Speed-time graph: (0, 0) (4, 16) (6, 20) (10, 20) (14, 0), find the… Step by Step

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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