 60 \times 30 = 900$ m — [1]
Stage 2 (rectangle): Convert 5 min = 300 s; $300 \times 30 = 9000$ m — [1]
Stage 3 (triangle): $\frac{1}{2} \times 40 \times 30 = 600$ m — [1]
Total distance = 900 + 9000 + 600 = 10,500 m — [1]
(b)
Total time = 60 + 300 + 40 = 400 s — [1]
Average speed = $\frac{10500}{400} = 26.25 \approx$ 26.3 m/s (3 s.f.) — [1]
(a)
Stage 1 (triangle): $\frac{1}{2} \times 60 \times 30 = 900$ m — [1]
Stage 2 (rectangle): Convert 5 min = 300 s; $300 \times 30 = 9000$ m — [1]
Stage 3 (triangle): $\frac{1}{2} \times 40 \times 30 = 600$ m — [1]
Total distance = 900 + 9000 + 600 = 10,500 m — [1]
(b)
Total time = 60 + 300 + 40 = 400 s — [1]
Average speed = $\frac{10500}{400} = 26.25 \approx$ 26.3 m/s (3 s.f.) — [1]
⭐ Grade 9 Model Answers
Full Annotated Answer — Q6 (6-mark velocity–time graph question)
WORKED EXAMPLE — Q6 Full Model Answer
Part (a): Total distance
A Grade 9 student identifies each section of the journey as a geometric shape in the v–t graph:
Stage 1 — triangle (acceleration): $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 60 \times 30 = 900 \text{ m}$
Stage 2 — rectangle (constant speed): Convert 5 min → 300 s first. $\text{Area} = 300 \times 30 = 9000 \text{ m}$
Stage 3 — triangle (deceleration): $\text{Area} = \frac{1}{2} \times 40 \times 30 = 600 \text{ m}$
Total distance $= 900 + 9000 + 600 = \mathbf{10\,500 \text{ m}}$
Part (b): Average speed
Total time $= 60 + 300 + 40 = 400$ s
Average speed $= \dfrac{\text{total distance}}{\text{total time}} = \dfrac{10500}{400} = 26.25 \approx \mathbf{26.3 \text{ m/s}}$ (3 s.f.)
Why This Scores Full Marks
| Mark Point | What the model answer does |
|---|---|
| Stage 1 distance [1] | Correctly identifies the triangle shape and applies $\frac{1}{2}bh$ |
| Stage 2 distance [1] | Converts minutes to seconds before multiplying — crucial unit step |
| Stage 3 distance [1] | Correctly handles the deceleration triangle |
| Total distance [1] | Adds all three distances correctly |
| Total time [1] | Sums all three time intervals (in seconds) |
| Average speed (3 s.f.) [1] | Uses $v = d/t$ with correct values, rounds properly to 3 s.f. |
IMPORTANT — Why Grade 9 Students Excel Here
A Grade 9 student does not just apply $v = d/t$ blindly. They identify that the graph is composed of geometric regions, that area = distance for v–t graphs, and that unit conversions (minutes to seconds) must happen before calculation. Each of these three insights is tested by the mark scheme.📋 Revision Sheet
Key Definitions
| Term | Definition |
|---|---|
| Speed | Distance per unit time |
| Average speed | Total distance ÷ total time |
| Instantaneous speed | Speed at a single moment (tangent gradient) |
| Density | Mass per unit volume |
| Pressure | Force per unit area |
| Compound measure | A rate comparing two different units |
Essential Formulae
$$v = \frac{d}{t} \quad d = vt \quad t = \frac{d}{v}$$
$$\rho = \frac{m}{V} \quad m = \rho V \quad V = \frac{m}{\rho}$$
$$P = \frac{F}{A} \quad F = PA \quad A = \frac{F}{P}$$
$$1 \text{ m/s} = 3.6 \text{ km/h}$$
$$1 \text{ m}^2 = 10{,}000 \text{ cm}^2$$
Gradient of d–t graph $= $ speed
Area under v–t graph $= $ distance
Memory Hooks
- SDT triangle: Cover the one you want — D over T gives Speed; D over S gives Time
- 3.6: m/s → km/h multiply by 3.6 (think: 3600 ÷ 1000)
- Density triangle: M over (D × V); cover M → density times volume
- Area² / Volume³: When converting units of area, square the factor; for volume, cube it
- Gradient → speed, Area → distance on v–t graphs
- Composite density: Don't average — total mass ÷ total volume
Exam Tips
- Always convert units to match before calculating
- For multi-step problems, write out what you know and what you need
- Show the rearrangement step — it earns marks
- Label area shapes on v–t graphs before adding
- For 3 s.f., count from the first non-zero digit
- Check: does your answer make physical sense?
- If the graph is curved, you need a tangent for instantaneous rate
🔄 Flashcards
Click a card to reveal the answer. Work through all 15 before your exam.
✗ Common Mistakes
MISTAKE 1 — Not converting time to hours
What students do wrong: Using 45 minutes directly in $v = d/t$ without converting to 0.75 hours.Why marks are lost: The speed formula requires consistent units; 45 ÷ 60 = 0.75 h must be written and used.
How to avoid: Always write out the unit of every quantity before substituting. If $v$ is in km/h, time must be in hours.
MISTAKE 2 — Confusing gradient and area on motion graphs
What students do wrong: Calculating the area under a distance–time graph and claiming it is the speed, or finding the gradient of a velocity–time graph and claiming it is the distance.Why marks are lost: Fundamental misapplication of graph interpretation.
How to avoid: Memorise the pair: gradient of d–t = speed; area under v–t = distance. Gradient of v–t = acceleration.
MISTAKE 3 — Averaging densities for composite objects
What students do wrong: Writing $\rho_{\text{composite}} = (\rho_1 + \rho_2)/2$.Why marks are lost: This is mathematically incorrect unless both volumes are equal.
How to avoid: Always find total mass and total volume separately, then divide: $\rho = m_{\text{total}}/V_{\text{total}}$.
MISTAKE 4 — Forgetting to square the factor when converting area
What students do wrong: Converting 1 m² to 100 cm² instead of 10,000 cm².Why marks are lost: The pressure calculation then uses the wrong area, giving wrong answer throughout.
How to avoid: Write $1 \text{ m} = 100 \text{ cm}$, then square: $1 \text{ m}^2 = 100^2 \text{ cm}^2 = 10{,}000 \text{ cm}^2$.
MISTAKE 5 — Multiplying instead of dividing when converting km/h to m/s
What students do wrong: Multiplying by 3.6 to get m/s from km/h (should divide by 3.6).Why marks are lost: The converted value is wrong by a factor of 12.96.
How to avoid: m/s is slower-sounding than km/h, so the number should be smaller. 72 km/h = 20 m/s (÷ 3.6). If your answer is bigger after "converting", you've gone the wrong way.
MISTAKE 6 — Not identifying all shapes in a velocity–time graph
What students do wrong: Only calculating the area of the rectangle (constant speed phase) and missing the triangles for acceleration and deceleration phases.Why marks are lost: Loses at least 2 marks — one for each missing triangle area.
How to avoid: Before calculating, mark every distinct phase on the graph with a coloured pen. Label each region as rectangle or triangle. Then calculate and add each area systematically.
✅ Final Checklist
Click each item as you master it. Your progress is saved automatically.
- I can state and use the formula $v = d/t$ and rearrange it for $d$ and $t$
- I can distinguish between average speed and instantaneous speed
- I can convert between m/s and km/h by multiplying or dividing by 3.6
- I can convert between cm² and m² by multiplying or dividing by 10,000
- I can calculate density using $\rho = m/V$ and rearrange for mass or volume
- I can find the density of a composite object using total mass and total volume
- I can calculate pressure using $P = F/A$ and rearrange for force or area
- I can find the gradient of a straight-line distance–time graph and state it as a speed
- I know that the gradient of a curved distance–time graph requires a tangent at that point
- I can find the area under a velocity–time graph by splitting it into rectangles and triangles
- I can calculate the total distance from a multi-phase velocity–time graph
- I can solve multi-step problems that require unit conversion as an intermediate step
- I know that gradient of v–t graph = acceleration (not distance)
- I can apply the trapezium formula $\frac{1}{2}(v_1 + v_2)t$ for distance under a uniform acceleration/deceleration
- I check that my final answer has the correct units and is physically reasonable