Speed-time graph: (0, 5) (4, 15) (8, 15) (12, 5) (14, 0), find the… Step by Step

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The question

Question: the speed-time graph joins (0, 5) (4, 15) (8, 15) (12, 5) (14, 0), find the distance.

speed-time graph: (0, 5) (4, 15) (8, 15) (12, 5) (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 rectangle and a triangle.

4 × 5 + a half × 4 × 10: 40.

A: area = 4 × 5 + ½ × 4 × 10 = 40

Under part B: a rectangle, 4 across and 15 up.

4 times 15 is 60.

B: area = 4 × 15 = 60

Under part C: a rectangle and a triangle.

4 × 5 + a half × 4 × 10: 40.

C: area = 4 × 5 + ½ × 4 × 10 = 40

Under part D: a triangle.

Half of 2 times 5 is 5.

D: area = ½ × 2 × 5 = 5

Careful!

15 is the top speed, read off the graph.

The distance is the area under it.

Careful: not distance = 15 m ✗

3 Add the areas

All the parts together: 145 m.

distance = 40 + 60 + 40 + 5 = 145 m

4 Check

Let's check: the average speed is 10.4 m/s.

That is between 0 and the top speed, 15.

So the answer makes sense!

average speed = 145 ÷ 14 ≈ 10.4 m/s: between 0 and 15

Let's look back.

First: read the graph.

Then: distance = area.

Then: add the areas.

Last: check.

Answer