Average Speed

Finding the average speed of a journey whose speed changes, in stages and with stops, as total distance ÷ total time; why it is not the average of the speeds; and working back from an average speed to one stage of the journey.

Before you start

Speed is the distance travelled per unit of time: speed = distance ÷ time, so distance = speed × time and time = distance ÷ speed. An object moves at a constant speed when it covers equal distances in equal intervals of time, however short the intervals are.

A time in minutes is written as a fraction of an hour before it is used with a speed in km/h: 40 min is $\frac{40}{60} = \frac{2}{3}$ h, and 1 h 15 min is $1\frac{1}{4}$ h (not 1.15 h).

For example, a car that travels 150 km at a constant speed of 60 km/h takes $150 \div 60 = 2\frac{1}{2}$ h, which is 2 h 30 min.

What Average Speed Is

Very few journeys are made at a constant speed. A bus speeds up after each stop, slows down at junctions and stands still at bus stops and traffic lights. An MRT train goes fastest between stations and stops at every one. Yet we still want one number that says how fast the whole journey was.

That number is the average speed: the constant speed at which the same total distance would be covered in the same total time. It is found from the whole journey at once, so we do not need to know the speed at every moment.

Average speed

Average speed = total distance travelled ÷ total time taken

So total distance travelled = average speed × total time taken, and total time taken = total distance travelled ÷ average speed.

The total time is the time from the start of the journey to its end, including any time spent standing still. If the speed is constant all the way, the average speed is that constant speed.

Example 1: A bus ride with stops

Ali takes a bus from his home to his school, a distance of 12 km. With all its stops, the ride takes 30 minutes. Find the average speed of the bus in km/h.

Solution

The speed is wanted in km/h, so write the time in hours: 30 min $= \frac{30}{60}$ h $= \frac{1}{2}$ h.

Average speed = total distance ÷ total time $= 12 \div \frac{1}{2} = 12 \times 2 = 24$ , so 24 km/h.

The bus did not go at 24 km/h all the time: at some moments it went much faster and at the bus stops it did not move at all. A bus going steadily at 24 km/h would cover the same 12 km in the same 30 minutes.

The average speed of the bus is 24 km/h.

A Journey in Stages

Many journeys are made in stages, each at its own speed. To find the average speed of the whole journey, find the distance and the time of each stage, add the distances and add the times, and then divide: total distance ÷ total time.

A stop is part of the journey. Its distance is 0, but its time is added to the total time, because the clock keeps running while you stand still. (If a question asks only for the average speed while moving, leave the stop out; read the question to see which it wants.)

Example 2: The times of the stages are given

Mrs Goh drives for 2 hours at an average speed of 60 km/h, then for 3 hours at an average speed of 70 km/h. Find her average speed for the whole journey.

Solution

The distance of each stage is speed × time.

First stage: $60 \times 2 = 120$ , so 120 km. Second stage: $70 \times 3 = 210$ , so 210 km.

Total distance $= 120 + 210 = 330$ , so 330 km. Total time $= 2 + 3 = 5$ , so 5 h.

Average speed $= 330 \div 5 = 66$ , so 66 km/h.

Notice that the average of the two speeds, $(60 + 70) \div 2 = 65$ , is not the answer. Mrs Goh drove longer at 70 km/h (3 h) than at 60 km/h (2 h), so her average speed is closer to 70.

Her average speed for the whole journey is 66 km/h.

Example 3: The distances of the stages are given

Mr Tan runs 5 km in 40 minutes and then walks 2 km at an average speed of 4 km/h. Find his average speed for the whole 7 km in km/h.

Solution

Find the time of each stage in hours.

Running: 40 min $= \frac{40}{60}$ h $= \frac{2}{3}$ h.

Walking: time = distance ÷ speed $= 2 \div 4 = \frac{1}{2}$ , so $\frac{1}{2}$ h.

Total time $= \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6}$ , so $\frac{7}{6}$ h (which is 1 h 10 min).

Total distance $= 5 + 2 = 7$ , so 7 km.

Average speed $= 7 \div \frac{7}{6} = 7 \times \frac{6}{7} = 6$ , so 6 km/h.

Mr Tan's average speed for the whole 7 km is 6 km/h.

Example 4: A journey with a stop

Siti cycles along a park connector. She cycles 12 km in 45 minutes, rests for 15 minutes, and then cycles another 6 km in 30 minutes.
(a) Find her average speed for the whole journey.
(b) Find her average speed for the time she was cycling.

Solution

Total distance $= 12 + 6 = 18$ , so 18 km. The rest adds no distance.

(a) The whole journey includes the rest. Total time $= 45 + 15 + 30 = 90$ min $= \frac{90}{60}$ h $= \frac{3}{2}$ h.

Average speed $= 18 \div \frac{3}{2} = 18 \times \frac{2}{3} = 12$ , so 12 km/h.

(b) The time spent cycling leaves out the rest: $45 + 30 = 75$ min $= \frac{75}{60}$ h $= \frac{5}{4}$ h.

Average speed while cycling $= 18 \div \frac{5}{4} = 18 \times \frac{4}{5} = \frac{72}{5} = 14.4$ , so 14.4 km/h.

The rest lowers the average speed of the whole journey, because it adds time but no distance.

(a) Her average speed for the whole journey is 12 km/h.
(b) Her average speed while cycling is 14.4 km/h.

Why It Is Not the Average of the Speeds

It is tempting to add the speeds of the stages and divide by the number of stages. This usually gives the wrong answer, because the stages need not take the same time. The journey spends longer at the slower speed when the distances are equal, so the average speed is pulled towards the slower speed.

Example 5: There and back at two speeds

Mr Tan drives 120 km from his home in Singapore to a town in Johor at an average speed of 60 km/h. On the way back, in heavy traffic, he drives the same 120 km at an average speed of 40 km/h. Find his average speed for the whole journey there and back.

Solution

Find the time of each stage: time = distance ÷ speed.

There: $120 \div 60 = 2$ , so 2 h. Back: $120 \div 40 = 3$ , so 3 h.

Total distance $= 120 + 120 = 240$ , so 240 km. Total time $= 2 + 3 = 5$ , so 5 h.

Average speed $= 240 \div 5 = 48$ , so 48 km/h.

The average of the two speeds is $(60 + 40) \div 2 = 50$ , which is wrong: he drove for 3 h at 40 km/h and only 2 h at 60 km/h, so his average speed is closer to 40.

His average speed for the whole journey is 48 km/h, not 50 km/h.

Compare the two journeys in the table below. In the first, the two distances are equal and the times are not. In the second, the two times are equal, and then the average speed comes out equal to the average of the speeds.

Journey	Distances	Times	Total distance ÷ total time	Average of the speeds
120 km at 60 km/h, then 120 km at 40 km/h	120 km and 120 km	2 h and 3 h	$240 \div 5 = 48$ , so 48 km/h	50 km/h (wrong)
1 h at 60 km/h, then 1 h at 40 km/h	60 km and 40 km	1 h and 1 h	$100 \div 2 = 50$ , so 50 km/h	50 km/h (equal, because the times are equal)

Do not average the speeds

The average speed of a journey is not, in general, the average of the speeds of its stages. When every stage takes the same time, the two are equal; when the times are different, they are usually different.

Always find the average speed as total distance ÷ total time.

Keeping Fractions of an Hour Exact

A stage often takes a time that is not a whole number of hours: 100 km at 60 km/h takes $\frac{100}{60} = 1\frac{2}{3}$ h, which is 1 h 40 min. As a decimal this is $1.666\ldots$ h, and it never ends. Writing it as 1.67 h is a small rounding, but the error is carried into every step that follows and can spoil the answer. Keep such times as fractions (or as hours and minutes) until the end, and round only the final answer.

Example 6: Times that are fractions of an hour

A coach takes a school group 100 km to a camp at an average speed of 60 km/h, and returns empty along the same road at an average speed of 90 km/h. Find the average speed of the coach for the whole journey.

Solution

There: $100 \div 60 = \frac{100}{60} = \frac{5}{3}$ , so $\frac{5}{3}$ h (1 h 40 min).

Back: $100 \div 90 = \frac{100}{90} = \frac{10}{9}$ , so $\frac{10}{9}$ h.

Total time $= \frac{5}{3} + \frac{10}{9} = \frac{15}{9} + \frac{10}{9} = \frac{25}{9}$ , so $\frac{25}{9}$ h.

Total distance $= 100 + 100 = 200$ , so 200 km.

Average speed $= 200 \div \frac{25}{9} = 200 \times \frac{9}{25} = 72$ , so 72 km/h.

Had the times been rounded to 1.67 h and 1.11 h, the total would be 2.78 h and $200 \div 2.78$ would give about 71.94 km/h, which is wrong. (The average of the two speeds, 75 km/h, is wrong too.)

The average speed of the coach for the whole journey is exactly 72 km/h.

Working Back from an Average Speed

Sometimes the average speed of the whole journey is known and a part of the journey is not. Use the average speed to find the total time (total distance ÷ average speed) or the total distance (average speed × total time). Then take away what is known about the other stages to find what is left.

Example 7: The speed needed for the rest of the journey

Mr Tan plans to drive 210 km to a resort at an average speed of 70 km/h for the whole journey. He drives the first 120 km at an average speed of 60 km/h. At what average speed must he drive the rest of the way to keep to his plan?

Solution

Total time allowed $= 210 \div 70 = 3$ , so 3 h.

Time for the first 120 km $= 120 \div 60 = 2$ , so 2 h.

Time left $= 3 - 2 = 1$ , so 1 h. Distance left $= 210 - 120 = 90$ , so 90 km.

Speed needed $= 90 \div 1 = 90$ , so 90 km/h.

Check: 210 km in $2 + 1 = 3$ h is $210 \div 3 = 70$ km/h. (Driving the rest at 80 km/h, "so that 60 and 80 average 70", would not do: the rest would take $90 \div 80 = 1\frac{1}{8}$ h, and he would arrive $\frac{1}{8}$ h, which is 7 min 30 s, late.)

He must drive the remaining 90 km at an average speed of 90 km/h.

Example 8: How long was the rest?

Siti cycles 30 km. While she is cycling, her average speed is 15 km/h. She stops once for a rest, and her average speed for the whole trip, including the rest, is 12 km/h. How long does she rest?

Solution

Total time for the whole trip $= 30 \div 12 = 2\frac{1}{2}$ , so $2\frac{1}{2}$ h.

Time spent cycling $= 30 \div 15 = 2$ , so 2 h.

Time resting $= 2\frac{1}{2} - 2 = \frac{1}{2}$ , so $\frac{1}{2}$ h, which is 30 min.

Siti rests for 30 minutes.

Common Mistakes

Averaging the speeds. 60 km/h there and 40 km/h back over the same distance is not 50 km/h. Always use total distance ÷ total time; here 48 km/h.
Leaving out a stop. The average speed for the whole journey counts every minute from start to finish, including rests and waits.
Writing minutes as a decimal of an hour. 40 min is $\frac{2}{3}$ h, not 0.4 h; 1 h 15 min is $1\frac{1}{4}$ h, not 1.15 h.
Rounding too early. 100 km at 60 km/h takes $1\frac{2}{3}$ h; writing 1.67 h in the middle of the working can make the final answer wrong. Keep fractions and round only at the end.
Adding times in different units. Write every time in hours (or every time in minutes) before adding.

Practice

Work each question on paper before you open its solution. Find the total distance and the total time first, keep fractions of an hour exact, and give answers that are not exact correct to 3 significant figures.

Question 1

A bus travels 18 km in 45 minutes, including its stops. Find its average speed in km/h.

ShowHide solution

45 min $= \frac{45}{60}$ h $= \frac{3}{4}$ h.

Average speed $= 18 \div \frac{3}{4} = 18 \times \frac{4}{3} = 24$ , so 24 km/h.

The average speed of the bus is 24 km/h.

Question 2

A lorry travels from A to B in 3 hours at an average speed of 60 km/h, and then from B to C in 2 hours at an average speed of 85 km/h. Find its average speed for the journey from A to C.

ShowHide solution

A to B: $60 \times 3 = 180$ , so 180 km. B to C: $85 \times 2 = 170$ , so 170 km.

Total distance $= 180 + 170 = 350$ , so 350 km. Total time $= 3 + 2 = 5$ , so 5 h.

Average speed $= 350 \div 5 = 70$ , so 70 km/h. (Not $(60 + 85) \div 2 = 72.5$ : the times are not equal.)

The average speed from A to C is 70 km/h.

Question 3

A car travels for the first $1\frac{1}{2}$ hours at an average speed of 64 km/h and for the next $2\frac{1}{2}$ hours at an average speed of 72 km/h. Find its average speed for the whole journey.

ShowHide solution

First part: $64 \times 1\frac{1}{2} = 96$ , so 96 km. Second part: $72 \times 2\frac{1}{2} = 180$ , so 180 km.

Total distance $= 96 + 180 = 276$ , so 276 km. Total time $= 1\frac{1}{2} + 2\frac{1}{2} = 4$ , so 4 h.

Average speed $= 276 \div 4 = 69$ , so 69 km/h.

The average speed for the whole journey is 69 km/h.

Question 4

Ravi runs 3 km at an average speed of 12 km/h and then walks 2 km at an average speed of 6 km/h. Find his average speed for the whole 5 km.

ShowHide solution

Running time $= 3 \div 12 = \frac{1}{4}$ , so $\frac{1}{4}$ h. Walking time $= 2 \div 6 = \frac{1}{3}$ , so $\frac{1}{3}$ h.

Total time $= \frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}$ , so $\frac{7}{12}$ h (35 min).

Average speed $= 5 \div \frac{7}{12} = 5 \times \frac{12}{7} = \frac{60}{7} = 8\frac{4}{7}$ , so $8\frac{4}{7}$ km/h $\approx 8.57$ km/h.

Ravi's average speed is $8\frac{4}{7}$ km/h, which is about 8.57 km/h (to 3 significant figures).

Question 5

A train travels 20 km at an average speed of 80 km/h, stops at a station for 5 minutes, and then travels a further 15 km at an average speed of 60 km/h. Find its average speed for the whole journey, including the stop.

ShowHide solution

First stage: $20 \div 80 = \frac{1}{4}$ h, which is 15 min. Second stage: $15 \div 60 = \frac{1}{4}$ h, which is 15 min.

Total time $= 15 + 5 + 15 = 35$ min $= \frac{35}{60}$ h $= \frac{7}{12}$ h.

Total distance $= 20 + 15 = 35$ , so 35 km.

Average speed $= 35 \div \frac{7}{12} = 35 \times \frac{12}{7} = 60$ , so 60 km/h.

The average speed of the train for the whole journey is 60 km/h.

Question 6

Mei Ling cycles 12 km to the beach at an average speed of 18 km/h, and cycles back along the same route at an average speed of 12 km/h.
(a) Find her average speed for the whole journey.
(b) Explain why the answer is not 15 km/h.

ShowHide solution

(a) There: $12 \div 18 = \frac{2}{3}$ , so $\frac{2}{3}$ h. Back: $12 \div 12 = 1$ , so 1 h.

Total time $= \frac{2}{3} + 1 = \frac{5}{3}$ , so $\frac{5}{3}$ h. Total distance $= 12 + 12 = 24$ , so 24 km.

Average speed $= 24 \div \frac{5}{3} = 24 \times \frac{3}{5} = \frac{72}{5} = 14.4$ , so 14.4 km/h.

(b) 15 km/h is the average of the two speeds, $(18 + 12) \div 2$ . But she spends 1 h at 12 km/h and only $\frac{2}{3}$ h at 18 km/h, so the slower speed counts for more and the average speed is below 15 km/h.

(a) Her average speed is 14.4 km/h.
(b) She spends longer at the slower speed, so the average speed is not the average of the two speeds.

Question 7

Hui Min says that the average speed of a journey is never the average of the speeds of its stages. Mrs Goh drives for 1 hour at 50 km/h and then for 1 hour at 70 km/h. Find Mrs Goh's average speed, and say whether Hui Min is right.

ShowHide solution

Distances: $50 \times 1 = 50$ km and $70 \times 1 = 70$ km; total 120 km.

Total time $= 1 + 1 = 2$ , so 2 h. Average speed $= 120 \div 2 = 60$ , so 60 km/h.

The average of the speeds is $(50 + 70) \div 2 = 60$ , which is the same, because the two stages take the same time.

Mrs Goh's average speed is 60 km/h, which equals the average of the speeds. Hui Min is not right: the two are equal when every stage takes the same time. They are different in general, so the safe method is always total distance ÷ total time.

Question 8

A coach makes a journey of 4 hours 30 minutes at an average speed of 64 km/h.
(a) How far does it travel?
(b) It covers the first 160 km in 2 hours. Find its average speed for the rest of the journey.

ShowHide solution

(a) 4 h 30 min $= 4\frac{1}{2}$ h. Total distance = average speed × total time $= 64 \times 4\frac{1}{2} = 288$ , so 288 km.

(b) Distance left $= 288 - 160 = 128$ , so 128 km. Time left $= 4\frac{1}{2} - 2 = 2\frac{1}{2}$ , so $2\frac{1}{2}$ h.

Average speed for the rest $= 128 \div 2\frac{1}{2} = 128 \times \frac{2}{5} = 51.2$ , so 51.2 km/h.

(a) The coach travels 288 km. (b) Its average speed for the rest of the journey is 51.2 km/h.

Question 9

Mr Tan drives 45 km to work at an average speed of 60 km/h. One morning he leaves home 9 minutes later than usual but wants to arrive at his usual time. At what average speed must he drive?

ShowHide solution

Usual time $= 45 \div 60 = \frac{3}{4}$ h, which is 45 min.

Time he has that morning $= 45 - 9 = 36$ min $= \frac{36}{60}$ h $= \frac{3}{5}$ h.

Average speed needed $= 45 \div \frac{3}{5} = 45 \times \frac{5}{3} = 75$ , so 75 km/h.

He must drive at an average speed of 75 km/h.

Question 10

Hui Min leaves home on her bicycle at 8.50 a.m. She cycles 16 km to a park, rests there for 20 minutes, and cycles the same 16 km home, arriving at 11.30 a.m.
(a) Find her average speed for the whole trip, including the rest.
(b) Find her average speed while she was cycling, correct to 3 significant figures.

ShowHide solution

From 8.50 a.m. to 11.30 a.m. is 2 h 40 min (8.50 a.m. to 10.50 a.m. is 2 h, and 10.50 a.m. to 11.30 a.m. is 40 min).

Total distance $= 16 + 16 = 32$ , so 32 km.

(a) Total time $= 2\frac{40}{60} = 2\frac{2}{3} = \frac{8}{3}$ , so $\frac{8}{3}$ h. Average speed $= 32 \div \frac{8}{3} = 32 \times \frac{3}{8} = 12$ , so 12 km/h.

(b) Cycling time $= 2$ h 40 min $- 20$ min $= 2$ h 20 min $= 2\frac{1}{3}$ h $= \frac{7}{3}$ h.

Average speed while cycling $= 32 \div \frac{7}{3} = 32 \times \frac{3}{7} = \frac{96}{7} = 13\frac{5}{7} \approx 13.7$ , so about 13.7 km/h.

(a) Her average speed for the whole trip is 12 km/h. (b) Her average speed while cycling is about 13.7 km/h.

Question 11

A bus travels 24 km. While it is moving, its average speed is 32 km/h. Its average speed for the whole trip, including its stops, is 24 km/h. How long does it spend at its stops?

ShowHide solution

Total time for the whole trip $= 24 \div 24 = 1$ , so 1 h.

Time spent moving $= 24 \div 32 = \frac{3}{4}$ , so $\frac{3}{4}$ h.

Time at the stops $= 1 - \frac{3}{4} = \frac{1}{4}$ , so $\frac{1}{4}$ h, which is 15 min.

Check: $24 \div 1 = 24$ km/h for the whole trip, and $24 \div \frac{3}{4} = 32$ km/h while moving.

The bus spends 15 minutes at its stops.

Question 12

A car travels for 2 hours and then for another 3 hours. Its average speed for the whole 5 hours is 40 km/h.
(a) If its average speed for the first 2 hours is 55 km/h, find its average speed for the last 3 hours.
(b) Give another pair of whole-number speeds, one for the first 2 hours and one for the last 3 hours, that also gives an average speed of 40 km/h, and show that it works.

ShowHide solution

Total distance = average speed × total time $= 40 \times 5 = 200$ , so 200 km. Whatever the two speeds are, the two stages must add up to 200 km.

(a) First 2 h: $55 \times 2 = 110$ , so 110 km. Distance left $= 200 - 110 = 90$ , so 90 km in 3 h.

Average speed for the last 3 h $= 90 \div 3 = 30$ , so 30 km/h.

(b) For example, 70 km/h for the first 2 h covers $70 \times 2 = 140$ km, leaving $200 - 140 = 60$ km for the last 3 h, which is $60 \div 3 = 20$ km/h.

Check: $(140 + 60) \div 5 = 200 \div 5 = 40$ . (Other pairs work too, such as 25 km/h and 50 km/h: $50 + 150 = 200$ .)

(a) 30 km/h. (b) For example, 70 km/h for the first 2 hours and 20 km/h for the last 3 hours; any pair whose two distances add up to 200 km works.

Question 13

Ali cycles 10 km to his friend's house at an average speed of 10 km/h. He wants his average speed for the whole 20 km there and back to be 20 km/h. His sister says he must cycle home at 30 km/h, "because 10 and 30 average 20". Is she right? How fast must he cycle home?

ShowHide solution

For an average speed of 20 km/h over 20 km, the total time must be $20 \div 20 = 1$ , so 1 h.

The ride there already takes $10 \div 10 = 1$ , so 1 h. That uses up all the time allowed, leaving no time at all for the ride home.

Check his sister's answer: at 30 km/h the ride home takes $10 \div 30 = \frac{1}{3}$ h, so the total time is $1\frac{1}{3} = \frac{4}{3}$ h and the average speed is $20 \div \frac{4}{3} = 20 \times \frac{3}{4} = 15$ km/h, not 20 km/h.

However fast he rides home, the total time is more than 1 h, so the average speed is always less than 20 km/h.

His sister is wrong: riding home at 30 km/h gives an average of only 15 km/h. No speed is fast enough, because the ride there already uses the whole hour that an average of 20 km/h allows.

Key Points

Average speed = total distance travelled ÷ total time taken. It is the constant speed that would cover the same distance in the same time.
Total distance = average speed × total time, and total time = total distance ÷ average speed.
For a journey in stages, find each stage's distance and time, add the distances, add the times, then divide.
A stop adds time but no distance, and counts in the total time of the whole journey.
Do not average the speeds: the average speed is equal to the average of the speeds when every stage takes the same time, but not in general.
Write minutes as a fraction of an hour (40 min $= \frac{2}{3}$ h), keep fractions exact, and round only the final answer.
To work back, find the total time (or total distance) from the average speed, then take away the stages you know.