Worked examples
The question
Question: the speed-time graph joins (0, 0) (4, 12) (10, 12) (14, 0), find the distance.
speed-time graph: (0, 0) (4, 12) (10, 12) (14, 0)
1 Read the graph
Across: the time.
Up: the speed, in m/s.
Going up: speeding up.
Flat: a steady speed.
Going down: slowing down.
2 Distance = area
The distance is the area under a speed-time graph.
Speed times time is a rectangle; a steady speed-up is half a rectangle.
distance = area under the graph
Under part A: a triangle.
Half of 4 times 12 is 24.
A: area = ½ × 4 × 12 = 24
Under part B: a rectangle, 6 across and 12 up.
6 times 12 is 72.
B: area = 6 × 12 = 72
Under part C: a triangle.
Half of 4 times 12 is 24.
C: area = ½ × 4 × 12 = 24
Careful!
12 is the top speed, read off the graph.
The distance is the area under it.
Careful: not distance = 12 m ✗
3 Add the areas
All the parts together: 120 m.
distance = 24 + 72 + 24 = 120 m
4 Check
Let's check: the average speed is 8.57 m/s.
That is between 0 and the top speed, 12.
So the answer makes sense!
average speed = 120 ÷ 14 ≈ 8.57 m/s: between 0 and 12
Let's look back.
First: read the graph.
Then: distance = area.
Then: add the areas.
Last: check.
Answer distance = 24 + 72 + 24 = 120 m
▶ Watch it solved
The question
Question: the speed-time graph joins (0, 0) (5, 10) (15, 10) (20, 0), find the distance.
speed-time graph: (0, 0) (5, 10) (15, 10) (20, 0)
1 Read the graph
Across: the time.
Up: the speed, in m/s.
Going up: speeding up.
Flat: a steady speed.
Going down: slowing down.
2 Distance = area
The distance is the area under a speed-time graph.
Speed times time is a rectangle; a steady speed-up is half a rectangle.
distance = area under the graph
Under part A: a triangle.
Half of 5 times 10 is 25.
A: area = ½ × 5 × 10 = 25
Under part B: a rectangle, 10 across and 10 up.
10 times 10 is 100.
B: area = 10 × 10 = 100
Under part C: a triangle.
Half of 5 times 10 is 25.
C: area = ½ × 5 × 10 = 25
Careful!
10 is the top speed, read off the graph.
The distance is the area under it.
Careful: not distance = 10 m ✗
3 Add the areas
All the parts together: 150 m.
distance = 25 + 100 + 25 = 150 m
4 Check
Let's check: the average speed is 7.5 m/s.
That is between 0 and the top speed, 10.
So the answer makes sense!
average speed = 150 ÷ 20 ≈ 7.5 m/s: between 0 and 10
Let's look back.
First: read the graph.
Then: distance = area.
Then: add the areas.
Last: check.
Answer distance = 25 + 100 + 25 = 150 m
▶ Watch it solved
The question
Question: the speed-time graph joins (0, 0) (4, 16) (6, 20) (10, 20) (14, 0), find the distance.
speed-time graph: (0, 0) (4, 16) (6, 20) (10, 20) (14, 0)
1 Read the graph
Across: the time.
Up: the speed, in m/s.
Going up: speeding up.
Flat: a steady speed.
Going down: slowing down.
2 Distance = area
The distance is the area under a speed-time graph.
Speed times time is a rectangle; a steady speed-up is half a rectangle.
distance = area under the graph
Under part A: a triangle.
Half of 4 times 16 is 32.
A: area = ½ × 4 × 16 = 32
Under part B: a rectangle and a triangle.
2 × 16 + a half × 2 × 4: 36.
B: area = 2 × 16 + ½ × 2 × 4 = 36
Under part C: a rectangle, 4 across and 20 up.
4 times 20 is 80.
C: area = 4 × 20 = 80
Under part D: a triangle.
Half of 4 times 20 is 40.
D: area = ½ × 4 × 20 = 40
Careful!
20 is the top speed, read off the graph.
The distance is the area under it.
Careful: not distance = 20 m ✗
3 Add the areas
All the parts together: 188 m.
distance = 32 + 36 + 80 + 40 = 188 m
4 Check
Let's check: the average speed is 13.4 m/s.
That is between 0 and the top speed, 20.
So the answer makes sense!
average speed = 188 ÷ 14 ≈ 13.4 m/s: between 0 and 20
Let's look back.
First: read the graph.
Then: distance = area.
Then: add the areas.
Last: check.
Answer distance = 32 + 36 + 80 + 40 = 188 m
▶ Watch it solved
