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Math 6 · Science
Interaction · Motion

Speed

At a track meet, a runner covers 100 m in 20 s. Her friend rides her bike 30 km in 2 h on a Saturday trip. Who is faster? To compare them, you need a single quantity that relates the distance they cover to the time they take: speed.

What speed is

Two swimmers train in a pool: the first swims 50 m in 25 s and the second swims 100 m in 60 s; how can we tell for certain which one swims faster if they have covered different distances in different times?

Straight path of 100 meters covered in 20 seconds, with the calculation of the resulting average speed of 5 meters per second. d = 100 m t = 0 s t = 20 s v = 100 m ÷ 20 s = 5 m/s
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A runner covers $100\ \text{m}$ in $20\ \text{s}$. Calculate their average speed in $\text{m/s}$.

A runner covers $100\ \text{m}$ in $20\ \text{s}$. Calculate their average speed in $\text{m/s}$.

Solve
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State which of these expressions are units of speed and which are not, and explain why: a) $\text{m/s}$ b) $\text{km}$ c) $\text{km/h}$ d) $\text{s}$

State which of these expressions are units of speed and which are not, and explain why:

  1. a) $\text{m/s}$
  2. b) $\text{km}$
  3. c) $\text{km/h}$
  4. d) $\text{s}$
Solve
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Who travels faster: Emma, who covers $60\ \text{m}$ in $12\ \text{s}$, or Liam, who covers $40\ \text{m}$ in $10\ \text{s}$? Calculate the speed of each and compare them.

Who travels faster: Emma, who covers $60\ \text{m}$ in $12\ \text{s}$, or Liam, who covers $40\ \text{m}$ in $10\ \text{s}$? Calculate the speed of each and compare them.

Solve

Changing units: m/s and km/h

Diagram of the conversion factor between kilometers per hour and meters per second, dividing or multiplying by 3.6. km/h m/s ÷ 3.6 × 3.6 72 km/h ÷ 3.6 = 20 m/s 15 m/s × 3.6 = 54 km/h
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A car travels along a highway at $90\ \text{km/h}$. Express its speed in $\text{m/s}$. Remember: $1\ \text{km} = 1000\ \text{m}$ and $1\ \text{h} = 3600\ \text{s}$.

A car travels along a highway at $90\ \text{km/h}$. Express its speed in $\text{m/s}$. Remember: $1\ \text{km} = 1000\ \text{m}$ and $1\ \text{h} = 3600\ \text{s}$.

Solve
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A greyhound runs at $20\ \text{m/s}$. Express its speed in $\text{km/h}$.

A greyhound runs at $20\ \text{m/s}$. Express its speed in $\text{km/h}$.

Solve
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The distance from the Earth to the Moon is $3.84\times10^{8}\ \text{m}$ and light travels at $3\times10^{8}\ \text{m/s}$. How long does it take for moonlight to reach…

The distance from the Earth to the Moon is $3.84\times10^{8}\ \text{m}$ and light travels at $3\times10^{8}\ \text{m/s}$. How long does it take for moonlight to reach us? Give the answer in seconds, to two decimal places.

Solve

Comparing speeds

A car at 72 km/h, equivalent to 20 m/s, and a cyclist at 15 m/s, with arrows whose lengths are proportional to their speeds: the car's arrow is longer. 72 km/h ÷ 3.6 20 m/s now in m/s 15 m/s 20 m/s > 15 m/s
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Who is faster: a car traveling at $72\ \text{km/h}$ or a cyclist going downhill at $15\ \text{m/s}$? Convert both speeds to the same unit to compare them.

Who is faster: a car traveling at $72\ \text{km/h}$ or a cyclist going downhill at $15\ \text{m/s}$? Convert both speeds to the same unit to compare them.

Solve
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Order the following speeds from lowest to highest, expressing them all in $\text{m/s}$ first: a) $18\ \text{km/h}$ b) $10\ \text{m/s}$ c) $900\ \text{m/min}$

Order the following speeds from lowest to highest, expressing them all in $\text{m/s}$ first:

  1. a) $18\ \text{km/h}$
  2. b) $10\ \text{m/s}$
  3. c) $900\ \text{m/min}$
Solve

Average speed on journeys with several stages

You set off on a road trip and drive the first 60 km on an interstate at 120 km/h, but then you are stuck in a traffic jam for 60 km at 20 km/h: what is your overall average speed, and why is it not 70 km/h, the simple average of the two figures?

To see why, first look at a different trip with two stages of the same distance (60 km each); at the end we will come back to the traffic jam.

Journey in two stages of 60 km: the first at 120 km/h for 0.5 h and the second at 60 km/h for 1 h; in total 120 km in 1.5 h, that is, an average of 80 km/h. Stage 1: 60 km at 120 km/h (0.5 h) Stage 2: 60 km at 60 km/h (1 h) 0.5 h 120 km in 1.5 h → average of 80 km/h
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A car travels $60\ \text{km}$ at $120\ \text{km/h}$ and another $60\ \text{km}$ at $60\ \text{km/h}$, as shown in the figure. a) Calculate the time taken for each…

A car travels $60\ \text{km}$ at $120\ \text{km/h}$ and another $60\ \text{km}$ at $60\ \text{km/h}$, as shown in the figure.

60 km 120 km/h 60 km 60 km/h
  1. a) Calculate the time taken for each section.
  2. b) Calculate the average speed for the whole trip.
  3. c) Explain why the average speed is not $90\ \text{km/h}$, the mean of the two speeds.
Solve
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A cyclist covers $3\ \text{km}$ in $10\ \text{min}$ with a tailwind and takes $15\ \text{min}$ to return along the same route. a) Calculate her speed on the outward and…

A cyclist covers $3\ \text{km}$ in $10\ \text{min}$ with a tailwind and takes $15\ \text{min}$ to return along the same route.

  1. a) Calculate her speed on the outward and return trips, in $\text{km/h}$.
  2. b) Calculate her average speed for the complete trip.
  3. c) Is the average speed equal to the mean of the two speeds from part a)? Explain your reasoning.
Solve
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Mateo walks to school: he covers $400\ \text{m}$ in $4\ \text{min}$, waits $2\ \text{min}$ at a traffic light, and covers another $400\ \text{m}$ in $4\ \text{min}$, as…

Mateo walks to school: he covers $400\ \text{m}$ in $4\ \text{min}$, waits $2\ \text{min}$ at a traffic light, and covers another $400\ \text{m}$ in $4\ \text{min}$, as shown in the figure.

t (min) 400 m traffic light 400 m 4 min 2 min 4 min
  1. a) What is his speed while walking, in $\text{m/min}$?
  2. b) What is his average speed for the whole journey?
  3. c) Why are the two results different?
Solve

Back to the beginning: The runner has a speed of $5\text{ m/s}$ ($18\text{ km/h}$) and the cyclist $15\text{ km/h}$ ($30\text{ km} \div 2\text{ h}$); therefore, the runner is faster. And on the road trip with the traffic jam you cover $60 + 60 = 120\text{ km}$ in $3.5\text{ h}$: an average speed of about $34.3\text{ km/h}$, not $70\text{ km/h}$.

Summary

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