How do you calculate speed using distance and time in Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, speed is calculated using the formula: Speed = Distance ÷ Time. This formula helps determine how fast an object is moving by dividing the distance traveled by the time taken to travel that distance.

What is the relationship between speed, distance, and time?

The relationship between speed, distance, and time is expressed through the formula: Distance = Speed × Time. This formula indicates that distance is the product of speed and time, showing how far an object travels at a certain speed over a given period.

How can I solve problems involving average speed in the IGCSE curriculum?

To solve problems involving average speed in the IGCSE curriculum, use the formula: Average Speed = Total Distance ÷ Total Time. This involves calculating the total distance traveled and dividing it by the total time taken, which gives the average speed over the entire journey.

What units are commonly used for speed, distance, and time in IGCSE Mathematics?

In IGCSE Mathematics, speed is typically measured in meters per second (m/s) or kilometers per hour (km/h), distance in meters (m) or kilometers (km), and time in seconds (s) or hours (h). It's important to ensure consistency in units when solving problems.

How do you convert between different units of speed, such as from m/s to km/h?

To convert speed from meters per second (m/s) to kilometers per hour (km/h), multiply the speed by 3.6. Conversely, to convert from km/h to m/s, divide the speed by 3.6. This conversion is essential for solving problems where different units are used.

Why is "Using mixed units in $d = s \times t$ without converting" a common mistake in Speed, Distance and Time?

Why it happens: Students plug in km/h and minutes together. How to avoid it: Convert the time to the same unit base as the speed (e.g. hours for km/h). Source: 0580 examiner reports — recurring

Why is "Averaging two speeds to get average speed" a common mistake in Speed, Distance and Time?

Why it happens: It feels intuitive: 60 + 90/2 = 75. How to avoid it: Always use total distance ÷ total time. The average happened to equal 75 in our example by coincidence — don't rely on it. Source: 0580/42 May/Jun 2023 — examiner report Q5

Why is "Misconverting decimal hours to minutes" a common mistake in Speed, Distance and Time?

Why it happens: 0.3 h is read as 30 min instead of 18 min. How to avoid it: Multiply the decimal part by 60 to convert to minutes.

Why is "Rearranging the speed–distance–time triangle incorrectly" a common mistake in Speed, Distance and Time?

Why it happens: Students don't memorise the triangle and try to reverse-engineer. How to avoid it: Memorise: D on top, S and T underneath. Cover the unknown to read off the formula.

Cambridge IGCSE Mathematics (0580)

Speed, Distance and Time

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Short Notes - Speed, Distance and Time

Detailed Study Notes

Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Speed, Distance and Time — Cambridge IGCSE 0580 Maths Extended (2026)

One formula triangle, three rearrangements, and constant unit conversions — that's the whole topic. Where students lose marks: km/h vs m/s, and using individual-leg speeds in 'average speed' questions.

At a glance

$\text{speed} = \dfrac{\text{distance}}{\text{time}}$ , $\text{distance} = \text{speed} \times \text{time}$ , $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ .
Use a triangle: $D$ on top, $S$ and $T$ at the bottom — cover what you want.
Convert units BEFORE substituting: km/h vs m/s, hours vs minutes.
$1 \text{ km/h} = \dfrac{1000}{3600} \text{ m/s} = \dfrac{1}{3.6} \text{ m/s}$ .
Average speed for a journey is TOTAL distance $\div$ TOTAL time. NEVER the average of the segment speeds.
Convert minutes to hours by dividing by $60$ . Convert hours to seconds by $\times 3600$ .
Read distance-time graphs: gradient = speed; horizontal segment = stationary.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.16 — Use given data to solve problems on personal and household finance, including earnings, simple interest and compound interest, and motion at constant speed.
Calculate average speed for a journey and convert between km/h and m/s.
Read and interpret distance-time and speed-time graphs.

The formula triangle

One picture handles all three rearrangements. Cover the unknown to read off the formula.

The relationship $D = S \times T$ (distance = speed × time) gives three formulas:

$S = \dfrac{D}{T}, \quad D = S \times T, \quad T = \dfrac{D}{S}.$

Picture them as a triangle with $D$ on top and $S, T$ side-by-side underneath:

    D
  -----
  S | T

Cover whichever you want; the rest is what you do with the others.

Cover the quantity you want — the other two show how to combine them.

Cover $D$ → $D = S \times T$ .
Cover $S$ → $S = D / T$ .
Cover $T$ → $T = D / S$ .

Always check the units match BEFORE substituting.

$D = S \times T$ .
$S = D \div T$ .
$T = D \div S$ .
Cover the unknown — the rest is the formula.

Unit conversions — km/h ↔ m/s, mins ↔ h

More marks are lost on units in this topic than on the formula itself.

Distance and speed must use the SAME unit before substitution.

Time conversions.

$1\,\text{h} = 60\,\text{min} = 3600\,\text{s}$ .
Convert minutes to hours: divide by $60$ . So $45\,\text{min} = 0.75\,\text{h}$ .
Convert seconds to hours: divide by $3600$ .

Speed conversions. $1\,\text{km/h} = \dfrac{1000\,\text{m}}{3600\,\text{s}} = \dfrac{1}{3.6}\,\text{m/s} \approx 0.278\,\text{m/s}.$

In practice:

km/h to m/s: divide by $3.6$ .
m/s to km/h: multiply by $3.6$ .

Worked. A car travels $90\,\text{km/h}$ . Express in m/s.

$90 \div 3.6 = 25\,\text{m/s}$ .

Worked. A runner's speed is $8\,\text{m/s}$ . Express in km/h.

$8 \times 3.6 = 28.8\,\text{km/h}$ .

Divide by 3.6 for km/h → m/s; multiply by 3.6 for m/s → km/h.

Mixed-unit traps. "A train travels $48\,\text{km}$ in $40\,\text{minutes}$ . Find its speed in km/h."

$40\,\text{min} = \dfrac{40}{60}\,\text{h} = \dfrac{2}{3}\,\text{h}$ .
$S = 48 \div \dfrac{2}{3} = 48 \times \dfrac{3}{2} = 72\,\text{km/h}$ .

$60\,\text{min} = 1\,\text{h}$ . $3600\,\text{s} = 1\,\text{h}$ .
km/h to m/s: $\div 3.6$ .
m/s to km/h: $\times 3.6$ .
Convert FIRST, substitute SECOND.

Average speed — the trap

Average speed for a journey is total distance over total time. Don't average the segment speeds.

Average speed for a complete journey is $\bar{S} = \dfrac{\text{total distance}}{\text{total time}}.$

This is a TIME-WEIGHTED average. It is NOT the simple mean of the segment speeds.

Worked example (the trap). A driver travels $60\,\text{km}$ at $60\,\text{km/h}$ then $60\,\text{km}$ at $30\,\text{km/h}$ . Find the average speed.

Wrong (simple average): $\dfrac{60 + 30}{2} = 45\,\text{km/h}$ . Wrong by 5 km/h.

Right (total distance / total time):

Time for leg 1: $60 / 60 = 1\,\text{h}$ .
Time for leg 2: $60 / 30 = 2\,\text{h}$ .
Total distance: $120\,\text{km}$ .
Total time: $3\,\text{h}$ .
Average speed: $120 / 3 = 40\,\text{km/h}$ .

The slower 30 km/h leg takes twice as long, so it pulls the average down to 40 km/h.

The slower leg counts MORE because more time was spent on it.

A quick sanity check. Average speed across a journey is always between the minimum and maximum segment speeds. If your answer is outside that range, recheck.

Average speed = TOTAL distance / TOTAL time.
NEVER average the segment speeds directly.
Compute time for each leg first, sum, then divide total distance.
Average speed is between the min and max of segment speeds.

Distance-time and speed-time graphs

Graphs translate motion into picture form. Read the gradients and areas to recover speed and distance.

Distance-time graphs. Time on the $x$ -axis, distance on the $y$ -axis.

A straight line = constant speed.
A horizontal line = stationary (no distance change).
The gradient equals the speed: $\dfrac{\Delta d}{\Delta t}$ .
A steeper line = faster.
A line returning toward the $x$ -axis = moving back toward the start.

On a distance-time graph the gradient is the speed; a flat segment means the object is stationary.

Worked. "On a $d$ - $t$ graph, a runner covers $400\,\text{m}$ in $80\,\text{s}$ as a straight line. Find the speed."

Gradient: $400 / 80 = 5\,\text{m/s}$ .

Speed-time graphs. Time on the $x$ -axis, speed on the $y$ -axis.

A straight horizontal line = constant speed.
A horizontal line at speed $0$ = stationary.
The gradient equals acceleration.
The area UNDER the line equals the distance travelled.

Worked. A vehicle runs at $20\,\text{m/s}$ for $5\,\text{s}$ , then decelerates uniformly to rest in $4\,\text{s}$ . Find the total distance.

Constant section is a rectangle: $20 \times 5 = 100\,\text{m}$ .
Deceleration is a triangle: $\dfrac{1}{2} \times 4 \times 20 = 40\,\text{m}$ .
Total: $140\,\text{m}$ .

Distance is the area under a speed-time graph: rectangle 100 m + triangle 40 m = 140 m.

$d$ - $t$ graph gradient = speed.
$s$ - $t$ graph gradient = acceleration; area under = distance.
Horizontal $d$ - $t$ = stationary.
Steeper line = faster motion.

Quick recap

$D = S \times T$ . Use the formula triangle.
Convert units BEFORE substituting. $40\,\text{min} = \tfrac{2}{3}\,\text{h}$ .
km/h to m/s: $\div 3.6$ . m/s to km/h: $\times 3.6$ .
Average speed = TOTAL distance / TOTAL time. Never average the speeds.
$d$ - $t$ gradient = speed. $s$ - $t$ area = distance.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Speed — distance travelled per unit time.
Average speed — total distance divided by total time for a complete journey.
$d = vt$ (distance = speed × time) — the core relationship.
Distance-time graph — gradient gives speed; horizontal segment is stationary.
Speed-time graph — gradient gives acceleration; area gives distance.

How it’s examined

Speed-distance-time questions appear on every paper. Paper 2 typically has a one-step calculation (1-2 marks), often with a unit-conversion sting. Paper 4 escalates to a multi-leg journey or a graph interpretation (3-4 marks). Examiner reports flag two recurring slips: forgetting to convert minutes to hours before dividing, and computing average speed as the arithmetic mean of segment speeds.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.16); 0580/22 May/Jun 2024 — Q7 (km/h to m/s); 0580/42 Oct/Nov 2024 — Q6 (average speed multi-leg); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Speed, Distance and Time

Step-by-step solutions to past-paper-style questions on speed, distance and time, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Speed, Distance and Time

Almost every speed, distance and time exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the distance/time, convert a speed, find the average speed — with the values supplied directly.

How to approach it

Pick the right rearrangement of $s=\frac{d}{t}$ , convert units so speed and time share a base (hours for km/h), and for average speed use total distance over total time.

Common trap

Examiner reports flag plugging mixed units into $d=s\times t$ without converting, and averaging two speeds instead of using $\frac{\text{total distance}}{\text{total time}}$ .

Word problem

Recognise it by

A real-world scenario — two vehicles meeting or overtaking, fuel consumption — with no formula stated.

How to approach it

Translate the context into the maths: vehicles approaching add their speeds, vehicles in the same direction subtract them, and 'distance = rate $\times$ amount' for consumption problems.

Common trap

Examiner reports flag candidates setting up separate simultaneous distance equations instead of using closing or relative speed — a single division is far quicker.

Graph or diagram

Recognise it by

A distance-time (travel) graph to read, with instructions to find a speed from a stage or the overall average speed.

How to approach it

Read each stage's distance and time off the axes; speed is the gradient of that segment, and total time includes every stationary period.

Common trap

Examiner reports flag candidates omitting the stationary period from total time when computing average speed.

Show that / prove

Recognise it by

The words show that with the target result given — for example that one train takes exactly one hour longer.

How to approach it

Compute each individual time explicitly, then state the difference and confirm it matches the required value.

Common trap

Examiner reports flag leaving an expression such as $\frac{240}{60}-\frac{240}{80}$ unevaluated — both times and the final difference must be shown.

Multi-step problem

Recognise it by

Two unknown speeds linked by two scenarios, requiring a pair of equations to be solved together.

How to approach it

Write a time equation for each scenario, substitute $a=\frac{1}{v}$ and $b=\frac{1}{w}$ to linearise, then solve the simultaneous equations.

Common trap

Examiner reports flag candidates solving the original non-linear system directly and making algebra slips — linearise first.

1Find distance from speed and time
CoreDirect calculation• distance
Question
A car travels at $72\,\text{km/h}$ for $2\,\text{h}\ 15\,\text{min}$ . How far does it travel?
Step-by-step solution
1. Step 1
  Convert time to hours: $2\,\text{h}\ 15\,\text{min} = 2.25\,\text{h}$ .
2. Step 2
  Use $d = s \times t$ .
  $d = 72 \times 2.25 = 162\,\text{km}$
Answer
$162\,\text{km}$
2Convert speed between m/s and km/h
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q12• unit conversion
Question
Convert $25\,\text{m/s}$ to km/h.
Step-by-step solution
1. Step 1
  $1\,\text{m/s} = 3.6\,\text{km/h}$ (multiply by $3.6$ ).
2. Step 2
  $25 \times 3.6 = 90\,\text{km/h}$ .
Answer
$90\,\text{km/h}$
Examiner tip
Memorise the multipliers: m/s → km/h multiply by $3.6$ ; km/h → m/s divide by $3.6$ .
3Find average speed for a multi-leg journey
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q5• average speed
Question
A train travels $120\,\text{km}$ at $60\,\text{km/h}$ then $180\,\text{km}$ at $90\,\text{km/h}$ . Find the average speed for the whole journey.
Step-by-step solution
1. Step 1
  Time leg 1: $\dfrac{120}{60} = 2\,\text{h}$ .
2. Step 2
  Time leg 2: $\dfrac{180}{90} = 2\,\text{h}$ .
3. Step 3
  Total distance: $300\,\text{km}$ . Total time: $4\,\text{h}$ .
4. Step 4
  Average speed.
  $\dfrac{300}{4} = 75\,\text{km/h}$
Answer
$75\,\text{km/h}$
Examiner tip
Average speed is never the average of the two speeds — always total distance over total time.
4Find time given distance and speed (mixed units)
CoreDirect calculation• time, unit conversion
Question
A cyclist covers $7.5\,\text{km}$ at an average speed of $25\,\text{km/h}$ . How long does the journey take, in minutes?
Step-by-step solution
1. Step 1
  Time in hours: $\dfrac{7.5}{25} = 0.3\,\text{h}$ .
2. Step 2
  Convert to minutes: $0.3 \times 60 = 18$ .
Answer
$18$ minutes
5Convert speed from km/h to m/s
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• unit conversion
Question
An athlete runs at $36\,\text{km/h}$ . Express this speed in m/s.
Step-by-step solution
1. Step 1
  km/h → m/s by dividing by $3.6$ (or multiplying by $\dfrac{1000}{3600}$ ).
  $v_{\text{m/s}} = \dfrac{36 \times 1000}{3600} = \dfrac{36{,}000}{3600}$
2. Step 2
  Simplify.
  $= 10\,\text{m/s}$
Answer
$10\,\text{m/s}$
Examiner tip
The 2024 mark scheme awards a method mark for writing the conversion factor $\dfrac{1000}{3600}$ or stating "divide by $3.6$ " before computing. Bare answers without working lose half the marks.
6Two vehicles travelling towards each other — when do they meet?
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2023 Q7• meeting problem, relative speed
Question
Two towns $A$ and $B$ are $360\,\text{km}$ apart. A car leaves $A$ at $09{:}00$ travelling at $80\,\text{km/h}$ towards $B$ . At the same time a coach leaves $B$ at $40\,\text{km/h}$ towards $A$ . (a) Find the time when they meet. (b) Find their distance from $A$ when they meet.
Step-by-step solution
1. Step 1
  Approach: the gap closes at the combined speed.
  $v_{\text{closing}} = 80 + 40 = 120\,\text{km/h}$
2. Step 2
  Time to meet.
  $t = \dfrac{360}{120} = 3\,\text{h}$
3. Step 3
  (a) They meet $3$ hours after $09{:}00$ , i.e. at $12{:}00$ .
4. Step 4
  (b) Distance the car has covered from $A$ : $80 \times 3 = 240\,\text{km}$ .
Answer
(a) $12{:}00$ (noon) (b) $240\,\text{km}$ from $A$
Examiner tip
Examiner reports flag that candidates often set up two separate distance equations and try to solve simultaneously. The closing-speed trick reduces this to a single division — much faster under exam pressure.
7Read a distance-time graph and compute average speed
ExtendedGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q3• travel graph, distance-time
Question
A distance-time graph shows a car's journey. From $0$ to $30\,\text{min}$ the car travels $20\,\text{km}$ . From $30$ to $50\,\text{min}$ it is stationary. From $50$ to $90\,\text{min}$ it travels a further $30\,\text{km}$ . Find (a) the speed during the first stage in km/h, (b) the speed during the third stage in km/h, (c) the average speed for the whole journey in km/h.
Step-by-step solution
1. Step 1
  (a) First stage: $20\,\text{km}$ in $30\,\text{min} = 0.5\,\text{h}$ . Speed $= \dfrac{20}{0.5} = 40\,\text{km/h}$ .
2. Step 2
  (b) Third stage: $30\,\text{km}$ in $40\,\text{min} = \tfrac{2}{3}\,\text{h}$ . Speed $= 30 \div \tfrac{2}{3} = 45\,\text{km/h}$ .
3. Step 3
  (c) Total distance $= 20 + 0 + 30 = 50\,\text{km}$ . Total time $= 90\,\text{min} = 1.5\,\text{h}$ . Average speed $= \dfrac{50}{1.5} = 33.3\,\text{km/h}$ (to 3 s.f.).
  $\dfrac{50}{1.5} = 33.\overline{3}\,\text{km/h}$
Answer
(a) $40\,\text{km/h}$ (b) $45\,\text{km/h}$ (c) $33.3\,\text{km/h}$ (3 s.f.)
Examiner tip
The examiner report flags that candidates routinely forget the stationary period counts towards total time when computing average speed. Total time = elapsed time on the time axis, including pauses.
8Use fuel consumption (km per litre) to find distance
ExtendedWord problem• fuel consumption, applied
Question
A car's fuel consumption is $14\,\text{km/L}$ . The fuel tank holds $45\,\text{L}$ and is initially $\tfrac{2}{3}$ full. How far can the car travel before needing to refuel?
Step-by-step solution
1. Step 1
  Fuel available: $\tfrac{2}{3} \times 45 = 30\,\text{L}$ .
2. Step 2
  Distance = consumption × fuel.
  $d = 14 \times 30 = 420\,\text{km}$
Answer
$420\,\text{km}$
Examiner tip
Examiners reward candidates who state the rate equation in words ("distance = km per litre × litres") before substituting. A bare numerical answer earns only the accuracy mark.
9Show that a time difference exceeds a stated value
ExtendedShow that / prove• Adapted from 0580/42 Feb/Mar 2024 Q5• show that, reasoning
Question
Train P travels $240\,\text{km}$ at $60\,\text{km/h}$ . Train Q travels the same distance at $80\,\text{km/h}$ . Show that Train P takes exactly $1$ hour longer than Train Q.
Step-by-step solution
1. Step 1
  Time for Train P: $\dfrac{240}{60} = 4\,\text{h}$ .
2. Step 2
  Time for Train Q: $\dfrac{240}{80} = 3\,\text{h}$ .
3. Step 3
  Difference: $4 - 3 = 1\,\text{h}$ , exactly as required. QED.
Answer
$t_P - t_Q = \dfrac{240}{60} - \dfrac{240}{80} = 4 - 3 = 1\,\text{h}$ . ✓
Examiner tip
The 2024 mark scheme is strict on "show that" questions: both individual times must be computed explicitly and the difference stated. Skipping a step or leaving the result as $\dfrac{240}{60} - \dfrac{240}{80}$ unevaluated loses the conclusion mark.
10Two trains, same direction — find when the faster overtakes the slower
ChallengeWord problem• Adapted from 0580/42 Oct/Nov 2024 Q12• relative speed, stretch, overtaking
Question
At $08{:}00$ a goods train leaves station $X$ travelling at $48\,\text{km/h}$ . At $09{:}30$ an express train leaves the same station, travelling in the same direction at $84\,\text{km/h}$ . At what time does the express overtake the goods train?
Step-by-step solution
1. Step 1
  Head-start: by $09{:}30$ the goods train has been travelling $1.5\,\text{h}$ at $48\,\text{km/h}$ , covering $48 \times 1.5 = 72\,\text{km}$ .
2. Step 2
  From $09{:}30$ onwards, the express closes the gap at the relative speed of $84 - 48 = 36\,\text{km/h}$ .
3. Step 3
  Time to close $72\,\text{km}$ at $36\,\text{km/h}$ .
  $t = \dfrac{72}{36} = 2\,\text{h}$
4. Step 4
  Overtake time: $09{:}30 + 2\,\text{h} = 11{:}30$ .
Answer
$11{:}30$
Examiner tip
The 2024 examiner report flags that fewer than $20\%$ of candidates spotted the relative-speed approach. Many tried simultaneous equations and made algebra slips. Memorise: same direction → subtract speeds; opposite directions → add speeds.
11Find two unknown speeds using simultaneous equations
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q12• simultaneous equations, stretch
Question
On a trip, Yusuf cycles $30\,\text{km}$ at speed $v\,\text{km/h}$ and then walks $6\,\text{km}$ at speed $w\,\text{km/h}$ . The total time is $4$ hours. If he had cycled at speed $w$ and walked at speed $v$ , the total time would have been $9$ hours. Find $v$ and $w$ .
Step-by-step solution
1. Step 1
  Write the two time equations.
  $\dfrac{30}{v} + \dfrac{6}{w} = 4 \quad (1)$
2. Step 2
  Second scenario.
  $\dfrac{30}{w} + \dfrac{6}{v} = 9 \quad (2)$
3. Step 3
  Let $a = \dfrac{1}{v}$ , $b = \dfrac{1}{w}$ to linearise. The equations become $30a + 6b = 4$ and $6a + 30b = 9$ .
4. Step 4
  Multiply (1) by $5$ : $150a + 30b = 20$ . Subtract (2): $144a = 11 \Rightarrow a = \dfrac{11}{144}$ .
5. Step 5
  So $v = \dfrac{144}{11} \approx 13.1\,\text{km/h}$ .
6. Step 6
  Substitute back: $30 \times \dfrac{11}{144} + 6b = 4 \Rightarrow \dfrac{330}{144} + 6b = 4 \Rightarrow 6b = 4 - 2.2917 = 1.7083 \Rightarrow b = 0.2847$ .
7. Step 7
  So $w = \dfrac{1}{0.2847} \approx 3.51\,\text{km/h}$ .
Answer
$v \approx 13.1\,\text{km/h}$ (cycling), $w \approx 3.5\,\text{km/h}$ (walking).
Examiner tip
The 2024 mark scheme awards 3 method marks for the substitution $a = 1/v$ , $b = 1/w$ . Candidates who try to solve the original system directly almost always make algebra slips. Linearise first — it's the standard A* technique.

Key Formulae — Speed, Distance and Time

The formulae you need to memorise for speed, distance and time on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Speed–distance–time triangle
$s = \dfrac{d}{t},\quad d = s \times t,\quad t = \dfrac{d}{s}$
$s$
speed
$d$
distance
$t$
time
When to use
Any speed/distance/time question — pick the rearrangement that matches what's asked.
Average speed
$\text{average speed} = \dfrac{\text{total distance}}{\text{total time}}$
When to use
Multi-leg journeys; never average the speeds directly.
m/s ↔ km/h conversion
$v\,\text{m/s} \times 3.6 = v\,\text{km/h}$
$v$
speed value
When to use
Whenever you must compare speeds across different units.

Key Definitions and Keywords — Speed, Distance and Time

Definitions to memorise and the exact keywords mark schemes credit for speed, distance and time answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Speed
Examiner keyword
Distance travelled per unit time (e.g. km/h, m/s).
Average speed
Examiner keyword
Total distance divided by total time for the whole journey.
Velocity
Speed in a specific direction; appears in motion graph problems.
Distance
The length travelled along the path.
Displacement
Straight-line distance from start to end (with direction).

Common Mistakes and Misconceptions — Speed, Distance and Time

The traps other students keep falling into on speed, distance and time questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using mixed units in $d = s \times t$ without converting
0580 examiner reports — recurring
Why it happens
Students plug in km/h and minutes together.
How to avoid it
Convert the time to the same unit base as the speed (e.g. hours for km/h).
✕Averaging two speeds to get average speed
0580/42 May/Jun 2023 — examiner report Q5
Why it happens
It feels intuitive: $\dfrac{60 + 90}{2} = 75$ .
How to avoid it
Always use total distance ÷ total time. The average happened to equal $75$ in our example by coincidence — don't rely on it.
✕Misconverting decimal hours to minutes
Why it happens
$0.3$ h is read as $30$ min instead of $18$ min.
How to avoid it
Multiply the decimal part by $60$ to convert to minutes.
✕Rearranging the speed–distance–time triangle incorrectly
Why it happens
Students don't memorise the triangle and try to reverse-engineer.
How to avoid it
Memorise: $D$ on top, $S$ and $T$ underneath. Cover the unknown to read off the formula.

Speed, Distance and Time — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Speed, Distance and Time

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Short Notes - Speed, Distance and Time

Detailed Study Notes

Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Speed, Distance and Time — Cambridge IGCSE 0580 Maths Extended (2026)

At a glance

$\text{speed} = \dfrac{\text{distance}}{\text{time}}$ , $\text{distance} = \text{speed} \times \text{time}$ , $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ .
Use a triangle: $D$ on top, $S$ and $T$ at the bottom — cover what you want.
Convert units BEFORE substituting: km/h vs m/s, hours vs minutes.
$1 \text{ km/h} = \dfrac{1000}{3600} \text{ m/s} = \dfrac{1}{3.6} \text{ m/s}$ .
Average speed for a journey is TOTAL distance $\div$ TOTAL time. NEVER the average of the segment speeds.
Convert minutes to hours by dividing by $60$ . Convert hours to seconds by $\times 3600$ .
Read distance-time graphs: gradient = speed; horizontal segment = stationary.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.16 — Use given data to solve problems on personal and household finance, including earnings, simple interest and compound interest, and motion at constant speed.
Calculate average speed for a journey and convert between km/h and m/s.
Read and interpret distance-time and speed-time graphs.

The formula triangle

One picture handles all three rearrangements. Cover the unknown to read off the formula.

The relationship $D = S \times T$ (distance = speed × time) gives three formulas:

$S = \dfrac{D}{T}, \quad D = S \times T, \quad T = \dfrac{D}{S}.$

Picture them as a triangle with $D$ on top and $S, T$ side-by-side underneath:

    D
  -----
  S | T

Cover whichever you want; the rest is what you do with the others.

Cover the quantity you want — the other two show how to combine them.

Cover $D$ → $D = S \times T$ .
Cover $S$ → $S = D / T$ .
Cover $T$ → $T = D / S$ .

Always check the units match BEFORE substituting.

$D = S \times T$ .
$S = D \div T$ .
$T = D \div S$ .
Cover the unknown — the rest is the formula.

Unit conversions — km/h ↔ m/s, mins ↔ h

More marks are lost on units in this topic than on the formula itself.

Distance and speed must use the SAME unit before substitution.

Time conversions.

$1\,\text{h} = 60\,\text{min} = 3600\,\text{s}$ .
Convert minutes to hours: divide by $60$ . So $45\,\text{min} = 0.75\,\text{h}$ .
Convert seconds to hours: divide by $3600$ .

Speed conversions. $1\,\text{km/h} = \dfrac{1000\,\text{m}}{3600\,\text{s}} = \dfrac{1}{3.6}\,\text{m/s} \approx 0.278\,\text{m/s}.$

In practice:

km/h to m/s: divide by $3.6$ .
m/s to km/h: multiply by $3.6$ .

Worked. A car travels $90\,\text{km/h}$ . Express in m/s.

$90 \div 3.6 = 25\,\text{m/s}$ .

Worked. A runner's speed is $8\,\text{m/s}$ . Express in km/h.

$8 \times 3.6 = 28.8\,\text{km/h}$ .

Divide by 3.6 for km/h → m/s; multiply by 3.6 for m/s → km/h.

Mixed-unit traps. "A train travels $48\,\text{km}$ in $40\,\text{minutes}$ . Find its speed in km/h."

$40\,\text{min} = \dfrac{40}{60}\,\text{h} = \dfrac{2}{3}\,\text{h}$ .
$S = 48 \div \dfrac{2}{3} = 48 \times \dfrac{3}{2} = 72\,\text{km/h}$ .

$60\,\text{min} = 1\,\text{h}$ . $3600\,\text{s} = 1\,\text{h}$ .
km/h to m/s: $\div 3.6$ .
m/s to km/h: $\times 3.6$ .
Convert FIRST, substitute SECOND.

Average speed — the trap

Average speed for a journey is total distance over total time. Don't average the segment speeds.

Average speed for a complete journey is $\bar{S} = \dfrac{\text{total distance}}{\text{total time}}.$

This is a TIME-WEIGHTED average. It is NOT the simple mean of the segment speeds.

Worked example (the trap). A driver travels $60\,\text{km}$ at $60\,\text{km/h}$ then $60\,\text{km}$ at $30\,\text{km/h}$ . Find the average speed.

Wrong (simple average): $\dfrac{60 + 30}{2} = 45\,\text{km/h}$ . Wrong by 5 km/h.

Right (total distance / total time):

Time for leg 1: $60 / 60 = 1\,\text{h}$ .
Time for leg 2: $60 / 30 = 2\,\text{h}$ .
Total distance: $120\,\text{km}$ .
Total time: $3\,\text{h}$ .
Average speed: $120 / 3 = 40\,\text{km/h}$ .

The slower 30 km/h leg takes twice as long, so it pulls the average down to 40 km/h.

The slower leg counts MORE because more time was spent on it.

A quick sanity check. Average speed across a journey is always between the minimum and maximum segment speeds. If your answer is outside that range, recheck.

Average speed = TOTAL distance / TOTAL time.
NEVER average the segment speeds directly.
Compute time for each leg first, sum, then divide total distance.
Average speed is between the min and max of segment speeds.

Distance-time and speed-time graphs

Graphs translate motion into picture form. Read the gradients and areas to recover speed and distance.

Distance-time graphs. Time on the $x$ -axis, distance on the $y$ -axis.

A straight line = constant speed.
A horizontal line = stationary (no distance change).
The gradient equals the speed: $\dfrac{\Delta d}{\Delta t}$ .
A steeper line = faster.
A line returning toward the $x$ -axis = moving back toward the start.

On a distance-time graph the gradient is the speed; a flat segment means the object is stationary.

Worked. "On a $d$ - $t$ graph, a runner covers $400\,\text{m}$ in $80\,\text{s}$ as a straight line. Find the speed."

Gradient: $400 / 80 = 5\,\text{m/s}$ .

Speed-time graphs. Time on the $x$ -axis, speed on the $y$ -axis.

A straight horizontal line = constant speed.
A horizontal line at speed $0$ = stationary.
The gradient equals acceleration.
The area UNDER the line equals the distance travelled.

Worked. A vehicle runs at $20\,\text{m/s}$ for $5\,\text{s}$ , then decelerates uniformly to rest in $4\,\text{s}$ . Find the total distance.

Constant section is a rectangle: $20 \times 5 = 100\,\text{m}$ .
Deceleration is a triangle: $\dfrac{1}{2} \times 4 \times 20 = 40\,\text{m}$ .
Total: $140\,\text{m}$ .

Distance is the area under a speed-time graph: rectangle 100 m + triangle 40 m = 140 m.

$d$ - $t$ graph gradient = speed.
$s$ - $t$ graph gradient = acceleration; area under = distance.
Horizontal $d$ - $t$ = stationary.
Steeper line = faster motion.

Quick recap

$D = S \times T$ . Use the formula triangle.
Convert units BEFORE substituting. $40\,\text{min} = \tfrac{2}{3}\,\text{h}$ .
km/h to m/s: $\div 3.6$ . m/s to km/h: $\times 3.6$ .
Average speed = TOTAL distance / TOTAL time. Never average the speeds.
$d$ - $t$ gradient = speed. $s$ - $t$ area = distance.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Speed — distance travelled per unit time.
Average speed — total distance divided by total time for a complete journey.
$d = vt$ (distance = speed × time) — the core relationship.
Distance-time graph — gradient gives speed; horizontal segment is stationary.
Speed-time graph — gradient gives acceleration; area gives distance.

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Speed, Distance and Time

Step-by-step solutions to past-paper-style questions on speed, distance and time, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Speed, Distance and Time

Almost every speed, distance and time exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the distance/time, convert a speed, find the average speed — with the values supplied directly.

How to approach it

Pick the right rearrangement of $s=\frac{d}{t}$ , convert units so speed and time share a base (hours for km/h), and for average speed use total distance over total time.

Common trap

Examiner reports flag plugging mixed units into $d=s\times t$ without converting, and averaging two speeds instead of using $\frac{\text{total distance}}{\text{total time}}$ .

Word problem

Recognise it by

A real-world scenario — two vehicles meeting or overtaking, fuel consumption — with no formula stated.

How to approach it

Translate the context into the maths: vehicles approaching add their speeds, vehicles in the same direction subtract them, and 'distance = rate $\times$ amount' for consumption problems.

Common trap

Examiner reports flag candidates setting up separate simultaneous distance equations instead of using closing or relative speed — a single division is far quicker.

Graph or diagram

Recognise it by

A distance-time (travel) graph to read, with instructions to find a speed from a stage or the overall average speed.

How to approach it

Read each stage's distance and time off the axes; speed is the gradient of that segment, and total time includes every stationary period.

Common trap

Examiner reports flag candidates omitting the stationary period from total time when computing average speed.

Show that / prove

Recognise it by

The words show that with the target result given — for example that one train takes exactly one hour longer.

How to approach it

Compute each individual time explicitly, then state the difference and confirm it matches the required value.

Common trap

Examiner reports flag leaving an expression such as $\frac{240}{60}-\frac{240}{80}$ unevaluated — both times and the final difference must be shown.

Multi-step problem

Recognise it by

Two unknown speeds linked by two scenarios, requiring a pair of equations to be solved together.

How to approach it

Write a time equation for each scenario, substitute $a=\frac{1}{v}$ and $b=\frac{1}{w}$ to linearise, then solve the simultaneous equations.

Common trap

Examiner reports flag candidates solving the original non-linear system directly and making algebra slips — linearise first.

1Find distance from speed and time
CoreDirect calculation• distance
Question
A car travels at $72\,\text{km/h}$ for $2\,\text{h}\ 15\,\text{min}$ . How far does it travel?
Step-by-step solution
1. Step 1
  Convert time to hours: $2\,\text{h}\ 15\,\text{min} = 2.25\,\text{h}$ .
2. Step 2
  Use $d = s \times t$ .
  $d = 72 \times 2.25 = 162\,\text{km}$
Answer
$162\,\text{km}$
2Convert speed between m/s and km/h
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q12• unit conversion
Question
Convert $25\,\text{m/s}$ to km/h.
Step-by-step solution
1. Step 1
  $1\,\text{m/s} = 3.6\,\text{km/h}$ (multiply by $3.6$ ).
2. Step 2
  $25 \times 3.6 = 90\,\text{km/h}$ .
Answer
$90\,\text{km/h}$
Examiner tip
Memorise the multipliers: m/s → km/h multiply by $3.6$ ; km/h → m/s divide by $3.6$ .
3Find average speed for a multi-leg journey
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q5• average speed
Question
A train travels $120\,\text{km}$ at $60\,\text{km/h}$ then $180\,\text{km}$ at $90\,\text{km/h}$ . Find the average speed for the whole journey.
Step-by-step solution
1. Step 1
  Time leg 1: $\dfrac{120}{60} = 2\,\text{h}$ .
2. Step 2
  Time leg 2: $\dfrac{180}{90} = 2\,\text{h}$ .
3. Step 3
  Total distance: $300\,\text{km}$ . Total time: $4\,\text{h}$ .
4. Step 4
  Average speed.
  $\dfrac{300}{4} = 75\,\text{km/h}$
Answer
$75\,\text{km/h}$
Examiner tip
Average speed is never the average of the two speeds — always total distance over total time.
4Find time given distance and speed (mixed units)
CoreDirect calculation• time, unit conversion
Question
A cyclist covers $7.5\,\text{km}$ at an average speed of $25\,\text{km/h}$ . How long does the journey take, in minutes?
Step-by-step solution
1. Step 1
  Time in hours: $\dfrac{7.5}{25} = 0.3\,\text{h}$ .
2. Step 2
  Convert to minutes: $0.3 \times 60 = 18$ .
Answer
$18$ minutes
5Convert speed from km/h to m/s
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• unit conversion
Question
An athlete runs at $36\,\text{km/h}$ . Express this speed in m/s.
Step-by-step solution
1. Step 1
  km/h → m/s by dividing by $3.6$ (or multiplying by $\dfrac{1000}{3600}$ ).
  $v_{\text{m/s}} = \dfrac{36 \times 1000}{3600} = \dfrac{36{,}000}{3600}$
2. Step 2
  Simplify.
  $= 10\,\text{m/s}$
Answer
$10\,\text{m/s}$
Examiner tip
The 2024 mark scheme awards a method mark for writing the conversion factor $\dfrac{1000}{3600}$ or stating "divide by $3.6$ " before computing. Bare answers without working lose half the marks.
6Two vehicles travelling towards each other — when do they meet?
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2023 Q7• meeting problem, relative speed
Question
Two towns $A$ and $B$ are $360\,\text{km}$ apart. A car leaves $A$ at $09{:}00$ travelling at $80\,\text{km/h}$ towards $B$ . At the same time a coach leaves $B$ at $40\,\text{km/h}$ towards $A$ . (a) Find the time when they meet. (b) Find their distance from $A$ when they meet.
Step-by-step solution
1. Step 1
  Approach: the gap closes at the combined speed.
  $v_{\text{closing}} = 80 + 40 = 120\,\text{km/h}$
2. Step 2
  Time to meet.
  $t = \dfrac{360}{120} = 3\,\text{h}$
3. Step 3
  (a) They meet $3$ hours after $09{:}00$ , i.e. at $12{:}00$ .
4. Step 4
  (b) Distance the car has covered from $A$ : $80 \times 3 = 240\,\text{km}$ .
Answer
(a) $12{:}00$ (noon) (b) $240\,\text{km}$ from $A$
Examiner tip
Examiner reports flag that candidates often set up two separate distance equations and try to solve simultaneously. The closing-speed trick reduces this to a single division — much faster under exam pressure.
7Read a distance-time graph and compute average speed
ExtendedGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q3• travel graph, distance-time
Question
A distance-time graph shows a car's journey. From $0$ to $30\,\text{min}$ the car travels $20\,\text{km}$ . From $30$ to $50\,\text{min}$ it is stationary. From $50$ to $90\,\text{min}$ it travels a further $30\,\text{km}$ . Find (a) the speed during the first stage in km/h, (b) the speed during the third stage in km/h, (c) the average speed for the whole journey in km/h.
Step-by-step solution
1. Step 1
  (a) First stage: $20\,\text{km}$ in $30\,\text{min} = 0.5\,\text{h}$ . Speed $= \dfrac{20}{0.5} = 40\,\text{km/h}$ .
2. Step 2
  (b) Third stage: $30\,\text{km}$ in $40\,\text{min} = \tfrac{2}{3}\,\text{h}$ . Speed $= 30 \div \tfrac{2}{3} = 45\,\text{km/h}$ .
3. Step 3
  (c) Total distance $= 20 + 0 + 30 = 50\,\text{km}$ . Total time $= 90\,\text{min} = 1.5\,\text{h}$ . Average speed $= \dfrac{50}{1.5} = 33.3\,\text{km/h}$ (to 3 s.f.).
  $\dfrac{50}{1.5} = 33.\overline{3}\,\text{km/h}$
Answer
(a) $40\,\text{km/h}$ (b) $45\,\text{km/h}$ (c) $33.3\,\text{km/h}$ (3 s.f.)
Examiner tip
The examiner report flags that candidates routinely forget the stationary period counts towards total time when computing average speed. Total time = elapsed time on the time axis, including pauses.
8Use fuel consumption (km per litre) to find distance
ExtendedWord problem• fuel consumption, applied
Question
A car's fuel consumption is $14\,\text{km/L}$ . The fuel tank holds $45\,\text{L}$ and is initially $\tfrac{2}{3}$ full. How far can the car travel before needing to refuel?
Step-by-step solution
1. Step 1
  Fuel available: $\tfrac{2}{3} \times 45 = 30\,\text{L}$ .
2. Step 2
  Distance = consumption × fuel.
  $d = 14 \times 30 = 420\,\text{km}$
Answer
$420\,\text{km}$
Examiner tip
Examiners reward candidates who state the rate equation in words ("distance = km per litre × litres") before substituting. A bare numerical answer earns only the accuracy mark.
9Show that a time difference exceeds a stated value
ExtendedShow that / prove• Adapted from 0580/42 Feb/Mar 2024 Q5• show that, reasoning
Question
Train P travels $240\,\text{km}$ at $60\,\text{km/h}$ . Train Q travels the same distance at $80\,\text{km/h}$ . Show that Train P takes exactly $1$ hour longer than Train Q.
Step-by-step solution
1. Step 1
  Time for Train P: $\dfrac{240}{60} = 4\,\text{h}$ .
2. Step 2
  Time for Train Q: $\dfrac{240}{80} = 3\,\text{h}$ .
3. Step 3
  Difference: $4 - 3 = 1\,\text{h}$ , exactly as required. QED.
Answer
$t_P - t_Q = \dfrac{240}{60} - \dfrac{240}{80} = 4 - 3 = 1\,\text{h}$ . ✓
Examiner tip
The 2024 mark scheme is strict on "show that" questions: both individual times must be computed explicitly and the difference stated. Skipping a step or leaving the result as $\dfrac{240}{60} - \dfrac{240}{80}$ unevaluated loses the conclusion mark.
10Two trains, same direction — find when the faster overtakes the slower
ChallengeWord problem• Adapted from 0580/42 Oct/Nov 2024 Q12• relative speed, stretch, overtaking
Question
At $08{:}00$ a goods train leaves station $X$ travelling at $48\,\text{km/h}$ . At $09{:}30$ an express train leaves the same station, travelling in the same direction at $84\,\text{km/h}$ . At what time does the express overtake the goods train?
Step-by-step solution
1. Step 1
  Head-start: by $09{:}30$ the goods train has been travelling $1.5\,\text{h}$ at $48\,\text{km/h}$ , covering $48 \times 1.5 = 72\,\text{km}$ .
2. Step 2
  From $09{:}30$ onwards, the express closes the gap at the relative speed of $84 - 48 = 36\,\text{km/h}$ .
3. Step 3
  Time to close $72\,\text{km}$ at $36\,\text{km/h}$ .
  $t = \dfrac{72}{36} = 2\,\text{h}$
4. Step 4
  Overtake time: $09{:}30 + 2\,\text{h} = 11{:}30$ .
Answer
$11{:}30$
Examiner tip
The 2024 examiner report flags that fewer than $20\%$ of candidates spotted the relative-speed approach. Many tried simultaneous equations and made algebra slips. Memorise: same direction → subtract speeds; opposite directions → add speeds.
11Find two unknown speeds using simultaneous equations
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q12• simultaneous equations, stretch
Question
On a trip, Yusuf cycles $30\,\text{km}$ at speed $v\,\text{km/h}$ and then walks $6\,\text{km}$ at speed $w\,\text{km/h}$ . The total time is $4$ hours. If he had cycled at speed $w$ and walked at speed $v$ , the total time would have been $9$ hours. Find $v$ and $w$ .
Step-by-step solution
1. Step 1
  Write the two time equations.
  $\dfrac{30}{v} + \dfrac{6}{w} = 4 \quad (1)$
2. Step 2
  Second scenario.
  $\dfrac{30}{w} + \dfrac{6}{v} = 9 \quad (2)$
3. Step 3
  Let $a = \dfrac{1}{v}$ , $b = \dfrac{1}{w}$ to linearise. The equations become $30a + 6b = 4$ and $6a + 30b = 9$ .
4. Step 4
  Multiply (1) by $5$ : $150a + 30b = 20$ . Subtract (2): $144a = 11 \Rightarrow a = \dfrac{11}{144}$ .
5. Step 5
  So $v = \dfrac{144}{11} \approx 13.1\,\text{km/h}$ .
6. Step 6
  Substitute back: $30 \times \dfrac{11}{144} + 6b = 4 \Rightarrow \dfrac{330}{144} + 6b = 4 \Rightarrow 6b = 4 - 2.2917 = 1.7083 \Rightarrow b = 0.2847$ .
7. Step 7
  So $w = \dfrac{1}{0.2847} \approx 3.51\,\text{km/h}$ .
Answer
$v \approx 13.1\,\text{km/h}$ (cycling), $w \approx 3.5\,\text{km/h}$ (walking).
Examiner tip
The 2024 mark scheme awards 3 method marks for the substitution $a = 1/v$ , $b = 1/w$ . Candidates who try to solve the original system directly almost always make algebra slips. Linearise first — it's the standard A* technique.

Key Formulae — Speed, Distance and Time

The formulae you need to memorise for speed, distance and time on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Speed–distance–time triangle
$s = \dfrac{d}{t},\quad d = s \times t,\quad t = \dfrac{d}{s}$
$s$
speed
$d$
distance
$t$
time
When to use
Any speed/distance/time question — pick the rearrangement that matches what's asked.
Average speed
$\text{average speed} = \dfrac{\text{total distance}}{\text{total time}}$
When to use
Multi-leg journeys; never average the speeds directly.
m/s ↔ km/h conversion
$v\,\text{m/s} \times 3.6 = v\,\text{km/h}$
$v$
speed value
When to use
Whenever you must compare speeds across different units.

Key Definitions and Keywords — Speed, Distance and Time

Definitions to memorise and the exact keywords mark schemes credit for speed, distance and time answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Speed
Examiner keyword
Distance travelled per unit time (e.g. km/h, m/s).
Average speed
Examiner keyword
Total distance divided by total time for the whole journey.
Velocity
Speed in a specific direction; appears in motion graph problems.
Distance
The length travelled along the path.
Displacement
Straight-line distance from start to end (with direction).

Common Mistakes and Misconceptions — Speed, Distance and Time

The traps other students keep falling into on speed, distance and time questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using mixed units in $d = s \times t$ without converting
0580 examiner reports — recurring
Why it happens
Students plug in km/h and minutes together.
How to avoid it
Convert the time to the same unit base as the speed (e.g. hours for km/h).
✕Averaging two speeds to get average speed
0580/42 May/Jun 2023 — examiner report Q5
Why it happens
It feels intuitive: $\dfrac{60 + 90}{2} = 75$ .
How to avoid it
Always use total distance ÷ total time. The average happened to equal $75$ in our example by coincidence — don't rely on it.
✕Misconverting decimal hours to minutes
Why it happens
$0.3$ h is read as $30$ min instead of $18$ min.
How to avoid it
Multiply the decimal part by $60$ to convert to minutes.
✕Rearranging the speed–distance–time triangle incorrectly
Why it happens
Students don't memorise the triangle and try to reverse-engineer.
How to avoid it
Memorise: $D$ on top, $S$ and $T$ underneath. Cover the unknown to read off the formula.

Speed, Distance and Time — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Speed, Distance and Time

Short Notes - Speed, Distance and Time

Detailed Study Notes

At a glance

What you’ll learn

The formula triangle

Unit conversions — km/h ↔ m/s, mins ↔ h

Average speed — the trap

Distance-time and speed-time graphs

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Speed, Distance and Time

1Find distance from speed and time

2Convert speed between m/s and km/h

3Find average speed for a multi-leg journey

4Find time given distance and speed (mixed units)

5Convert speed from km/h to m/s

6Two vehicles travelling towards each other — when do they meet?

7Read a distance-time graph and compute average speed

8Use fuel consumption (km per litre) to find distance

9Show that a time difference exceeds a stated value

10Two trains, same direction — find when the faster overtakes the slower

11Find two unknown speeds using simultaneous equations

Speed–distance–time triangle

Average speed

m/s ↔ km/h conversion

Speed

Average speed

Velocity

Distance

Displacement

✕Using mixed units in d=s×td = s \times td=s×t without converting

✕Averaging two speeds to get average speed

✕Misconverting decimal hours to minutes

✕Rearranging the speed–distance–time triangle incorrectly

Speed, Distance and Time — frequently asked questions

Speed, Distance and Time

Short Notes - Speed, Distance and Time

Detailed Study Notes

At a glance

What you’ll learn

The formula triangle

Unit conversions — km/h ↔ m/s, mins ↔ h

Average speed — the trap

Distance-time and speed-time graphs

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Speed, Distance and Time

1Find distance from speed and time

2Convert speed between m/s and km/h

3Find average speed for a multi-leg journey

4Find time given distance and speed (mixed units)

5Convert speed from km/h to m/s

6Two vehicles travelling towards each other — when do they meet?

7Read a distance-time graph and compute average speed

8Use fuel consumption (km per litre) to find distance

9Show that a time difference exceeds a stated value

10Two trains, same direction — find when the faster overtakes the slower

11Find two unknown speeds using simultaneous equations

Speed–distance–time triangle

Average speed

m/s ↔ km/h conversion

Speed

Average speed

Velocity

Distance

Displacement

✕Using mixed units in d=s×td = s \times td=s×t without converting

✕Averaging two speeds to get average speed

✕Misconverting decimal hours to minutes

✕Rearranging the speed–distance–time triangle incorrectly

Speed, Distance and Time — frequently asked questions

✕Using mixed units in $d = s \times t$ without converting

✕Using mixed units in $d = s \times t$ without converting