Mathematics questions, chapter 7

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Graphs of Motion

45 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

The gradient of a distance-time graph represents the

acceleration
speed
total distance
total time

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Gradient = change in distance ÷ change in time, which is speed. Acceleration is the gradient of a speed-time graph.

For the teacher: Remembering, low difficulty, standard 7.1.1

Textbook p. 183

Question 2

The area under a speed-time graph represents the

speed
acceleration
time taken
distance travelled

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Area = speed × time = distance travelled.

For the teacher: Remembering, low difficulty, standard 7.2.2

Textbook p. 198

Question 3

A horizontal straight line on a distance-time graph shows that the object is

moving at a uniform speed
accelerating
stationary
returning to the start

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A zero gradient means no change in distance, so the object is stationary. On a speed-time graph a horizontal line means uniform speed instead.

For the teacher: Understanding, low difficulty, standard 7.1.2

Textbook p. 187

Question 4

Diagram 1 shows a distance-time graph of a car. Calculate the speed of the car from Q to R.

Distance-time graph through (0,0), P (1 h, 45 km), Q (2 h, 45 km) and R (4 h, 125 km) — Diagram 1

31.25 km/h
62.5 km/h
40 km/h
80 km/h

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Speed = (125 − 45) ÷ (4 − 2) = 80 ÷ 2 = 40 km/h. 31.25 is 125 ÷ 4 (the average speed) and 80 forgets to divide by the time.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 188

Question 5

On a distance-time graph, a part of the line with a negative gradient shows that the object

is decelerating
is moving back towards its starting point
has stopped
is moving at a faster speed

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The distance from the starting point is decreasing, so the object is returning (or moving in the opposite direction). Deceleration is a speed-time graph idea.

For the teacher: Understanding, low difficulty, standard 7.1.2

Textbook p. 187

Question 6

Diagram 2 shows a speed-time graph. Calculate the acceleration in the first 4 seconds.

Speed-time graph from (0, 5) rising to (4 s, 17 m/s) then flat to 10 s — Diagram 2

3 m/s²
4.25 m/s²
5 m/s²
12 m/s²

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Acceleration = (17 − 5) ÷ (4 − 0) = 12 ÷ 4 = 3 m/s². 4.25 is 17 ÷ 4 (it ignores the starting speed).

For the teacher: Applying, low difficulty, standard 7.2.3

Textbook p. 200

Question 7

Diagram 3 shows the speed-time graph of a van for 10 seconds. Calculate the total distance travelled.

Speed-time graph from (0, 10 m/s) rising to (6 s, 30 m/s), then flat at 30 m/s to 10 s — Diagram 3

180 m
300 m
240 m
120 m

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Trapezium: ½ × 6 × (10 + 30) = 120 m. Rectangle: 4 × 30 = 120 m. Total = 240 m. 300 m is 10 × 30 (it treats the whole journey as uniform speed).

For the teacher: Applying, medium difficulty, standard 7.2.2

Textbook p. 199

Question 8

A car travels at a uniform speed of 72 km/h for 20 minutes. What distance does it travel?

24 km
1 440 km
3.6 km
72 km

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20 minutes = 1/3 hour. Distance = 72 × 1/3 = 24 km. 1 440 km uses 20 hours.

For the teacher: Applying, low difficulty, standard 7.2.2

Textbook p. 198

Question 9

A cyclist rides 12 km in 40 minutes, rests for 20 minutes, then ends the ride. What is the average speed for the whole 1 hour?

18 km/h
9 km/h
20 km/h
12 km/h

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Average speed = total distance ÷ total time = 12 km ÷ 1 h = 12 km/h. The rest time must be included. 18 km/h is the speed while riding.

For the teacher: Analysing, medium difficulty, standard 7.1.3

Textbook p. 189

Question 10

Two cars are drawn on the same distance-time graph. Car P has a steeper line than car Q. Which statement is true?

P is faster than Q
Q is faster than P
P is stationary
They have the same speed

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A steeper line has a bigger gradient, so a greater speed. Equal speeds would give parallel lines.

For the teacher: Analysing, low difficulty, standard 7.1.2

Textbook p. 187

Question 11

A cyclist's distance from the finishing line is s = 20 − 0.5t, where s is in km and t is in minutes. Find t when s = 8.

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8 = 20 − 0.5t, so 0.5t = 12 and t = 24. 16 would come from 0.5t = 8 (using s instead of 20 − s); t = 40 is when s = 0.

For the teacher: Applying, high difficulty, standard 7.1.1

Textbook p. 185

Question 12

A train slows down from 24 m/s to rest in 8 s. Its deceleration is

0.33 m/s²
8 m/s²
192 m/s²
3 m/s²

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Change of speed = (0 − 24) ÷ 8 = −3 m/s², so the deceleration is 3 m/s². 0.33 is 8 ÷ 24 (the fraction inverted) and 192 is 24 × 8.

For the teacher: Applying, low difficulty, standard 7.2.3

Textbook p. 200

Section B: structured questions

33 marks

Answer all questions. Show your working.

Question 13

Diagram 4 shows Sahana's walk from her house to a library and back. A is her house.

Distance-time graph: A at origin, B (15 min, 3 km), C (30 min, 3 km), D (40 min, 0 km) — Diagram 4

[1 mark] (a)For how many minutes was she at the library?

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30 − 15 = 15 minutes [1]1

Tick each point you got.

[2 marks] (b)Calculate her speed, in km/h, from A to B.

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15 minutes = 0.25 h [1]1
3 ÷ 0.25 = 12 km/h [1]1

Tick each point you got.

[2 marks] (c)Calculate her speed, in km/h, from C to D.

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10 minutes = 1/6 h [1]1
3 ÷ (1/6) = 18 km/h [1]1

Tick each point you got.

[2 marks] (d)Calculate her average speed, in km/h, for the whole journey.

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Total distance = 3 + 3 = 6 km [1]1
6 ÷ (40/60) = 9 km/h [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 189

Question 14

Diagram 5 shows a car travelling from Town X to Town Y, and a bus from Town Y to Town X. The towns are 120 km apart.

Car line from (0, 0) to (2 h, 120 km); bus line from (0, 120 km) to (3 h, 0) — Diagram 5

[1 mark] (a)Calculate the speed of the car.

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120 ÷ 2 = 60 km/h [1]1

Tick each point you got.

[1 mark] (b)Calculate the speed of the bus.

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120 ÷ 3 = 40 km/h [1]1

Tick each point you got.

[2 marks] (c)The two vehicles meet when 60t = 120 − 40t. Find the time they meet and their distance from Town X.

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100t = 120, t = 1.2 hours [1]1
60 × 1.2 = 72 km [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 7.1.3

Textbook p. 191

Question 15

A car brakes so that v = 40 − 5t, where v is the speed in m/s and t the time in seconds, for 0 ≤ t ≤ 8. Diagram 6 shows the speed-time graph.

Speed-time graph: straight line from (0, 40 m/s) to (8 s, 0) — Diagram 6

[2 marks] (a)State the speed when t = 0 and when t = 8.

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40 m/s [1]1
0 m/s [1]1

Tick each point you got.

[1 mark] (b)State the deceleration.

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5 m/s² [1]1

Tick each point you got.

[2 marks] (c)Calculate the distance travelled before the car stops.

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½ × 8 × 40 [1]1
= 160 m [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 7.2.1

Textbook p. 196

Question 16

Diagram 7 shows the speed-time graph of a motorcycle for 20 seconds.

Speed-time graph: (0, 0) to (6 s, 24 m/s), flat to 16 s, down to 0 at 20 s — Diagram 7

[1 mark] (a)Calculate the acceleration in the first 6 seconds.

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24 ÷ 6 = 4 m/s² [1]1

Tick each point you got.

[1 mark] (b)Calculate the deceleration in the last 4 seconds.

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24 ÷ 4 = 6 m/s² [1]1

Tick each point you got.

[1 mark] (c)Calculate the distance travelled at uniform speed.

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10 × 24 = 240 m [1]1

Tick each point you got.

[2 marks] (d)Calculate the total distance for the 20 seconds.

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½ × 6 × 24 = 72 and ½ × 4 × 24 = 48 [1]1
72 + 240 + 48 = 360 m [1]1

Tick each point you got.

[1 mark] (e)Calculate the average speed.

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360 ÷ 20 = 18 m/s [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 7.2.3

Textbook p. 201

Question 17

Diagram 8 shows the speed of a car for 10 seconds. The speed rises from 6 m/s to v m/s in 4 seconds, then stays at v m/s. The total distance travelled is 124 m.

Speed-time graph from (0, 6 m/s) rising to v at 4 s, then flat to 10 s — Diagram 8

[3 marks] (a)Find the value of v.

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½ × 4 × (6 + v) + 6 × v = 124 [1]1
12 + 2v + 6v = 124 [1]1
8v = 112, v = 14 [1]1

Tick each point you got.

[1 mark] (b)Calculate the acceleration in the first 4 seconds.

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(14 − 6) ÷ 4 = 2 m/s² [1]1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 7.2.4

Textbook p. 203

Question 18

Diagram 9 shows a distance-time graph. Aiman says, "The horizontal part shows the car moving at a constant speed."

Distance-time graph: (0,0) to (10 min, 4 km), flat to 20 min, then up to (30 min, 10 km) — Diagram 9

[1 mark] (a)Is Aiman correct? Explain.

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No. A horizontal line means the distance does not change, so the car is stationary [1]1

Tick each point you got.

[2 marks] (b)Calculate the speed, in km/h, of the car in the last 10 minutes.

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(10 − 4) ÷ (10/60) [1]1
= 36 km/h [1]1

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 7.1.2

Textbook p. 187

Question 19

Hana cycles 6 km from her house to a park in 30 minutes at a uniform speed. She rests for 15 minutes, then cycles home in 45 minutes.

[1 mark] (a)Write the coordinates (time in minutes, distance in km) of the four points you would plot to draw her distance-time graph.

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(0, 0), (30, 6), (45, 6), (90, 0) [1]1

Tick each point you got.

[1 mark] (b)Calculate her speed, in km/h, on the way to the park.

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6 ÷ 0.5 = 12 km/h [1]1

Tick each point you got.

[1 mark] (c)Calculate her speed, in km/h, on the way home.

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6 ÷ 0.75 = 8 km/h [1]1

Tick each point you got.

[1 mark] (d)Calculate her average speed, in km/h, for the whole journey.

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12 ÷ 1.5 = 8 km/h [1]1

Tick each point you got.

For the teacher: Creating, medium difficulty, standard 7.1.3

Textbook p. 191

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