Speed, Distance and Time

Speed as the distance travelled per unit of time; constant speed; finding the speed, the distance or the time from the other two, with times in hours and minutes and times of day.

Before you start

A rate tells how much of one quantity there is per unit of another kind of quantity, such as 12 litres per minute. To find a rate, divide to find the amount for one unit: a tap that fills 60 L in 5 min fills $60 \div 5 = 12$ L in each minute, so its rate is 12 L/min.

There are 60 minutes in an hour, so 1 min is $\frac{1}{60}$ h, and 60 seconds in a minute. Dividing by a fraction is multiplying by its reciprocal: $3 \div \frac{1}{4} = 3 \times 4 = 12$ .

Speed Is a Rate

A cyclist who rides 30 km in 2 hours without slowing down or speeding up rides $30 \div 2 = 15$ km in each hour. We say her speed is 15 kilometres per hour, written 15 km/h. Speed is a rate: it tells how far something travels per unit of time.

The unit of a speed is a unit of distance "per" a unit of time: kilometres per hour (km/h), metres per second (m/s), metres per minute (m/min), kilometres per minute (km/min) or centimetres per second (cm/s). The unit tells you what the number means: 15 km/h is 15 km in each hour.

Speed

The speed of an object is the distance it travels per unit of time.

Speed = distance ÷ time

The unit of the speed comes from the units of the distance and the time: metres ÷ seconds gives m/s; kilometres ÷ hours gives km/h.

Example 1: Finding a speed

A car travels 90 m in 5 s at a constant speed. Find its speed.

Solution

The distance is in metres and the time in seconds, so the speed will be in metres per second.

Speed = distance ÷ time $= 90 \div 5 = 18$ , so 18 m/s.

In each second the car travels 18 m. Check: in 5 s it travels $18 \times 5 = 90$ m.

The speed of the car is 18 m/s.

Example 2: The same speed in two units

Ali cycles 3 km along a park connector in 15 minutes at a constant speed. Find his speed
(a) in km/min,
(b) in km/h.

Solution

(a) The time is in minutes, so divide by 15 minutes: speed $= 3 \div 15 = \frac{1}{5} = 0.2$ , so 0.2 km/min.

(b) For km/h the time must be in hours. 15 min $= \frac{15}{60}$ h $= \frac{1}{4}$ h.

Speed $= 3 \div \frac{1}{4} = 3 \times 4 = 12$ , so 12 km/h.

Check by thinking of what a speed means: 15 minutes is a quarter of an hour, and an hour is 4 lots of 15 minutes, so in an hour Ali cycles $4 \times 3 = 12$ km.

(a) Ali's speed is 0.2 km/min. (b) His speed is 12 km/h.

Constant Speed

The cyclist in the table rides at a constant speed of 12 km/h. In every 15 minutes she rides another 3 km, and whichever column you take after 0 min, distance ÷ time gives the same speed.

Time from the start	0 min	15 min	30 min	45 min	60 min
Distance from the start	0 km	3 km	6 km	9 km	12 km

Constant speed

An object moves at a constant speed when it travels equal distances in equal intervals of time, however short the intervals are. Its speed does not change.

At a constant speed, distance ÷ time gives the same speed for the whole journey and for every part of it.

Most real journeys are not at a constant speed: a bus stops at bus stops and slows down at traffic lights. For such a journey, total distance ÷ total time gives the average speed, which is the subject of a later lesson. In this lesson every speed is constant unless the question says otherwise.

Example 3: Is the speed constant?

The table shows how far Mei Ling, running, and a bus have gone from the start at the end of each minute. Which of them could be moving at a constant speed? Find that speed.

Time from the start	0 min	1 min	2 min	3 min	4 min
Mei Ling	0 m	250 m	500 m	750 m	1000 m
Bus	0 m	400 m	1100 m	1300 m	2000 m

Solution

Find the distance travelled in each minute by subtracting each distance from the one after it.

Mei Ling: $250 - 0 = 250$ , $500 - 250 = 250$ , $750 - 500 = 250$ , $1000 - 750 = 250$ . She runs 250 m in every minute: equal distances in equal times.

Bus: $400 - 0 = 400$ , $1100 - 400 = 700$ , $1300 - 1100 = 200$ , $2000 - 1300 = 700$ . The distances in the four minutes are not equal, so the bus's speed is not constant.

Mei Ling's speed $= 1000 \div 4 = 250$ , so 250 m/min (any one minute gives the same: 250 m in 1 min).

Mei Ling covers 250 m in every minute, so she could be running at a constant speed of 250 m/min. The bus cannot be moving at a constant speed.

Finding the Distance or the Time

A speed of 12 km/h means 12 km in each hour. In 3 hours the cyclist rides 3 lots of 12 km, which is $12 \times 3 = 36$ km. So to find a distance, multiply the speed by the time.

Now turn it round: how long does she take to ride 36 km? Each hour gives her 12 km, so the number of hours is the number of 12s in 36, which is $36 \div 12 = 3$ . So to find a time, divide the distance by the speed.

The three forms

Speed = distance ÷ time

Distance = speed × time

Time = distance ÷ speed

The units must agree. With a speed in km/h, the distance is in kilometres and the time in hours; with a speed in m/s, the distance is in metres and the time in seconds.

Which way round?

If you are not sure whether to multiply or divide, say what the speed means in words. "60 km/h" is "60 km in every hour". More hours means more kilometres, so a distance is speed × time. The number of hours is how many lots of 60 km fit into the distance, so a time is distance ÷ speed.

Then check that the answer is sensible: a walk of 450 m does not take 675 seconds at 1.5 m/s (that is speed × distance, which has no meaning), but 300 seconds, since $1.5 \times 300 = 450$ .

Example 4: Finding a distance

An MRT train travels between two stations at a constant speed of 60 km/h for 2 minutes. How far apart are the stations?

Solution

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

Distance = speed × time $= 60 \times \frac{1}{30} = 2$ , so 2 km.

Check by reasoning: 60 km in 60 minutes is 1 km in each minute, so in 2 minutes the train travels 2 km.

The stations are 2 km apart.

Example 5: Finding a time

Mei Ling walks 600 m to the bus stop at a constant speed of 1.5 m/s. How long does she take? Give your answer in minutes and seconds.

Solution

The speed is in m/s and the distance is in metres, so the time will be in seconds.

Time = distance ÷ speed $= 600 \div 1.5 = 400$ , so 400 s.

400 s $= 6 \times 60 + 40$ s, which is 6 min 40 s.

Check: $1.5 \times 400 = 600$ , so in 400 s she walks 600 m.

Mei Ling takes 6 minutes 40 seconds.

Example 6: Two parts of a ride at the same speed

Ravi cycles at a constant speed along a straight path from checkpoint A, through checkpoint B, to checkpoint C. A to B is 400 m and he takes 80 s. A to C is 1000 m. How long does he take to cycle from B to C?

Solution

Draw the ride as a line and mark what is known.

His speed from A to B: speed $= 400 \div 80 = 5$ , so 5 m/s. The speed is constant, so it is 5 m/s from B to C too.

Distance from B to C $= 1000 - 400 = 600$ , so 600 m.

Time from B to C $= 600 \div 5 = 120$ , so 120 s, which is 2 minutes.

Check: 600 m is $1\frac{1}{2}$ times 400 m, so it takes $1\frac{1}{2}$ times 80 s, which is 120 s.

Ravi takes 120 seconds (2 minutes) to cycle from B to C.

Times in Hours and Minutes

With a speed in km/h the time must be in hours, but times are usually given in hours and minutes. An hour has 60 minutes, not 100, so 1 h 15 min is $1\frac{15}{60} = 1\frac{1}{4}$ h, which is 1.25 h. It is not 1.15 h.

Some numbers of minutes give a fraction of an hour with no exact decimal: 20 min is $\frac{20}{60} = \frac{1}{3}$ h $= 0.333\ldots$ h. Keep such a time as a fraction until the end, so that nothing is lost by rounding.

Minutes	6 min	12 min	15 min	20 min	30 min	40 min	45 min
Hours	$\frac{1}{10}$ h = 0.1 h	$\frac{1}{5}$ h = 0.2 h	$\frac{1}{4}$ h = 0.25 h	$\frac{1}{3}$ h	$\frac{1}{2}$ h = 0.5 h	$\frac{2}{3}$ h	$\frac{3}{4}$ h = 0.75 h

Changing between minutes and hours

Minutes to hours: divide by 60. 1 h 15 min $= 1\frac{15}{60}$ h $= 1\frac{1}{4}$ h.

A part of an hour to minutes: multiply by 60. 0.4 h $= 0.4 \times 60$ min $= 24$ min, so 2.4 h is 2 h 24 min, and $\frac{2}{3}$ h $= \frac{2}{3} \times 60$ min $= 40$ min.

Example 7: A time given in hours and minutes

A coach on the North–South Expressway in Malaysia travels at a constant speed of 80 km/h for 1 hour 15 minutes. How far does it travel?

Solution

Write the time in hours: 1 h 15 min $= 1\frac{15}{60}$ h $= 1\frac{1}{4}$ h $= 1.25$ h.

Distance = speed × time $= 80 \times 1.25 = 100$ , so 100 km.

Check: in 1 hour it travels 80 km, and in a quarter of an hour $80 \div 4 = 20$ km; $80 + 20 = 100$ km.

Using 1.15 h would give $80 \times 1.15 = 92$ km, which is wrong: 15 minutes is not 0.15 of an hour.

The coach travels 100 km.

Example 8: An answer in hours and minutes

Mr Tan drives 210 km at a constant speed of 72 km/h. How long does the drive take? Give your answer in hours and minutes.

Solution

Time = distance ÷ speed $= 210 \div 72 = \frac{210}{72} = \frac{35}{12} = 2\frac{11}{12}$ , so $2\frac{11}{12}$ h.

The part of an hour in minutes: $\frac{11}{12} \times 60 = 55$ , so 55 min.

Check: $72 \times 2\frac{11}{12} = 72 \times \frac{35}{12} = 6 \times 35 = 210$ .

On a calculator, $210 \div 72 = 2.9166\ldots$ . Here 0.9166… is a part of an hour; it is not 91 minutes. Multiply it by 60 to get the minutes.

The drive takes 2 hours 55 minutes.

Example 9: A speed from a time in hours and minutes

Siti cycles 18 km in 1 hour 20 minutes at a constant speed. Find her speed in km/h.

Solution

Write the time in hours, as a fraction: 1 h 20 min $= 1\frac{20}{60}$ h $= 1\frac{1}{3}$ h $= \frac{4}{3}$ h.

Speed = distance ÷ time $= 18 \div \frac{4}{3} = 18 \times \frac{3}{4} = 13.5$ , so 13.5 km/h.

By the unitary method: 80 min → 18 km, so 20 min → $18 \div 4 = 4.5$ km, and 60 min → $4.5 \times 3 = 13.5$ km.

Check: $13.5 \times \frac{4}{3} = 18$ .

Siti's speed is 13.5 km/h.

Times of Day

A journey that starts at a time of day ends at the time of departure plus the time taken. Add the hours first, then the minutes, remembering that 60 minutes make the next hour. It often helps to count on to the next whole hour.

To find the time taken between two times of day, count on from the earlier time: from 2.40 p.m. to 4.40 p.m. is 2 hours, and from 4.40 p.m. to 4.55 p.m. is 15 minutes, so 2 hours 15 minutes in all.

Example 10: When does it arrive?

A coach leaves Singapore at 9.45 a.m. for a town 186 km away. It travels at a constant speed of 72 km/h. At what time does it arrive?

Solution

Draw the journey and mark what is known.

Time = distance ÷ speed $= 186 \div 72 = \frac{186}{72} = \frac{31}{12} = 2\frac{7}{12}$ , so $2\frac{7}{12}$ h.

$\frac{7}{12} \times 60 = 35$ , so the journey takes 2 h 35 min.

Count on from 9.45 a.m.: 2 hours later is 11.45 a.m.; 15 more minutes is 12 noon; 20 more minutes is 12.20 p.m. (The 35 minutes were taken as $15 + 20$ .)

The coach arrives at 12.20 p.m.

Example 11: A distance from two times of day

Ravi leaves home at 7.05 a.m. and cycles to school at a constant speed of 12 km/h. He arrives at 7.25 a.m. How far is his school from his home?

Solution

Time taken: from 7.05 a.m. to 7.25 a.m. is 20 minutes, which is $\frac{20}{60} = \frac{1}{3}$ h.

Distance = speed × time $= 12 \times \frac{1}{3} = 4$ , so 4 km.

Check: 12 km in 60 minutes is 4 km in each 20 minutes.

His school is 4 km from his home.

Common Mistakes

Writing minutes as a decimal of an hour. 1 h 15 min is 1.25 h, not 1.15 h, and 2 h 30 min is 2.5 h, not 2.3 h. Divide the minutes by 60.
Reading the decimal part of an hour as minutes. 2.4 h is 2 h 24 min, not 2 h 40 min. Multiply the decimal part by 60.
Mixing units. At 60 km/h for 2 minutes, $60 \times 2 = 120$ km is wrong: the speed is per hour, so the time must be in hours ( $\frac{1}{30}$ h), giving 2 km.
Dividing the wrong way round. Time is distance ÷ speed, not speed ÷ distance. Say what the speed means ("1.5 m in every second") and check that the answer is sensible.
Rounding a fraction of an hour too early. $\frac{1}{3}$ h is not 0.33 h exactly. Keep the fraction, or work in minutes, until the end.
Adding times of day as decimals. 9.45 a.m. and 2 h 35 min do not make "11.80": 45 + 35 minutes is 80 minutes, which is 1 hour 20 minutes, so the arrival time is 12.20 p.m.

Practice

Work each question on paper before you open its solution. All speeds are constant. Write the time in the unit that matches the speed, and check each answer by working it back.

Question 1

Ravi runs 100 m in 16 s. Find his speed in m/s.

ShowHide solution

Speed = distance ÷ time $= 100 \div 16 = 6.25$ , so 6.25 m/s.

Check: $6.25 \times 16 = 100$ .

Ravi's speed is 6.25 m/s.

Question 2

Hui Min cycles at 14 km/h for 2.5 hours. How far does she cycle?

ShowHide solution

Distance = speed × time $= 14 \times 2.5 = 35$ , so 35 km.

Check: 14 km in each of 2 hours is 28 km, and half an hour gives 7 km more; $28 + 7 = 35$ .

She cycles 35 km.

Question 3

A lift in a tall building rises at a constant speed of 5 m/s. How long does it take to rise 50 m?

ShowHide solution

Time = distance ÷ speed $= 50 \div 5 = 10$ , so 10 s.

Check: 5 m in each second for 10 seconds is 50 m.

The lift takes 10 seconds.

Question 4

The table shows how far a remote-control car has gone from the start.
(a) Explain how the table shows that the car could be moving at a constant speed.
(b) Find its speed.
(c) At this speed, how far will it have gone after 8 s, and after 15 s?
(d) How long does it take to travel 63 m?

Time from the start	0 s	2 s	4 s	6 s
Distance from the start	0 m	7 m	14 m	21 m

ShowHide solution

(a) In each 2 seconds the car goes another 7 m ( $7 - 0$ , $14 - 7$ , $21 - 14$ ): equal distances in equal intervals of time.

(b) Speed $= 21 \div 6 = 3.5$ , so 3.5 m/s. (Any column gives the same: $7 \div 2 = 3.5$ , $14 \div 4 = 3.5$ .)

(d) Time $= 63 \div 3.5 = 18$ , so 18 s. Check: 63 m is 3 times 21 m, so it takes 3 times 6 s.

(a) It goes 7 m in every 2 s. (b) 3.5 m/s. (c) 28 m after 8 s and 52.5 m after 15 s. (d) 18 seconds.

Question 5

(a) Write 2 h 45 min in hours.
(b) Write 50 min as a fraction of an hour.
(c) Write 1.35 h in hours and minutes.
(d) Write $2\frac{5}{12}$ h in hours and minutes.

ShowHide solution

(a) $45 \div 60 = \frac{3}{4} = 0.75$ , so 2 h 45 min $= 2\frac{3}{4}$ h $= 2.75$ h.

(b) $\frac{50}{60} = \frac{5}{6}$ , so 50 min $= \frac{5}{6}$ h.

(d) $\frac{5}{12} \times 60 = 25$ , so $2\frac{5}{12}$ h $= 2$ h 25 min.

(a) 2.75 h (b) $\frac{5}{6}$ h (c) 1 h 21 min (d) 2 h 25 min

Question 6

Hui Min runs 12 m in 4 s at a constant speed. How long does she take to run 66 m at the same speed?

ShowHide solution

Draw the run as a line and mark what is known.

Speed $= 12 \div 4 = 3$ , so 3 m/s.

Time for 66 m $= 66 \div 3 = 22$ , so 22 s.

Check: $3 \times 22 = 66$ .

She takes 22 seconds.

Question 7

A coach travels at a constant speed of 76 km/h for 3 hours 45 minutes. How far does it travel?

ShowHide solution

3 h 45 min $= 3\frac{45}{60}$ h $= 3\frac{3}{4}$ h $= 3.75$ h.

Distance $= 76 \times 3.75 = 285$ , so 285 km.

Check: $76 \times 3 = 228$ km in 3 hours, and $\frac{3}{4}$ of 76 km is 57 km in the last three quarters of an hour; $228 + 57 = 285$ .

The coach travels 285 km.

Question 8

A car drives 7.5 km at a constant speed of 50 km/h. How many minutes does it take?

ShowHide solution

Time = distance ÷ speed $= 7.5 \div 50 = 0.15$ , so 0.15 h.

In minutes: $0.15 \times 60 = 9$ , so 9 min.

Check: 50 km in 60 minutes is $50 \div 60 = \frac{5}{6}$ km in each minute, and $\frac{5}{6} \times 9 = 7.5$ km.

The car takes 9 minutes.

Question 9

An MRT train travels from one station to the next in 2 minutes 30 seconds at a constant speed of 65 km/h. Find the distance between the two stations, correct to 3 significant figures.

ShowHide solution

2 min 30 s $= 2.5$ min $= \frac{2.5}{60}$ h $= \frac{1}{24}$ h.

Distance $= 65 \times \frac{1}{24} = \frac{65}{24} = 2.7083\ldots$

Correct to 3 significant figures, this is 2.71 km.

The stations are about 2.71 km apart, correct to 3 significant figures.

Question 10

Mrs Goh leaves home at 10.50 a.m. and drives 150 km at a constant speed of 60 km/h. At what time does she arrive?

ShowHide solution

Time = distance ÷ speed $= 150 \div 60 = \frac{5}{2} = 2\frac{1}{2}$ , so $2\frac{1}{2}$ h, which is 2 h 30 min.

Count on from 10.50 a.m.: 2 hours later is 12.50 p.m.; 10 more minutes is 1.00 p.m.; 20 more minutes is 1.20 p.m. (The 30 minutes were taken as $10 + 20$ .)

Check: $60 \times 2\frac{1}{2} = 150$ km.

Mrs Goh arrives at 1.20 p.m.

Question 11

A coach leaves at 1.50 p.m. and arrives at 4.05 p.m. the same afternoon. It travels 171 km at a constant speed. Find its speed.

ShowHide solution

Time taken: from 1.50 p.m. to 3.50 p.m. is 2 h, and from 3.50 p.m. to 4.05 p.m. is 15 min, so 2 h 15 min.

2 h 15 min $= 2\frac{1}{4}$ h $= \frac{9}{4}$ h.

Speed $= 171 \div \frac{9}{4} = 171 \times \frac{4}{9} = 19 \times 4 = 76$ , so 76 km/h.

Check: $76 \times 2\frac{1}{4} = 152 + 19 = 171$ .

The coach's speed is 76 km/h.

Question 12

Ali's school is 3.5 km from his home. He cycles there at a constant speed of 15 km/h and must arrive by 7.30 a.m. What is the latest time he can leave home?

ShowHide solution

Time = distance ÷ speed $= 3.5 \div 15 = \frac{7}{30}$ , so $\frac{7}{30}$ h.

In minutes: $\frac{7}{30} \times 60 = 14$ , so 14 min.

Count back 14 minutes from 7.30 a.m.: 7.30 a.m. − 14 min = 7.16 a.m.

Check: 15 km in 60 minutes is 1 km in every 4 minutes, so 3.5 km takes $3.5 \times 4 = 14$ minutes, and 7.16 a.m. + 14 min = 7.30 a.m.

The latest time Ali can leave home is 7.16 a.m.

Question 13

Ravi usually runs 5 km in 30 minutes at a constant speed. If he ran 2 km/h faster, how many minutes less would he take to run the same 5 km?

ShowHide solution

His usual speed: 30 min $= \frac{1}{2}$ h, so speed $= 5 \div \frac{1}{2} = 10$ , so 10 km/h.

The faster speed is $10 + 2 = 12$ km/h.

Time at 12 km/h $= 5 \div 12 = \frac{5}{12}$ h, and $\frac{5}{12} \times 60 = 25$ , so 25 min.

Time saved $= 30 - 25 = 5$ min.

Check: $12 \times \frac{5}{12} = 5$ km.

He would take 5 minutes less.

Key Points

Speed is a rate: the distance travelled per unit of time, in units such as km/h, m/s, m/min and cm/s.
Speed = distance ÷ time; distance = speed × time; time = distance ÷ speed.
A speed of 60 km/h means 60 km in every hour: more time means more distance, and the time is how many lots of 60 km fit into the distance.
An object moves at a constant speed when it travels equal distances in equal intervals of time, however short the intervals are.
The units must agree: with a speed in km/h, use kilometres and hours; with m/s, metres and seconds.
Minutes to hours: divide by 60 (15 min $= \frac{1}{4}$ h, so 1 h 15 min is 1.25 h, not 1.15 h). A part of an hour to minutes: multiply by 60.
Keep a time such as $\frac{1}{3}$ h as a fraction until the end; give an answer that is not exact to 3 significant figures.
Time of arrival = time of departure + time taken; count on through the next whole hour.