Lesson 1 of 6
Distance, speed and velocity
I can explain speed and velocity.Syllabus
Syllabus K326 / K327, 2(a). Explain speed and velocity.
I can calculate average speed from total distance and time.Syllabus
Syllabus K326 / K327, 2(b). Calculate average speed from total distance and time.
Make a guess. You are not marked.
A runner covers the first 100 m of a track at 5 m/s and the next 100 m at 10 m/s. Is the average speed for the 200 m equal to 7.5 m/s?
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No. It is less than 7.5 m/s.
The slow half takes 20 s and the fast half only 10 s, so the runner spends twice as long at 5 m/s. Average speed = 200 m / 30 s = 6.7 m/s.
Distance is the total length of the path travelled. Displacement is the change in position from start to finish, including direction.
If you walk 3 m to the right, then 1 m back to the left, you have walked 4 m in total, but finish 2 m to the right of where you started. Your distance is 4 m; your displacement is 2 m to the right.
- Speed
- Distance travelled per unit time. It tells you how fast, without a direction. For example, 2 m/s.
- Velocity
- Rate of change of displacement. It tells you how fast and in which direction. For example, 2 m/s to the right.
Speed is a scalar; velocity is a vector. A scalar has magnitude (size) only, while a vector has magnitude and direction. Along a straight line, choose one direction as positive. Motion in the opposite direction has negative velocity, even though speed is always non-negative.
The time includes any stops. Average speed describes the whole journey; the object does not have to travel at that speed at every instant. Convert units before substituting: 1 minute = 60 s and 1 km = 1000 m. To convert km/h to m/s, divide by 3.6.
Distance in centimetres divided by time in seconds gives cm/s. For example, 30 cm in 2.0 s gives 15 cm/s = 0.15 m/s, because each centimetre is 0.01 m.
Worked example
Average speed including a stop
A student walks 60 m in 40 s, waits for 20 s, then walks another 40 m in 40 s. Find the average speed.
- Add the distances: 60 + 40 = 100 m.
- Include all the time: 40 + 20 + 40 = 100 s.
- Divide: average speed = 100 m / 100 s = 1 m/s.
This means that travelling steadily at 1 m/s for 100 s would cover the same total distance.
Try it step by step
Work it out in steps: Total distance in m, then Total time in s, then Average speed in m/s.
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Problem 1.
Total distance in m: Distance = 300 + 180 = 480 m
Total time in s: Time = 50 + 10 + 40 = 100 s (the stop counts)
Average speed in m/s: Average speed = 480 / 100 = 4.8 m/s (2 sf)
Problem 2.
Total distance in m: Distance = 120 + 80 = 200 m
Total time in s: Time = 15 + 25 + 20 = 60 s
Average speed in m/s: Average speed = 200 / 60 = 3.3 m/s (2 sf)
Problem 3.
Total distance in m: Distance = 900 + 600 = 1500 m
Total time in s: Time = 60 + 30 + 60 = 150 s
Average speed in m/s: Average speed = 1500 / 150 = 10 m/s (2 sf)
Include the whole journey. Add all the distances and all the times, including stops. Taking a simple mean of the speeds only works when the time spent at each speed is equal.
Check your understanding
A cyclist travels 90 m in 15 s, waits for 5 s, then travels 30 m in 10 s. What is the average speed for the whole journey?
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4 m/s
Total distance is 120 m and total time is 30 s, so average speed = 120 / 30 = 4 m/s.