Real Life Graphs · Study Notes · Edexcel IGCSE 4MA1 Higher

Distance-time, speed-time, and conversion graphs. Higher Tier expects gradient and area interpretations.

At a glance

Distance-time: gradient = SPEED.
Speed-time: gradient = ACCELERATION; AREA = DISTANCE.
Horizontal distance-time = stationary.
Conversion graphs: linear, read off values.
Always check units (km/h vs m/s).
Break complex areas into simpler shapes.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.5 — Interpret distance-time graphs.
3.5 — Interpret speed-time graphs (gradient + area).
3.5 — Use conversion graphs.
3.5 — Apply graphs to real-world contexts.

Distance-time graphs

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Gradient = speed. Horizontal = stationary. Steeper = faster.

Axes: time on $x$ , distance on $y$ .

Reading:

Steeper gradient = faster speed.
Horizontal line = stationary (not moving).
Negative gradient = returning.

Speed = gradient = $\dfrac{\Delta d}{\Delta t}$ .

Example. Travel $30$ km in $0.5$ hours.

Speed = $30 / 0.5 = 60$ km/h.

Multi-stage journeys. Each section has its own gradient = its own speed.

Worked qualitative. A distance-time graph shows:

$0 \to 0.5$ h: rises to $30$ km.
$0.5 \to 1$ h: horizontal at $30$ km.
$1 \to 1.5$ h: rises to $50$ km.

What's the journey?

First $0.5$ h: $60$ km/h.
Middle $0.5$ h: stopped (rest).
Last $0.5$ h: $20/0.5 = 40$ km/h.
Total: $50$ km in $1.5$ h, average $33.3$ km/h.

On a distance-time graph the gradient of each segment is the speed; a flat line means at rest.

Edexcel tip. Always state UNITS with speed (km/h or m/s). Mark schemes credit unit consistency.

Time x, distance y.
Gradient = speed.
Horizontal = stationary.
Average speed = total distance / total time.

Speed-time graphs

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Gradient = acceleration. Area under = distance.

Axes: time on $x$ , speed on $y$ .

Two key relationships:

Gradient = acceleration: $a = \dfrac{\Delta v}{\Delta t}$

Units: m/s² (or km/h²).

Area under graph = distance:

Distance = sum of areas of rectangles, triangles, trapeziums.

Example. Speed accelerates from $0$ to $20$ m/s in $5$ s, then constant for $10$ s.

Acceleration phase (triangle):

Area = $\dfrac{1}{2} \times 5 \times 20 = 50$ m.
Gradient = $20/5 = 4$ m/s².

Constant phase (rectangle):

Area = $10 \times 20 = 200$ m.
Gradient = $0$ (no acceleration).

Total distance: $50 + 200 = 250$ m.

Split the graph into a triangle and rectangle; their areas add to the total distance travelled.

Constant deceleration. Negative gradient. Triangle on the right showing speed dropping.

Worked qualitative. A car decelerates from $30$ m/s to $0$ in $6$ s. Distance covered?

Average speed during deceleration: $(30 + 0)/2 = 15$ m/s.
Distance: $15 \times 6 = 90$ m.
Or as triangle area: $\dfrac{1}{2} \times 6 \times 30 = 90$ m. ✓

Edexcel tip. ALWAYS identify whether you're using gradient or area. State 'area = distance' or 'gradient = acceleration' explicitly.

Gradient = acceleration.
Area = distance.
Triangle: $\dfrac{1}{2} \times$ base × height.
Rectangle: base × height.
Trapezium: $\dfrac{1}{2}(a + b)h$ .

Conversion graphs

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Linear graphs converting between units. Read off values directly.

Conversion graphs are usually straight lines through origin.

Examples:

Currency (£ to USD).
Temperature (Celsius to Fahrenheit, though this has y-intercept).
Distance (km to miles).
Mass (kg to pounds).

Method:

Identify what each axis represents.
Read off the value from the line.
(Or use the GRADIENT as a multiplier.)

Example. £1 = $1.20$ USD shown as a line through origin and $(1, 1.20)$ .

Read up from a pounds value to the line, then across to the dollar value — the gradient is the rate.

Convert £ $80$ to USD:

Use: USD = $1.20 \times$ GBP.
$1.20 \times 80 = 96$ USD.

Reverse conversion. Read the other direction.

$50$ USD to GBP:

$50 / 1.20 = £41.67$ .

Worked qualitative. Why is Celsius-to-Fahrenheit NOT through origin?

Formula: F = (9/5)C + 32.
y-intercept is 32, not 0.
0°C = 32°F.
Linear, but with shift.

Edexcel tip. Distinguish between graphs through origin (just multiply) vs graphs with y-intercept (use full line equation).

Linear conversions.
Through origin: multiply by gradient.
Otherwise: use line equation.
Read values directly from graph.

Quick recap

Distance-time: gradient = speed.
Speed-time: gradient = acceleration, area = distance.
Horizontal = stationary (distance-time) or constant speed (speed-time).
Conversion graphs: linear, multiply by rate.
Always state units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Distance-time: gradient = speed.
Speed-time: gradient = a; AREA = distance.
Conversion: rate as multiplier.
Triangle area = $\frac{1}{2}bh$ .
Trapezium = $\frac{1}{2}(a + b)h$ .

How it’s examined

Real-life graphs every Higher Tier paper (4-7 marks). Often combined with area calculations. Examiner reports flag (1) confusing graph types, (2) wrong units, (3) not breaking complex shapes into simpler parts.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on real life graphs, written exactly the way a tutor would explain them at the board.

1Distance-time graph: speed from gradient
Foundation• distance-time
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Question
A distance-time graph shows a journey: $0$ to $30$ km in $0.5$ hours. Find the speed.
Step-by-step solution
1. Step 1
  On a distance-time graph, GRADIENT = SPEED.
2. Step 2
  $\text{Speed} = \dfrac{\Delta d}{\Delta t} = \dfrac{30}{0.5} = 60$ km/h.
Answer
$60$ km/h
2Speed-time graph: distance from area
Higher• Adapted from 4MA1/2H May/Jun 2024 Q17• speed-time
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Question
A speed-time graph shows: speed accelerates from $0$ to $20$ m/s in $5$ s, then constant for $10$ s. Find the total distance travelled.
Step-by-step solution
1. Step 1
  On speed-time graph, AREA = DISTANCE.
2. Step 2
  Acceleration phase (triangle): $\dfrac{1}{2} \times 5 \times 20 = 50$ m.
3. Step 3
  Constant phase (rectangle): $10 \times 20 = 200$ m.
4. Step 4
  Total: $250$ m.
Answer
$250$ m
Examiner tip
Distance-time → gradient = speed. Speed-time → area = distance, gradient = acceleration. Memorise.
3Acceleration from speed-time graph
Higher• acceleration
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Question
A speed-time graph shows speed change from $5$ m/s to $20$ m/s in $3$ seconds. Find the acceleration.
Step-by-step solution
1. Step 1
  Acceleration = gradient of speed-time = $\dfrac{\Delta v}{\Delta t}$ .
2. Step 2
  $a = \dfrac{20 - 5}{3} = \dfrac{15}{3} = 5$ m/s².
Answer
$5$ m/s²
4Conversion graph
Foundation• conversion
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Question
A conversion graph shows £1 = $1.20$ USD. Convert £ $80$ to USD using the graph principle.
Step-by-step solution
1. Step 1
  Linear conversion: USD = $1.20 \times$ GBP.
2. Step 2
  $1.20 \times 80 = 96$ USD.
Answer
$96$ USD
5Matching a flask to its depth–time graph [1 mark]
Getting started• Paper 1H/2H style — 1 mark• real-life graphs, filling containers
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Question
Water is poured at a constant rate into a conical flask. The flask is wide at the bottom and narrow at the top.

Which description fits the graph of the depth of water against time?

A: a straight line through the origin

B: a curve that gets steeper as time goes on

C: a curve that gets less steep as time goes on

Write down the letter of the correct description. [1]
Step-by-step solution
1. Step 1
  The same volume goes in every second. Near the bottom the flask is wide, so that volume makes only a small rise in depth. Higher up the flask is narrower, so the same volume makes a bigger rise.
2. Step 2
  B1 The depth rises faster and faster, so the graph gets steeper.
Answer
B
Examiner tip
Examiner tip: Think about how wide the container is at each height. Wide means the depth rises slowly (shallow graph); narrow means it rises quickly (steep graph). Straight lines only come from parts with vertical sides.

Common mistake: Choosing C because the flask "gets smaller". It is the width that gets smaller, so the rate of rise gets larger.
6Reading a step graph for parking charges [2 marks]
Getting started• Paper 1H/2H style — 2 marks• real-life graphs, step graphs
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Question
A car park shows its charges as a step graph of the charge, £ $C$ , against the time parked, $t$ hours. The graph is made of three horizontal line segments:

$C = 2$ for $0 < t \leqslant 1$

$C = 5$ for $1 < t \leqslant 3$

$C = 9$ for $3 < t \leqslant 6$

Each segment has a filled dot at its right-hand end (included) and an open dot at its left-hand end (not included).

(a) Write down the charge for parking for exactly $3$ hours. [1]

(b) Priya parks for $50$ minutes on Monday and for $3$ hours $20$ minutes on Tuesday. Work out her total charge for the two days. [1]
Step-by-step solution
1. Step 1
  B1 (a) At $t = 3$ the filled dot is at the right-hand end of the segment $C = 5$ . The open dot at $t = 3$ on the segment $C = 9$ means $3$ hours is not in that step.
  $C = 5$
2. Step 2
  B1 (b) $50$ minutes is less than $1$ hour, so it costs £2. $3$ hours $20$ minutes is more than $3$ hours, so it costs £9.
  $2 + 9 = 11$
Answer
(a) £5 (b) £11
Examiner tip
Examiner tip: On a step graph, the filled dot tells you which step a boundary value belongs to. Compare minutes with $60$ (or change them into hours) before you read the graph.

Common mistake: Giving £9 for exactly $3$ hours, because the higher step starts at $t = 3$ . The open dot there means $3$ hours is not in that step.
7Completing a distance–time graph to find an arrival time [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• real-life graphs, distance-time
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Question
Kofi leaves home at 14:00 and cycles to a friend's house. His distance–time graph goes from (14:00, 0 km) to (14:40, 12 km). He stays until 14:55, then cycles straight home at a constant speed of $18$ km/h.

Work out the time at which Kofi arrives home. [3]
Step-by-step solution
1. Step 1
  M1 Time for the $12$ km home journey at $18$ km/h.
  $\frac{12}{18} = \frac{2}{3} \text{ hour}$
2. Step 2
  M1 Change to minutes.
  $\frac{2}{3} \times 60 = 40 \text{ minutes}$
3. Step 3
  A1 Add $40$ minutes to the time he leaves his friend's house.
  $14{:}55 + 40 \text{ min} = 15{:}35$
Answer
15:35
Examiner tip
Examiner tip: On the graph, the journey home is a straight line from (14:55, 12) down to (15:35, 0). Its gradient has size $12$ km in $\tfrac{2}{3}$ hour, which is $18$ km/h, so the answer checks.

Common mistake: Writing $\tfrac{2}{3}$ hour as $0.67$ hours and then as $67$ minutes, or as $0.67$ added to the clock time. Convert a decimal of an hour by multiplying by $60$ .
8Extending a spring graph back to its intercept [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• real-life graphs, intercept, gradient
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Question
Masses are hung on a spring. The graph of the length of the spring, $L$ cm, against the mass, $m$ grams, is a straight line through $(100, 14)$ and $(400, 20)$ .

(a) Work out the length of the spring when no mass is hung on it. [2]

(b) Work out the mass that makes the spring $25$ cm long. [1]
Step-by-step solution
1. Step 1
  M1 (a) Find the gradient: the spring stretches $6$ cm for $300$ g.
  $\frac{20 - 14}{400 - 100} = \frac{6}{300} = 0.02 \text{ cm per gram}$
2. Step 2
  A1 Go back $100$ g from $(100, 14)$ to $m = 0$ . The length is $100 \times 0.02 = 2$ cm less.
  $14 - 2 = 12 \text{ cm}$
3. Step 3
  B1 (b) From $12$ cm the spring must stretch $25 - 12 = 13$ cm.
  $13 \div 0.02 = 650 \text{ g}$
Answer
(a) $12$ cm (b) $650$ g
Examiner tip
Examiner tip: The intercept is not always printed. Find the gradient first, then work back from a known point to $m = 0$ . Check (b) from the other point: from $(400, 20)$ the spring needs $5$ cm more, which is $250$ g more, and $400 + 250 = 650$ .

Common mistake: Taking $14$ cm as the natural length because it is the first value given, or assuming the line goes through the origin and working out $25 \div 0.02 = 1250$ g. The spring already has a length when nothing is hung on it.
9Deceleration and distance while slowing down [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• real-life graphs, speed-time, acceleration
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Question
A speed–time graph for a car has a straight-line section from $(0, 25)$ to $(6, 10)$ , where time is in seconds and speed is in m/s.

(a) Work out the deceleration of the car. [2]

(b) Work out the distance the car travels in these $6$ seconds. [2]
Step-by-step solution
1. Step 1
  M1 (a) The speed falls by $25 - 10 = 15$ m/s in $6$ seconds.
  $\frac{15}{6}$
2. Step 2
  A1
  $= 2.5 \text{ m/s}^2$
3. Step 3
  M1 (b) The area under this section is a trapezium with parallel sides $25$ and $10$ and width $6$ .
  $\tfrac{1}{2}(25 + 10) \times 6$
4. Step 4
  A1
  $= 105 \text{ m}$
Answer
(a) $2.5$ m/s² (b) $105$ m
Examiner tip
Examiner tip: Gradient gives acceleration; area gives distance. The car does not stop, so the shape under the line is a trapezium, not a triangle.

Common mistake: Using a triangle, $\tfrac{1}{2} \times 6 \times 25 = 75$ m, or multiplying the final speed by the time, $10 \times 6 = 60$ m. Neither uses the whole area under the line.
10When an accelerating car catches a van [5 marks]
Stretch• Paper 1H/2H style — 5 marks• real-life graphs, speed-time, problem solving
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Question
At time $t = 0$ a car starts from rest at a junction just as a van passes it. The van travels at a constant $16$ m/s. The car's speed–time graph is a straight line from $(0, 0)$ to $(10, 20)$ , and then the car travels at a constant $20$ m/s. Both vehicles travel in the same direction along the same road.

Work out the time at which the car catches up with the van. [5]
Step-by-step solution
1. Step 1
  M1 Distance travelled by the car in the first $10$ seconds is the area of a triangle.
  $\tfrac{1}{2} \times 10 \times 20 = 100 \text{ m}$
2. Step 2
  M1 After $10$ seconds the car adds $20$ m every second. For $t \geqslant 10$ , the car has travelled:
  $100 + 20(t - 10) = 20t - 100$
3. Step 3
  M1 The van has travelled $16t$ metres. Set the distances equal.
  $20t - 100 = 16t$
4. Step 4
  M1 Solve.
  $4t = 100$
5. Step 5
  A1 Check the car has not caught up in the first $10$ s: at $t = 10$ the car has gone $100$ m and the van $160$ m, so it has not.
  $t = 25 \text{ seconds}$
Answer
$25$ seconds after the start
Examiner tip
Examiner tip: "Catches up" means the two distances are equal, so write each distance as an expression in $t$ . At $t = 25$ both have gone $400$ m, which confirms the answer.

Common mistake: Finding when the speeds are equal ( $t = 8$ , when the car reaches $16$ m/s). At that moment the van is furthest ahead, not level with the car.

Question 1
Paper 1H/2H style — 1 mark1 mark
Part of a speed–time graph for a cyclist is a horizontal line segment.

Write down what this part of the graph shows about the cyclist. [1]
Model answer
The speed does not change during this time.

Answer: the cyclist is travelling at a constant speed (the acceleration is zero)
Why this scores
Mark scheme:
- B1 for constant speed oe
Why this scores full marks: The candidate reads the vertical axis as speed. "The cyclist has stopped" is the answer for a horizontal section on a distance–time graph, and it scores 0 here.
Question 2
Paper 1H/2H style — 2 marks2 marks
A conversion graph for speeds is a straight line through $(0, 0)$ and $(90, 25)$ , with km/h on the horizontal axis and m/s on the vertical axis.

(a) Change $54$ km/h to m/s. [1]

(b) Change $30$ m/s to km/h. [1]
Model answer
(a) $54 \times \frac{25}{90} = 15$

Answer: $15$ m/s

(b) $30 \times \frac{90}{25} = 108$

Answer: $108$ km/h
Why this scores
Mark scheme:
- B1 for $15$
- B1 for $108$
Why this scores full marks: The line passes through the origin, so every value scales in the ratio $90 : 25$ . For (b), $30$ m/s is beyond the end of the graph, and scaling handles it. A check: $1$ m/s is $3.6$ km/h, and $30 \times 3.6 = 108$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
A car approaches a junction. Its distance–time graph is a smooth curve through $(0, 0)$ , $(4, 72)$ , $(10, 150)$ , $(16, 192)$ and $(20, 200)$ , where time is in seconds and distance is in metres. The curve gets less steep as time goes on, and after $t = 20$ the graph is a horizontal line.

(a) Describe how the speed of the car changes between $t = 0$ and $t = 20$ . [1]

(b) Write down what the car is doing after $t = 20$ . [1]
Model answer
(a) The gradient of a distance–time graph is the speed. The curve gets less steep, so the speed decreases.

Answer: the car is slowing down

(b) The graph is horizontal, so the distance does not change.

Answer: the car has stopped (it is stationary)
Why this scores
Mark scheme:
- B1 for the speed decreases oe (e.g. the car is slowing down or decelerating)
- B1 for stopped oe (stationary, or speed is $0$ )
Why this scores full marks: The candidate reads the steepness of the curve, not its height. "The car goes further" describes the distance, not the speed, and scores 0 in (a). The given points confirm it: the car covers $72$ m in the first $4$ seconds but only $8$ m in the last $4$ seconds.
Question 4
Paper 1H/2H style — 3 marks3 marks
A candle is $24$ cm tall when it is lit. After $2$ hours it is $18$ cm tall. The graph of its height against time is a straight line.

(a) Work out the gradient of the line. [1]

(b) Write down a formula for the height, $h$ cm, of the candle $t$ hours after it is lit. [1]

(c) Work out how long the candle burns for in total. [1]
Model answer
(a) $\frac{18 - 24}{2 - 0} = -3$

Answer: $-3$

(b) The line starts at $24$ and has gradient $-3$ .

Answer: $h = 24 - 3t$

(c) $24 - 3t = 0 \Rightarrow t = 8$

Answer: $8$ hours
Why this scores
Mark scheme:
- B1 for $-3$
- B1 for $h = 24 - 3t$ oe
- B1 for $8$ hours
Why this scores full marks: The sign of the gradient matches the context (the candle gets shorter), the formula uses the intercept $24$ and the gradient $-3$ , and (c) finds where the line meets the time axis, when the height is $0$ . Writing $h = 24 + 3t$ would make the candle grow.
Question 5
Paper 1H/2H style — 3 marks3 marks
A train travels between two stations. Its speed–time graph is made of straight lines joining $(0, 0)$ , $(4, 90)$ , $(10, 90)$ and $(14, 0)$ , where time is in minutes and speed is in km/h.

Work out the distance between the two stations. Give your answer in kilometres. [3]
Model answer
The area under the graph is a trapezium with parallel sides $10 - 4 = 6$ minutes and $14$ minutes, and height $90$ km/h.

The speed is per hour, so change the times into hours: $6$ minutes $= 0.1$ hours and $14$ minutes $= \frac{14}{60}$ hours.

$\tfrac{1}{2}\left(0.1 + \frac{14}{60}\right) \times 90 = 15$

Answer: $15$ km
Why this scores
Mark scheme:
- M1 for a correct area method, e.g. $\tfrac{1}{2}(6 + 14) \times 90$ oe, or two triangles and a rectangle
- M1 for time in hours throughout, or their area $\div 60$
- A1 for $15$ cao
Why this scores full marks: The time axis is in minutes but the speed is per hour, so the candidate changes minutes into hours before multiplying. The area in the graph's own units is $900$ ; giving $900$ km is the usual slip, and a train cannot travel $900$ km in $14$ minutes.
Question 6
Paper 1H/2H style — 3 marks3 marks
A currency exchange charges a fixed fee of £4 and then changes the rest of the money at £1 = €1.15. The graph of euros received against pounds paid is a straight line through $(4, 0)$ and $(104, 115)$ .

(a) Give a reason why the graph does not pass through the origin. [1]

(b) Work out the number of euros received for £300. [2]
Model answer
(a) The first £4 is the fee, so paying £4 gives €0.

(b) $300 - 4 = 296$ pounds is changed.

$296 \times 1.15 = 340.4$

Answer: €340.40
Why this scores
Mark scheme:
- B1 for a reason referring to the fee, e.g. £4 buys no euros
- M1 for $(300 - 4) \times 1.15$ oe
- A1 for €340.40 (accept €340.4)
Why this scores full marks: The candidate takes the fee off before converting. $300 \times 1.15 = 345$ ignores the fee, and $345 - 4$ takes the fee off in the wrong currency.
Question 7
Paper 1H/2H style — 4 marks4 marks
A car is travelling at $18$ m/s. The driver sees a hazard and takes $0.8$ seconds to react, still travelling at $18$ m/s. The car then decelerates uniformly at $4.5$ m/s² until it stops.

By considering the speed–time graph, work out the total distance the car travels from when the driver sees the hazard until the car stops. [4]
Model answer
Braking time: $18 \div 4.5 = 4 \text{ s}$

The speed–time graph is a rectangle ( $0.8$ s at $18$ m/s) followed by a triangle ( $4$ s from $18$ m/s to $0$ ).

Reaction: $18 \times 0.8 = 14.4$

Braking: $\tfrac{1}{2} \times 4 \times 18 = 36$

$14.4 + 36 = 50.4$

Answer: $50.4$ m
Why this scores
Mark scheme:
- M1 for $18 \div 4.5$ oe (braking time $4$ s)
- M1 for $18 \times 0.8$ or $14.4$ seen
- M1 for $\tfrac{1}{2} \times$ their $4 \times 18$
- A1 for $50.4$ cao
Why this scores full marks: The candidate uses the deceleration (the size of the gradient) to find the braking time before finding any area. Leaving out the reaction distance gives $36$ m, and one triangle over the whole $4.8$ s gives $43.2$ m.
Question 8
Paper 1H/2H style — 4 marks4 marks
An empty tank is a cuboid with a base $80$ cm by $60$ cm. Water flows in at a constant rate. The graph of the depth of water, in cm, against time, in minutes, is a straight line from $(0, 0)$ to $(25, 50)$ .

Work out the rate at which water flows into the tank, in litres per minute. ( $1$ litre $= 1000$ cm³.) [4]
Model answer
Depth rises $\frac{50}{25} = 2 \text{ cm per minute}$

Volume per minute $80 \times 60 \times 2 = 9600 \text{ cm}^3$

$9600 \div 1000 = 9.6$

Answer: $9.6$ litres per minute
Why this scores
Mark scheme:
- M1 for $50 \div 25$ oe
- M1 for $80 \times 60 \times$ their $2$ , or $80 \times 60 \times 50 \div 25$
- M1 for their volume $\div 1000$
- A1 for $9.6$ cao
Why this scores full marks: The gradient of the depth–time graph is only the rate of rise; multiplying by the base area turns it into a volume per minute. The candidate then converts cm³ to litres, as the question asks.
Question 9
Paper 1H/2H style — 5 marks5 marks
Towns $P$ and $Q$ are $30$ km apart. Priya leaves $P$ at 10:00 and cycles towards $Q$ at $15$ km/h. At 11:00 she stops for $20$ minutes. Quentin leaves $Q$ at 10:30 and cycles towards $P$ along the same road at $20$ km/h.

On a distance–time graph, both journeys are drawn as distance from $P$ against time.

Work out the time at which they meet and how far from $P$ they are. [5]
Model answer
At 11:00, Priya is $15 \times 1 = 15$ km from $P$ .

At 11:00, Quentin has cycled $20 \times 0.5 = 10$ km, so he is $30 - 10 = 20$ km from $P$ .

They have not met yet. Priya is stopped from 11:00 to 11:20, so Quentin must cover the $5$ km gap alone:

$\frac{5}{20} = 0.25 \text{ hours} = 15 \text{ minutes}$

11:15 is before 11:20, so Priya is still stopped when they meet.

Answer: 11:15, $15$ km from $P$
Why this scores
Mark scheme:
- M1 for Priya at $15$ km from $P$ at 11:00
- M1 for Quentin at $20$ km from $P$ (or $10$ km from $Q$ ) at 11:00
- M1 for $5 \div 20$ oe with Priya stationary
- A1 for 11:15
- A1 for $15$ km
Why this scores full marks: The candidate checks where each person is when Priya stops. Using a closing speed of $15 + 20 = 35$ km/h for the gap would be wrong, because Priya is not moving during that time; on the graph her line is horizontal where the two lines cross.
Question 10
Paper 1H/2H style — 5 marks5 marks
A car travelling at $24$ m/s brakes and slows down uniformly, stopping in $8$ seconds.

(a) Work out the deceleration of the car. [1]

(b) Work out the percentage of the total braking distance that is covered in the last $3$ seconds. Give your answer correct to $3$ significant figures. [4]
Model answer
(a) $24 \div 8 = 3$

Answer: $3$ m/s²

(b) The last $3$ seconds start at $t = 5$ . Speed at $t = 5$ :

$24 - 3 \times 5 = 9 \text{ m/s}$

Last $3$ s: $\tfrac{1}{2} \times 3 \times 9 = 13.5 \text{ m}$

Total: $\tfrac{1}{2} \times 8 \times 24 = 96 \text{ m}$

$\frac{13.5}{96} \times 100 = 14.0625$

Answer: $14.1$ %
Why this scores
Mark scheme:
- B1 for $3$
- M1 for the speed at $t = 5$ as $9$ oe
- M1 for $\tfrac{1}{2} \times 3 \times 9$ or $13.5$ seen, and $96$ seen
- M1 for $\dfrac{13.5}{96} \times 100$ oe
- A1 for awrt $14.1$
Why this scores full marks: The candidate finds the speed where the last $3$ seconds begin, so the small triangle has the right height. Assuming it is $\tfrac{3}{8}$ of the distance ( $37.5$ %) ignores that the car is much slower at the end.

4MA1 Real Life Graphs

At a glance

What you’ll learn

Distance-time graphs

Speed-time graphs

Conversion graphs

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Distance-time graph: speed from gradient

2Speed-time graph: distance from area

3Acceleration from speed-time graph

4Conversion graph

5Matching a flask to its depth–time graph [1 mark]

6Reading a step graph for parking charges [2 marks]

7Completing a distance–time graph to find an arrival time [3 marks]

8Extending a spring graph back to its intercept [3 marks]

9Deceleration and distance while slowing down [4 marks]

10When an accelerating car catches a van [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10