Speed, velocity and acceleration
Key idea: Speed tells how fast an object moves; velocity also gives its direction. Acceleration is change in velocity per unit time, so changing direction is acceleration even at constant speed.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Compare speed and velocity when the direction changes.
- Decide how the train’s speed and velocity change.
- Follow how to average a journey with unequal stage times.
- Find the total distance and time, then the average speed.
- Read how to convert between kilometres per hour and metres per second.
- Include the stop when finding the whole trip’s average speed.
- Find the velocity change before dividing by the time.
- Find the motorcycle’s average acceleration.
- Convert the speed before finding acceleration.
- Read why slowing down gives a negative acceleration here.
- Convert the lorry’s speed and find its acceleration.
- Compare velocity changes in equal time intervals.
- Find when acceleration stays constant and how it changes later.
- Find the interval averages and decide what happens between readings.
- Compare the calculations for average speed and acceleration.
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ContinueA bus timetable, a speed camera and a car advert all describe motion with numbers: how far, how fast, and how quickly the speed changes. Those numbers are distance, speed and acceleration. All the calculations here are for motion along one straight line without a change of direction, with the direction of travel taken as positive.
Speed, velocity and average speed
Speed is distance travelled per unit time
Speed is the distance travelled per unit time. It tells you how fast an object moves. Speed has a size but no direction, so it is a scalar. Its SI unit is metres per second (m/s).
A car moving at a steady speed of 15 m/s travels 15 m every second.
Velocity is speed in a stated direction
Velocity is speed in a stated direction. It has a size and a direction, so it is a vector. “20 m/s north” is a velocity; “20 m/s” on its own is a speed.
| Situation | Speed | Velocity |
|---|---|---|
| Car A moves at 20 m/s north; car B moves at 20 m/s south. | the same: 20 m/s | different: the directions differ |
| A cyclist rides round a circular track at a steady 8 m/s. | constant | changing: the direction changes all the time |
| A lift rises at 2 m/s, then later at 3 m/s, still upwards. | changes | changes: same direction, different size |
Velocity changes when the speed changes, when the direction changes, or both.
Check your understanding 1
A train travels north at 30 m/s. Later it travels south at 25 m/s. Has its speed changed? Has its velocity changed?
Show answer
The speed has changed, from 30 m/s to 25 m/s. The velocity has changed in both size and direction: 30 m/s north became 25 m/s south.
Average speed uses the whole journey
Most journeys do not happen at one steady speed. The average speed is the steady speed that would cover the same total distance in the same total time.
Include every part of the journey in the totals, including any time spent at rest.
Worked example 1
Average speed over two stages
Problem
A bus travels 12 km at 60 km/h, then another 12 km at 20 km/h. Find its average speed.
Show full solution
Avoid the trap
Method
Do not average the two speeds.Reason
The bus does not spend equal times at the two speeds, so (60 + 20) ÷ 2 = 40 km/h is wrong.Working
Answer
Use total distance ÷ total time
Find each time
Method
Divide each distance by its speed.Reason
Time = distance ÷ speed for each stage.Working
Answer
12 ÷ 60 = 0.20 h and 12 ÷ 20 = 0.60 h
Find the totals
Method
Add the distances and the times.Reason
Average speed uses the whole journey.Working
Answer
24 km in 0.80 h
Find the average speed
Method
Divide total distance by total time.Reason
This is the definition of average speed.Working
Answer
24 ÷ 0.80 = 30 km/h
Check
Working
Answer
The bus spends three times as long at 20 km/h, so the average should be closer to 20 km/h than to 60 km/h. It is.
Check your understanding 2
A cyclist rides at a steady 6.0 m/s for 2.0 minutes, then at 3.0 m/s for 1.0 minute. Calculate the average speed.
Show the steps Optional working
Show the steps Optional working
Show answer
Distances: 6.0 × 120 = 720 m and 3.0 × 60 = 180 m. Average speed = 900 ÷ 180 = 5.0 m/s. It is not 4.5 m/s, because the cyclist spends twice as long at the higher speed.
Convert between km/h and m/s
Road speeds are often given in kilometres per hour. Physics calculations with times in seconds need metres per second.
- km/h to m/s: divide by 3.6.
- m/s to km/h: multiply by 3.6.
| km/h | m/s |
|---|---|
| 18 | 5 |
| 36 | 10 |
| 72 | 20 |
| 90 | 25 |
| 108 | 30 |
For the same non-zero speed, the numerical value in m/s is smaller than the value in km/h. Use this to check whether you multiplied or divided correctly.
Check your understanding 3
A cyclist rides 3.0 km in 10 minutes, rests for 5 minutes, then rides 1.5 km in 5 minutes. Find the average speed for the whole trip in m/s. Start by converting both totals to base units.
Show the steps Optional working
Show the steps Optional working
Show the steps Optional working
Show answer
Total distance = 3.0 + 1.5 = 4.5 km = 4500 m. Total time = 10 + 5 + 5 = 20 minutes = 1200 s, including the rest. Average speed = 4500 ÷ 1200 = 3.75 m/s, about 3.8 m/s.
Acceleration and changing motion
Acceleration is change in velocity per unit time
Acceleration is the change in velocity per unit time. For these calculations, motion is along a straight line in the chosen positive direction, so the change in velocity is the final speed minus the initial speed.
Write speeds in m/s and times in s before you calculate an acceleration. Mixing km/h with seconds gives the wrong number and the wrong unit.
The unit is metres per second squared (m/s²): a change in speed in m/s divided by a time in s. The calculation gives the average acceleration over that time. If the acceleration is uniform, it is the same throughout: a constant acceleration of 3 m/s² means the speed increases by 3 m/s every second.
Worked example 2
Acceleration from two speeds
Problem
A car accelerates uniformly from 8.0 m/s to 20 m/s in 4.0 s. Find its acceleration.
Show full solution
Check the units
Method
Look at the units of the speeds and the time.Reason
m/s and s give m/s², so no conversion is needed.Working
Answer
m/s and s
Change in velocity
Method
Subtract the initial velocity from the final velocity.Reason
Acceleration depends on the change, not on the speed itself.Working
Answer
20 - 8.0 = 12 m/s
Acceleration
Method
Divide by the time.Reason
Acceleration is the change in velocity per second.Working
Answer
12 ÷ 4.0 = 3.0 m/s²
Check
Working
Answer
Gaining 3.0 m/s each second gives 8, 11, 14, 17 and 20 m/s after 4 s.
Check your understanding 4
A motorcycle speeds up from 5.0 m/s to 21 m/s in 8.0 s. Calculate its acceleration.
Show the steps Optional working
Show answer
(21 - 5.0) ÷ 8.0 = 2.0 m/s².
Worked example 3
Acceleration with a conversion
Problem
A train starts from rest and reaches 54 km/h in 30 s. Find its average acceleration.
Show full solution
Convert the speed
Method
Divide km/h by 3.6.Reason
The time is in seconds, so the speed must be in m/s.Working
Answer
54 ÷ 3.6 = 15 m/s
Find the acceleration
Working
Answer
(15 - 0) ÷ 30 = 0.50 m/s²
Acceleration is not the same as high speed. An aircraft cruising in a straight line at a steady 250 m/s has zero acceleration because neither its speed nor its direction is changing. A cyclist pulling away from traffic lights at only 3 m/s is accelerating. Ask how much the speed changes each second, not how fast the object is already moving.
Slowing down: deceleration
For motion in the chosen positive direction, slowing down gives a final speed smaller than the initial speed, so the acceleration comes out negative. Deceleration means a decrease in speed. Here the negative sign indicates that the acceleration is opposite to the motion.
Worked example 4
Slowing down
Problem
A car slows uniformly from 25 m/s to 10 m/s in 5.0 s. Find its acceleration.
Show full solution
Change in velocity
Method
Subtract the initial velocity from the final velocity.Reason
Final minus initial keeps the sign of the change.Working
Answer
10 - 25 = -15 m/s
Acceleration
Method
Divide by the time.Reason
Acceleration is the change in velocity per second.Working
Answer
-15 ÷ 5.0 = -3.0 m/s²
Interpret the sign
Working
Answer
The minus sign shows the car is slowing down. You can state the answer as “an acceleration of −3.0 m/s²” or “a deceleration of 3.0 m/s²”, but not as both a minus sign and the word deceleration.
Check your understanding 5
A lorry travelling at 63 km/h brakes to rest in 7.0 s. Convert its speed to m/s, then calculate its acceleration, keeping the sign.
Show the steps Optional working
Show answer
63 ÷ 3.6 = 17.5 m/s. Acceleration = (0 - 17.5) ÷ 7.0 = -2.5 m/s², a deceleration of 2.5 m/s².
Uniform acceleration: equal changes in equal times
Uniform acceleration means the velocity changes by the same amount in each equal time interval. For motion in one direction, the speed rises (or falls) by the same amount every second.
| time / s | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| speed / m/s | 3 | 5 | 7 | 9 | 11 |
This table models a trolley with uniform acceleration. It gains 2 m/s in every second: (11 - 3) ÷ 4 = 2.0 m/s².
Non-uniform acceleration
Acceleration is non-uniform when the speed changes by different amounts in equal time intervals.
| time / s | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| speed / m/s | 0 | 7 | 12 | 15 | 16 |
This car keeps speeding up, but gains less speed in each second: 7, 5, 3 and then 1 m/s. Its accelerations over those seconds are 7, 5, 3 and 1 m/s². They differ, so the acceleration is not uniform.
To decide, compare the changes in speed over equal time intervals. Different changes show that the acceleration is non-uniform.
Check your understanding 6
A scooter’s speed is measured every 2.0 s: 0, 6, 12, 15 and 16 m/s at 0, 2, 4, 6 and 8 s. Assume constant acceleration between each pair of readings. For which part is the acceleration uniform? Find it, then describe the rest.
Show the steps Optional working
Show answer
From 0 s to 4 s the speed rises by 6 m/s in each 2.0 s interval, so the acceleration is uniform: 6 ÷ 2.0 = 3.0 m/s². From 4 s to 8 s the speed rises by 3 m/s and then by 1 m/s in equal intervals, so the acceleration is non-uniform and decreasing. The scooter is still speeding up.
Check your understanding 7 What do three speed readings tell you?
A trolley moves along a straight track in one direction. Its measured speeds are 0, 4 and 8 m/s at 0, 2 and 4 s. Find the average acceleration over each two-second interval. A learner says the readings prove the acceleration was constant at every instant. Is that conclusion justified? Choose the extra assumption that would allow straight lines between the points on a speed–time graph.
Show the steps Optional working
Show answer
Each interval has average acceleration (4 − 0) ÷ 2 = (8 − 4) ÷ 2 = 2 m/s². Equal interval averages do not tell you how speed changed between the readings: the speed could have changed unevenly. Straight lines require assuming uniform acceleration within each interval, or further evidence for that motion. Without that assumption, the three readings alone do not prove constant acceleration throughout.
A journey with two stages
A delivery robot travels at a steady 2.5 m/s for 12 s. It then accelerates uniformly from 2.5 m/s to 7.5 m/s in 8.0 s, covering 40 m in that time. Find its average speed for the whole journey and its acceleration in the second stage.
- First-stage distance = 2.5 × 12 = 30 m.
- Total distance = 30 + 40 = 70 m; total time = 12 + 8.0 = 20 s.
- Average speed = 70 ÷ 20 = 3.5 m/s.
- Acceleration in the second stage = (7.5 - 2.5) ÷ 8.0 = 0.625 m/s², about 0.63 m/s².
- Check: the average speed lies between the slowest (2.5 m/s) and fastest (7.5 m/s) speeds, and m/s divided by s gives m/s².
The 40 m comes from the area under the speed–time graph for the second stage.
Common mistakes
- Averaging the speeds. The object usually spends different times at each speed, so divide the total distance by the total time.
- Calling 20 m/s a velocity. A velocity needs a direction, such as 20 m/s north.
- Saying a fast object must have a large acceleration. An aircraft at a steady 250 m/s in a straight line has zero acceleration.
- Saying a car at rest has zero acceleration. A car pulling away from rest is accelerating even at the instant its speed is zero.
- Mixing units: 72 km/h for 10 s is not 72 × 10 = 720 m. Convert first: 72 ÷ 3.6 = 20 m/s, so the car covers 200 m.
- Writing a positive acceleration for slowing down. From 25 m/s to 10 m/s in 5.0 s, final minus initial divided by time gives −3.0 m/s², a deceleration of 3.0 m/s².
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Syllabus and review details
- SEC G2 Science Physics component 2027 · 2027
Physics section, PDF pages 8–23
Last reviewed: