BTT Mathematics / Primary Mathematics Learning Hub / Guide 11
Speed compares distance with time. It tells us how much distance is covered for each unit of time under the stated model. Distance can be found from speed and time; time can be found from distance and speed. The three quantities form one rate relationship rather than three unrelated formulas.
A journey of 120 kilometres in two hours has average speed sixty kilometres per hour. That statement does not prove the traveller moved at exactly sixty kilometres per hour at every moment. It compresses the total journey into an equivalent constant rate.
This guide develops direct speed questions, unit conversion, average speed, multi-stage journeys, relative reasoning and common interpretation errors. Use the later material only when it matches the learner’s current upper-primary scope in the MOE Primary Mathematics syllabus.
Meaning of speed · One relationship, three quantities · Units · Average speed · Multi-stage journeys · 24 questions · Worked answers
1. Speed is a unit rate
If a cyclist travels thirty-six kilometres in three hours at a constant average rate over that period, divide distance by time: 36 ÷ 3 = 12 kilometres per hour.
The unit km/h says “kilometres for each hour.” It is not decorative. It identifies which quantity has been divided by which.
Same distance, different time
Two runners each cover eight kilometres. Runner A takes forty minutes and Runner B takes fifty minutes. Runner A has the greater average speed because the same distance was covered in less time.
This comparison can be made without calculating exact speeds. When distance is fixed, less time means greater speed.
Same time, different distance
Two cyclists both ride for two hours. One covers forty kilometres and the other fifty. The second has greater average speed because more distance was covered in the same time.
When both distance and time change
If one traveller covers sixty kilometres in one hour and another covers ninety kilometres in two hours, neither raw distance nor raw time alone decides the faster rate. Calculate a common rate: 60 km/h versus 45 km/h.
This is the same reason unit price is useful when packet sizes differ.
Speed does not measure distance
A speed of eighty kilometres per hour is not a distance of eighty kilometres unless the travel time is exactly one hour. In thirty minutes at that constant speed, the distance would be forty kilometres.
Keep speed, distance and time labels beside numbers until the relationship is secure.
2. One relationship connects speed, distance and time
The relationship is distance = speed × time. From it, speed = distance ÷ time and time = distance ÷ speed.
A formula triangle can be a memory aid, but the units provide a stronger check. Kilometres per hour multiplied by hours leaves kilometres. Kilometres divided by kilometres per hour leaves hours.
Worked example A: Find distance
A bus travels at 55 km/h for four hours. Distance = 55 × 4 = 220 km.
Check by division: 220 km ÷ 4 h = 55 km/h.
Worked example B: Find time
A train covers 270 km at an average speed of 90 km/h. Time = 270 ÷ 90 = 3 hours.
The result should be a time, not a speed. Writing 3 km/h would reveal a unit mismatch even if the numerical calculation were correct.
Worked example C: Find speed
A cyclist covers 72 km in four hours. Speed = 72 ÷ 4 = 18 km/h.
Do not multiply because two quantities are given. The operation follows from which quantity is missing.
Worked example D: Fractional time
A car travels at 80 km/h for one and a half hours. One and a half hours is 1.5 hours. Distance = 80 × 1.5 = 120 km.
Another route separates the journey: eighty kilometres in the first hour and forty kilometres in the half hour. Both routes should agree.
3. Convert units before combining quantities
Minutes to hours
Thirty minutes is half an hour. Fifteen minutes is one quarter of an hour. Forty-five minutes is three quarters of an hour.
If speed is in kilometres per hour, time should be expressed in hours before direct multiplication unless the rate itself is converted to kilometres per minute.
Worked example E: Ninety minutes
A cyclist travels at 24 km/h for ninety minutes. Ninety minutes equals 1.5 hours. Distance = 24 × 1.5 = 36 km.
Multiplying twenty-four by ninety would mix kilometres per hour with minutes and produce a number without the intended unit meaning.
Metres per second
A runner covers 100 metres in 20 seconds. Average speed = 100 ÷ 20 = 5 m/s.
To find how far the runner would travel in twelve seconds at the same constant speed, multiply 5 m/s by 12 s to obtain sixty metres.
Convert metres per second to kilometres per hour
One metre per second means 3,600 metres in one hour, or 3.6 kilometres per hour. Therefore 5 m/s equals 18 km/h.
This conversion is an upper-primary extension. A learner can solve most school speed questions by first placing all quantities into one compatible unit system rather than memorising 3.6.
Distance conversion
2.4 kilometres equals 2,400 metres. If time is given in seconds and the desired speed is metres per second, converting the distance to metres first is natural.
Choose the unit system that matches the question’s requested answer.
4. Average speed uses total distance divided by total time
Average speed for a whole journey is not usually found by simply averaging the speeds of the separate stages. The stages may last different lengths of time or cover different distances.
Worked example F: Equal times
A car travels one hour at 40 km/h and one hour at 60 km/h. It covers forty plus sixty = one hundred kilometres in two hours. Average speed = 100 ÷ 2 = 50 km/h.
Here the simple average of forty and sixty happens to work because the times are equal.
Worked example G: Unequal times
A car travels one hour at 40 km/h and three hours at 60 km/h. The distances are forty and one hundred eighty kilometres. Total distance = 220 km, total time = four hours.
Average speed = 220 ÷ 4 = 55 km/h. The simple average of forty and sixty would be fifty, which gives the two speeds equal time weight even though the car spent three times as long at sixty.
Worked example H: Equal distances
A cyclist travels sixty kilometres out at 30 km/h and returns sixty kilometres at 60 km/h. The outward time is two hours; the return time is one hour. Total distance is 120 km and total time is three hours.
Average speed = 120 ÷ 3 = 40 km/h, not forty-five. Equal distances do not make the two speeds equally weighted by time.
Stops belong to total elapsed time when stated
A van covers ninety kilometres in two hours, rests for one hour, then covers another ninety kilometres in two hours. If average speed for the entire elapsed journey is requested, total distance is 180 km and total elapsed time is five hours, giving 36 km/h.
If the question asks for average speed while moving, exclude the rest and use four moving hours, giving forty-five kilometres per hour. The wording decides which time belongs in the denominator.
5. Multi-stage journeys require a stage record
Worked example I: Distance remaining
A journey is 350 km. A car travels for two hours at 80 km/h, then for one hour at 70 km/h. The first stage covers 160 km and the second seventy, total 230 km.
Distance remaining = 350 − 230 = 120 km.
Do not add the speeds and multiply by total time. Different speeds apply to different durations.
Worked example J: Time remaining
A bus must travel 300 km. It has already covered 180 km. The remaining 120 km is travelled at 60 km/h. Time remaining = 120 ÷ 60 = 2 hours.
The average speed from the earlier part is irrelevant unless the question asks about the whole journey.
Worked example K: Compare two routes
Route A is 180 km at 60 km/h, taking three hours. Route B is 210 km at 70 km/h, also taking three hours. The longer route is not slower in elapsed time because its speed is proportionally greater.
This illustrates why distance alone cannot decide travel time.
Worked example L: Arrival time
A journey begins at 8:35 a.m. and lasts two hours forty-five minutes. Add two hours to reach 10:35 a.m., then forty-five minutes to reach 11:20 a.m..
Clock-time arithmetic should use sixty-minute hours, not decimal subtraction.
Worked example M: Catch-up as distance gap
Runner A begins with a 200 m lead. Runner B runs 2 m/s faster. The gap closes by two metres each second. Catch-up time = 200 ÷ 2 = 100 seconds.
This relative-speed reasoning is an extension. It works because both runners move in the same direction and the speed difference remains constant.
Worked example N: Moving towards each other
Two cyclists are sixty kilometres apart and ride towards each other at 12 km/h and 18 km/h. Their separation closes at 30 km/h. Meeting time = 60 ÷ 30 = 2 hours.
The speeds add because both movements reduce the same gap. If they moved in the same direction, the gap would change by the difference in speeds instead.
6. Practice: 24 questions
Keep the worked answers covered. Write compatible units before using a speed relationship.
Questions 1–8: Direct relationships
1. A cyclist travels 36 km in 3 h. Find average speed.
2. A bus travels at 55 km/h for 4 h. Find distance.
3. A train covers 270 km at 90 km/h. Find time.
4. A runner covers 100 m in 20 s. Find average speed in m/s.
5. A car travels at 80 km/h for 1.5 h. Find distance.
6. A cyclist travels at 24 km/h for 90 min. Find distance.
7. How far does an object moving at 5 m/s travel in 12 s?
8. Convert 2.4 km to metres.
Questions 9–16: Average speed and stages
9. A car travels 1 h at 40 km/h and 1 h at 60 km/h. Find average speed.
10. A car travels 1 h at 40 km/h and 3 h at 60 km/h. Find average speed.
11. A cyclist travels 60 km at 30 km/h and returns 60 km at 60 km/h. Find average speed for the whole journey.
12. A van travels 90 km in 2 h, rests 1 h, then travels 90 km in 2 h. Find average speed over the entire elapsed journey.
13. A journey is 350 km. A car travels 2 h at 80 km/h and 1 h at 70 km/h. How far remains?
14. A bus must travel 300 km and has covered 180 km. The remainder is travelled at 60 km/h. How much travel time remains?
15. Route A is 180 km at 60 km/h. Route B is 210 km at 70 km/h. Which route takes longer?
16. A trip begins at 8:35 a.m. and lasts 2 h 45 min. Find the arrival time.
Questions 17–24: Reasoning and comparison
17. Runner A covers 8 km in 40 min. Runner B covers 8 km in 50 min. Who has the greater average speed?
18. Cyclist A rides 40 km in 2 h. Cyclist B rides 50 km in 2 h. Who is faster?
19. Traveller A covers 60 km in 1 h. Traveller B covers 90 km in 2 h. Who is faster?
20. A car covers 150 km in 2.5 h. Find average speed.
21. Runner A has a 200 m lead. Runner B runs 2 m/s faster. How long does B take to catch A, assuming constant speeds?
22. Two cyclists are 60 km apart and ride toward each other at 12 km/h and 18 km/h. When do they meet?
23. A vehicle travels 120 km at 40 km/h, stops 30 min, then travels 120 km at 60 km/h. Find average speed over the full elapsed journey.
24. Explain why the average speed for equal distances travelled at 30 km/h and 60 km/h is not generally (30+60)÷2.
7. Worked answers
Answers 1–8
1. 12 km/h. 36 ÷ 3 = 12.
2. 220 km. 55 × 4 = 220.
3. 3 h. 270 ÷ 90 = 3.
4. 5 m/s. 100 ÷ 20 = 5.
5. 120 km. 80 × 1.5 = 120.
6. 36 km. Ninety minutes is 1.5 hours. 24 × 1.5 = 36.
7. 60 m. 5 m/s × 12 s = 60 m.
8. 2,400 m. Multiply kilometres by one thousand.
Answers 9–16
9. 50 km/h. Total distance is 100 km and total time two hours.
10. 55 km/h. Distances are forty and one hundred eighty, total 220 km over four hours.
11. 40 km/h. The times are two hours outward and one hour return. Total distance 120 km over three hours.
12. 36 km/h. Total distance 180 km; total elapsed time five hours including rest.
13. 120 km. The completed stages cover 160+70=230 km. Subtract from 350.
14. 2 h. Remaining distance is 120 km. Divide by 60 km/h.
15. Neither; both take 3 h. 180÷60=3 and 210÷70=3.
16. 11:20 a.m. Add two hours to 10:35, then forty-five minutes.
Answers 17–24
17. Runner A. Same distance in less time means greater average speed.
18. Cyclist B. Same time, greater distance means greater average speed.
19. Traveller A. A travels at 60 km/h; B at 45 km/h.
20. 60 km/h. 150 ÷ 2.5 = 60.
21. 100 s. The 200 m gap closes at 2 m/s, so 200÷2=100.
22. 2 h. Their closing speed is 12+18=30 km/h. 60÷30=2.
23. About 43.64 km/h. The first 120 km takes 120÷40=3 h. The stop lasts 0.5 h. The second 120 km takes 120÷60=2 h. Total distance is 240 km and total elapsed time is 5.5 h, so average speed is 240÷5.5 ≈ 43.64 km/h.
24. Equal distances usually take different times at different speeds. Average speed depends on total distance divided by total time. For 60 km at 30 km/h and 60 km at 60 km/h, total time is three hours and total distance 120 km, giving 40 km/h rather than 45 km/h.
8. Diagnose units and stages before formula fluency
If a learner multiplies speed by ninety minutes without converting a per-hour rate, ask what the unit product would mean. Keep speed and time labels visible until they are compatible.
If average-speed questions fail, record each stage separately: speed, time, distance. Then total distance and total time. This prevents a simple average of the printed speeds from replacing the actual journey.
Use size checks
An average speed for a journey made entirely between forty and sixty kilometres per hour while moving should usually lie between those moving speeds when no reverse-direction complication changes the interpretation. If rest time is included, the elapsed average can fall below the slowest moving speed.
Such bounds do not produce the exact answer but can expose a mistaken calculation.
Return the answer to the journey
A time answer should fit the distance and speed. A distance answer should be plausible for the duration. An arrival time should be checked against the start time and elapsed duration.
Connect to rate and proportion
Speed is one family of rate. The same unit-rate reasoning used for dollars per item appears here as kilometres per hour or metres per second.
Continue through the Primary Mathematics series
For unit-rate foundations, use Ratio, Rate and Percentage. For time and unit conversion, use Measurement, Units, Perimeter, Area and Volume. For checking whether a speed result is plausible, continue to Estimation, Reasoning and Checking.
Return to the BTT Primary Mathematics Learning Hub.
Original learning guide. Curriculum reference checked 6 September 2026. Examples are teaching illustrations, and constant-speed or continuation assumptions apply only where stated.