Math 6 · 6.RP.A.2 · 6.RP.A.3B

Speed, distance and time

A snail, a balloon and a lift all move, at very different paces. What one number says how fast each goes, and once you have it, how do you find how far it gets or how long it takes?

You need

Rates and unit ratesMath 6

Where you are 5 in Rate and unit rate

Fractions, ratio and percent3

Ratio6

Rate and unit rate5

Rates and unit rates Changing the units of a rate

Speed, distance and time

Unit price and comparing rates Unit rates when the amounts are fractions

Why this exists

A garden snail crosses a paving slab. A hot air balloon drifts over fields. A lift (an elevator) carries people up a tall building. Which one is fast? “It went 30 cm” does not say, and neither does “it took 5 minutes”. A snail that crawls 30 cm in an hour is slow; one that crawls 30 cm in 5 minutes is moving well, for a snail. How fast something goes depends on two amounts at once: the distance it covers and the time it takes.

A Rates and unit rates squeezes a pair like that into one number: how much of the first amount goes with exactly one of the second. Speed is that idea applied to moving things. It gives every snail, balloon and lift one number, so they can be compared, and once that number is known, the distance after any time and the time for any distance follow from it.

Definition

Day 1: speed, distance and time

The speed of a moving thing is the distance it covers for each one unit of time. A snail that crawls 30 cm in 5 minutes, at a steady pace, covers 6 cm in each minute, so its speed is 6 cm per minute. “Per” means “for each”. A speed is always written as a distance unit per a time unit: centimeters per minute, meters per second, kilometers per hour.

Something moves at a constant speed (a steady speed) when it covers the same distance in every stretch of time of the same length, however short the stretches are. The problems of the first day all assume a constant speed.

Three amounts belong together: the distance traveled along the path, the time taken, and the speed. Knowing any two of them fixes the third.

Day 2: average speed over a trip

Real trips speed up, slow down and stop. The average speed of a trip is its total distance divided by its total time, with any stops counted in the time. It is the one constant speed that would cover the same total distance in the same total time.

Intuition

Figure 1 follows Gwen’s snail minute by minute. Constant speed means equal steps: every minute adds the same 6 cm, so the marks are evenly spaced. It is a The double number line with time on one side and distance on the other, both starting from zero.

Figure 1Gwen’s snail at a steady 6 cm per minute. Every minute adds the same 6 cm, so the marks are evenly spaced.

The picture can be read three ways, one for each amount.

Distance from speed and time. After 4 minutes the snail is 4 steps along, and each step is 6 cm: 4 \times 6 = 24 cm.
Time from distance and speed. To reach 18 cm, count how many 6 cm steps it takes: 3, so 3 minutes.
Speed from distance and time. 30 cm spread evenly over 5 minutes puts 6 cm in each.

Adair’s balloon in Figure 2 does not move steadily. The wind changes, so its hourly steps are uneven: 8 km, then 12, then 6, then 10. Below it is a balloon drifting at a steady 9 km per hour. The two start together and finish together, 36 km away after 4 hours, but in between they part company. That steady 9 km per hour is the real balloon’s average speed.

Figure 2Adair’s balloon (top) drifts unevenly, but it ends at 36 km after 4 hours, exactly where a steady 9 km per hour trip (bottom) ends.

To say which of two things is faster, give them a fair contest. Over the same time, the one that goes further is faster. Over the same distance, the one that takes less time is faster. When neither the times nor the distances match, speed puts both on the same footing: it is the distance each covers in the same time, one unit of it.

Results

One relation, written three ways. For something moving at a constant speed, with the units matching:

\text{speed} = \text{distance} \div \text{time}

\text{distance} = \text{speed} \times \text{time}

\text{time} = \text{distance} \div \text{speed}

Units must match. A speed in kilometers per hour goes with distances in kilometers and times in hours; a speed in meters per second with meters and seconds.

Average speed of a trip. With every stop counted in the time:

\begin{aligned}\text{average speed} &= \text{total distance} \\ &\quad \div \text{total time}\end{aligned}

It equals the Mean of the speeds on the separate parts of the trip when every part takes the same time, and in general not otherwise.

Comparing speeds. The larger speed is the faster. Equivalently, over the same time the faster thing goes further, and over the same distance it takes less time.

Why it is true

Distance is speed times time. At 6 cm per minute, each minute adds 6 cm. After 4 minutes that is 4 equal copies, 6 + 6 + 6 + 6, and adding equal copies is what multiplication counts: 4 \times 6 = 24. Part of a minute works too. Constant speed means the two halves of a minute carry the snail equally far, so half a minute adds 3 cm, and 2.5 minutes add 6 + 6 + 3 = 15 cm, which is 2.5 \times 6.

Time is distance divided by speed. Asking how long the snail takes to crawl 18 cm is asking how many 6 cm steps fit into 18 cm, and “how many of these fit into that” is division: 18 \div 6 = 3.

Speed is distance divided by time. The snail crawls 30 cm in 5 minutes at a steady pace, so each of the 5 minutes carries an equal share of the 30 cm: 30 \div 5 = 6.

These are one fact seen three ways, because division undoes multiplication: 6 \times 5 = 30, 30 \div 5 = 6 and 30 \div 6 = 5 all say the same thing.

Average speed. Something moving steadily at the average speed for the whole time covers the average speed times the total time, which is the total distance back again. So it ends exactly where the real trip ends, at the same moment. The mean of the separate speeds gives each part of the trip an equal say. Average speed gives each part a say in proportion to how long it lasts, so when a trip spends longer at its slower speed, its average speed comes out below the mean of the speeds.

Worked examples

Examples 1 to 4 go with the first day, 5 to 7 with the second.

Example 1

The snail’s speed

Gwen times a garden snail crossing a paving slab. At a steady pace it crawls 30 cm in 5 minutes. What is its speed? How far does it crawl in 8 minutes at that speed?

Solution

Speed is the distance for each one minute, so share the 30 cm equally among the 5 minutes:

30 \div 5 = 6 \text{ cm per minute}

In 8 minutes the snail takes 8 steps of 6 cm, so distance is speed times time:

6 \times 8 = 48 \text{ cm}

Check. 48 cm in 8 minutes is 48 \div 8 = 6 cm per minute, the same steady speed.

Example 2

Up to the observation deck

Hamza rides a lift that climbs at a steady 4 meters per second. How long does it take to climb to the observation deck, 156 m above the ground floor? How high is it 12.5 seconds after leaving the ground floor?

Solution

Each second lifts Hamza 4 m, so the time is how many 4 m steps fit into 156 m:

156 \div 4 = 39 \text{ seconds}

For the height, 12.5 seconds is 12 whole seconds and a half. Twelve seconds give 12 \times 4 = 48 m, and at a constant speed half a second gives half of 4 m, which is 2 m. Together that is 50 m, the same as speed times time:

4 \times 12.5 = 50 \text{ m}

Distance is speed times time even when the time is not a whole number.

Example 3

A time that is not a whole number of hours

Darya’s hot air balloon drifts at a steady 12 km per hour. How far does it drift in 2.5 hours? How long does it take to drift 21 km?

Solution

The speed is in kilometers per hour and the time is in hours, so the units match and distance is speed times time:

12 \times 2.5 = 30 \text{ km}

For the time, count how many 12 km steps fit into 21 km:

21 \div 12 = 1.75 \text{ hours}

That is \displaystyle 1\frac{3}{4} hours. It is sensible: 21 km is more than one hour’s 12 km and less than two hours’ 24 km. Check. 12 \times 1.75 = 12 + 9 = 21.

Example 4

When the speed is not given

Sacha’s lift climbs 45 m in 9 seconds at a steady speed. How long does it take to climb 120 m?

Solution

The pair “45 m in 9 seconds” is a rate, but not yet a speed, since it is not per one second. Find the speed first, sharing the 45 m among the 9 seconds:

45 \div 9 = 5 \text{ meters per second}

Then the time for 120 m is how many 5 m steps fit into it:

120 \div 5 = 24 \text{ seconds}

A ratio table reaches the same answer without the speed: divide both amounts of 45 m in 9 seconds by 3 to get 15 m in 3 seconds, then multiply both by 8 to get 120 m in 24 seconds. The speed route works for any distance with one division; the table route is quick when a handy step like 15 m happens to fit.

Check. 5 \times 24 = 120.

Example 5

The balloon’s average speed

Adair’s balloon flight lasts 4 hours. As the wind changes, it drifts 8 km in the first hour, 12 km in the second, 6 km in the third and 10 km in the fourth (Figure 2). What is its average speed, and what does that number tell you?

Solution

Average speed is total distance divided by total time. The total distance is

8 + 12 + 6 + 10 = 36 \text{ km}

and the total time is 4 hours, so the average speed is

36 \div 4 = 9 \text{ km per hour}

A balloon drifting steadily at 9 km per hour covers 9 \times 4 = 36 km in 4 hours, so it would land beside Adair’s at the same moment: that is the lower track of Figure 2. What the number does not tell you is how fast the balloon went in any one hour. It drifted 12 km in the second hour and only 6 km in the third, and after the first hour it was at 8 km, not 9.

Example 6

A lift that stops on the way

Tullio’s lift climbs 48 m at 4 meters per second, stops for 8 seconds while people get on, then climbs another 60 m at 6 meters per second. What is its average speed from start to finish?

Solution

Average speed needs the total distance and the total time. The distances are given; the time of each climb is distance divided by speed:

48 \div 4 = 12 \text{ seconds}

60 \div 6 = 10 \text{ seconds}

The stop counts too, so the total time is 12 + 8 + 10 = 30 seconds. The total distance is 48 + 60 = 108 m. So the average speed is

108 \div 30 = 3.6 \text{ meters per second}

The mean of the two climbing speeds, (4 + 6) \div 2 = 5, is not the average speed here, for two reasons. The stop takes time but covers no distance, which drags the average down: 3.6 meters per second is slower than either climb. And even leaving the stop out, the lift spends longer at the slower speed (12 seconds) than at the faster (10 seconds). Moving steadily at 5 meters per second for the 22 seconds of climbing would cover 5 \times 22 = 110 m, more than the 108 m it really climbs, so its average while moving is a little under 5.

Example 7

Which snail is faster

Gwen’s snail (Example 1) crawls 30 cm in 5 minutes. Tullio’s snail crawls 39 cm in 6 minutes. Which snail is faster, and by how much each minute?

Solution

The distances differ and so do the times, so compare speeds, the distance each covers in one minute:

30 \div 5 = 6 \text{ cm per minute}

39 \div 6 = 6.5 \text{ cm per minute}

Tullio’s snail is faster, by 6.5 - 6 = 0.5 cm each minute.

A fair contest over the same time agrees. Both 5 and 6 fit into 30 minutes a whole number of times, so in 30 minutes Gwen’s snail crawls 6 \times 30 = 180 cm and Tullio’s crawls 39 \times 5 = 195 cm. Tullio’s goes further in the same time, and the gap, 15 cm, is 30 lots of 0.5 cm.

Boundaries

Constant speed is an assumption. The three relations of the first day describe steady motion. A real snail pauses and a real balloon meets gusts. For an uneven trip, distance and time give only the average speed, and the average says nothing about any one moment: in Figure 2 the steady 9 km per hour trip is at 9 km after an hour, but the real balloon is at 8 km.

The units must match. A balloon at 12 km per hour for 30 minutes does not drift 12 \times 30 = 360 km; the time has to be in hours to go with kilometers per hour. Every problem here keeps one set of units. Changing a speed into other units is Changing the units of a rate.

A speed of zero. A snail that has stopped has speed 0. It covers 0 \times 10 = 0 cm in 10 minutes, which is fine, but the time to crawl 18 cm would be 18 \div 0, and division by zero has no answer: no number of 0 cm steps makes 18 cm. The snail never arrives while it stays still. In the same way, a speed needs a time greater than zero to divide by.

Distance means distance along the path, and speed has no direction. A lift that climbs 60 m and comes straight back down has traveled 120 m, even though it ends where it began. And a lift going up at 4 meters per second has the same speed as one coming down at 4 meters per second; speed says how fast, not which way.

Averaging speeds. The average speed of a trip is the mean of its separate speeds only when each part takes the same time (Exercises 16 and 17).

Where this goes next

When distance and time are fractions, the same divisions give the speed: that is Unit rates when the amounts are fractions. Speeds stated in different units, meters per second against kilometers per hour, are compared by changing one of them, in Changing the units of a rate, and the same “compare per one unit” idea compares prices in Unit price and comparing rates. Speed is the first and most familiar Constant rate of change and linear tables: plotted on a graph, distance against time at a constant speed makes a straight line through zero, the subject of Recognizing a proportional relationship in a graph. Algebra later writes the three relations as one formula with letters and rearranges it (Literal equations: solving for one variable), and average speed grows into Average rate of change.

Exercises

Day 1: speed, distance and time

Every exercise in this group is at a constant speed.

A garden snail crawls 35 cm in 7 minutes. What is its speed?
At the speed in Exercise 1, how far does the snail crawl in 15 minutes?
At the same speed, how long does the snail take to crawl the 45 cm to a lettuce leaf?
A lift climbs at 3 meters per second. How far does it climb in 14 seconds?
How long does the lift in Exercise 4 take to climb 75 m?
A hot air balloon drifts at 8 km per hour. How far does it drift in 3.5 hours?
How long does the balloon in Exercise 6 take to drift 30 km?
A balloon drifts 27 km in 4.5 hours. What is its speed?
A lift climbs 84 m in 12 seconds. How long does it take to climb 56 m?
A slow snail crawls 2.5 cm each minute. How far does it crawl in 12 minutes, and how long does it take to crawl 40 cm?
Find the missing amount in each: (a) the distance covered at 5 meters per second for 18 seconds; (b) the speed of a balloon that drifts 45 km in 6 hours; (c) the time a snail moving at 9 cm per minute takes to crawl 63 cm.
A lift climbs 150 m to the top floor in 25 seconds. How high is it after 10 seconds?

Day 2: average speed and comparing speeds

A balloon drifts 5 km, 9 km, 7 km and 11 km in four hours, one after another. What is its average speed?
A snail crawls 10 cm in 2 minutes, rests for 3 minutes, then crawls 15 cm in 5 minutes. What is its average speed over the whole 10 minutes?
A balloon drifts 18 km at 9 km per hour, then 18 km at 6 km per hour. How long does each part take, and what is the average speed of the trip?
A balloon drifts for 2 hours at 10 km per hour, then for 2 hours at 5 km per hour. What is its average speed?
A balloon drifts for 2 hours at 10 km per hour, then for 3 hours at 5 km per hour. What is its average speed?
A lift climbs 40 m at 5 meters per second, stops for 10 seconds, then climbs 42 m at 6 meters per second. What is its average speed from start to finish?
Darya’s snail crawls 36 cm in 8 minutes, and Sacha’s crawls 50 cm in 10 minutes. Which is faster, and by how much each minute?
Adair’s balloon drifts 14 km in the first 2 hours of a 5 hour flight. For the whole flight to average 8 km per hour, how far must it drift in the last 3 hours, and at what steady speed?
One balloon drifts 21 km in 3 hours and another drifts 30 km in 4 hours. Find each balloon’s speed, and also the time each takes for one kilometer. Show that both ways name the same balloon as faster.
Two snails start side by side and crawl the same way, one at 5 cm per minute and one at 7 cm per minute. How far apart are they after 12 minutes? After how many minutes is the faster one 10 cm ahead?

Reasoning

Exercises 16 and 17 use the same two speeds, 10 and 5 km per hour. Explain why one trip’s average speed is the mean of the two speeds and the other’s is not.
A lift climbs 60 m from the ground floor and comes straight back down, taking 40 seconds in all without stopping. Find its average speed, and explain why it is not zero even though the lift ends where it started.
A snail has stopped under a leaf. Explain what happens when you try to find how long it takes to crawl 18 cm by dividing distance by speed.

Answers

5 cm per minute. Share 35 cm among 7 minutes: 35 \div 7 = 5.
75 cm. 5 \times 15 = 75.
9 minutes. 45 \div 5 = 9 steps of 5 cm.
42 m. 3 \times 14 = 42.
25 seconds. 75 \div 3 = 25.
28 km. 8 \times 3.5 = 24 + 4 = 28.
3.75 hours, which is \displaystyle 3\frac{3}{4} hours. 30 \div 8 = 3.75.
6 km per hour. 27 \div 4.5 = 6; check, 6 \times 4.5 = 27.
8 seconds. The speed is 84 \div 12 = 7 meters per second, and 56 \div 7 = 8.
30 cm and 16 minutes. 2.5 \times 12 = 30, and 40 \div 2.5 = 16.
(a) 90 m, since 5 \times 18 = 90. (b) 7.5 km per hour, since 45 \div 6 = 7.5. (c) 7 minutes, since 63 \div 9 = 7.
60 m. The speed is 150 \div 25 = 6 meters per second, and 6 \times 10 = 60.
8 km per hour. Total 5 + 9 + 7 + 11 = 32 km in 4 hours: 32 \div 4 = 8.
2.5 cm per minute. Total 10 + 15 = 25 cm in 10 minutes, the rest included: 25 \div 10 = 2.5.
2 hours and 3 hours; 7.2 km per hour. 18 \div 9 = 2 and 18 \div 6 = 3, so 36 km in 5 hours: 36 \div 5 = 7.2.
7.5 km per hour. 20 + 10 = 30 km in 4 hours: 30 \div 4 = 7.5.
7 km per hour. 20 + 15 = 35 km in 5 hours: 35 \div 5 = 7.
3.28 meters per second. The climbs take 40 \div 5 = 8 and 42 \div 6 = 7 seconds, so the total time is 8 + 10 + 7 = 25 seconds for 82 m: 82 \div 25 = 3.28.
Sacha’s, by 0.5 cm each minute. 36 \div 8 = 4.5 and 50 \div 10 = 5 cm per minute.
26 km, at \displaystyle 8\frac{2}{3} km per hour. The whole flight needs 8 \times 5 = 40 km, so 40 - 14 = 26 km remain, and \displaystyle 26 \div 3 = 8\frac{2}{3}.
The second balloon is faster. Speeds: 21 \div 3 = 7 and 30 \div 4 = 7.5 km per hour. Time per kilometer: \displaystyle 3 \div 21 = \frac{1}{7} hour and \displaystyle 4 \div 30 = \frac{2}{15} hour. Over a common denominator these are \displaystyle \frac{15}{105} and \displaystyle \frac{14}{105}, so the second balloon also takes less time for each kilometer.
24 cm; after 5 minutes. Each minute the faster snail gains 7 - 5 = 2 cm, so after 12 minutes the gap is 2 \times 12 = 24 cm, and a 10 cm gap takes 10 \div 2 = 5 minutes.
In Exercise 16 the balloon spends the same 2 hours at each speed, so each speed counts equally and the average speed, 7.5 km per hour, is the mean of 10 and 5. In Exercise 17 it spends 3 hours at the slower speed and only 2 at the faster, so the slower speed counts for more and the average, 7 km per hour, is below the mean.
3 meters per second. The lift travels 60 + 60 = 120 m along its path in 40 seconds, and 120 \div 40 = 3. Speed measures distance traveled along the path, not how far the lift ends up from where it started, so the trip back down counts just as much as the trip up.
Its speed is 0 cm per minute, so the division would be 18 \div 0, which has no answer: no number of 0 cm steps adds up to 18 cm. While the snail stays still, it never covers the 18 cm, so there is no time to find.