The problems are on the motion graphs worksheet. Try them on your own first, then compare with the answers below.
Quick answer: On a distance-time graph the gradient is the speed and a flat line means stopped. These worked answers correct the 20 problems of the grade 8 science worksheet on motion graphs, from axes and meaning to a tram journey: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the graph reading, the formula and the substitution before the result with its unit. When you read a graph, state the axes first and then whether you are using a gradient or an area.
Level: Grade 8 science (Year 9 · Class 8) · Curriculum: MS-PS2-2
Problem 1 Axes and meaning
1. Firstly, time is on the horizontal axis and distance on the vertical axis.2. Then the object is stationary.
3. Finally, a higher speed.
Problem 2 Read Aarav’s walk
1. Firstly, 20 m.2. Then from 10 s to 20 s: 10 s.
3. Finally, 60 m.
Problem 3 Speed from a distance
1. Firstly, speed = distance ÷ time.2. 150 ÷ 30 = 5.0 m/s.
Problem 4 Acceleration of a car
1. Firstly, a = (v − u) ÷ t.2. a = (12 − 0) ÷ 4.0 = 3.0 m/s².
Problem 5 Match the shapes
1. Firstly, graph B.2. Then graph A.
3. Finally, graph C.
Problem 6 Change of units
1. Firstly, 72 ÷ 3.6 = 20 m/s.2. Then 15 × 3.6 = 54 km/h.
Problem 7 Gradient of a line
1. Firstly, 90 ÷ 12 = 7.5 m/s.2. Then the speed of the object.
Problem 8 Out and back
1. Firstly, 60 ÷ 15 = 4.0 m/s.2. Then from 15 s to 25 s: 10 s.
3. Afterwards, 60 m in 20 s = 3.0 m/s.
4. total distance = 60 + 60 = 120 m. 120 ÷ 45 = 2.7 m/s.
Problem 9 A train speeds up
1. a = (25 − 10) ÷ 30 = 15 ÷ 30 = 0.50 m/s².Problem 10 Find the final speed
1. Firstly, v = u + a × t.2. v = 2.0 + 1.5 × 8.0 = 2.0 + 12 = 14 m/s.
Problem 11 Scooter journey
1. Firstly, a = 12 ÷ 6 = 2.0 m/s².2. Then from 6 s to 16 s: 10 s.
3. a = (0 − 12) ÷ 4 = −3.0 m/s², a deceleration of 3.0 m/s².
Problem 12 Who is faster?
1. Firstly, Diego: 100 ÷ 16.0 = 6.25 m/s. Emma: 100 ÷ 12.5 = 8.0 m/s.2. Then 8.0 − 6.25 = 1.75 m/s.
Problem 13 A curved graph
1. Firstly, the car is speeding up.2. Then the gradient of a distance-time graph is the speed, and the gradient is getting bigger.
3. Finally, a straight sloping line rising from the origin (constant acceleration).
Problem 14 Table of distances
1. Firstly, 80 ÷ 20 = 4.0 m/s.2. Then it is stationary, because the distance stays at 80 m.
3. (120 − 80) ÷ 10 = 4.0 m/s.
Problem 15 A long walk
1. Firstly, distance = 3.0 + 6.0 = 9.0 km. Time = 60 min = 1.0 h.2. Then 9.0 ÷ 1.0 = 9.0 km/h.
3. Finally, 9.0 ÷ 3.6 = 2.5 m/s.
Problem 16 Area under a graph
1. Firstly, a = 16 ÷ 8 = 2.0 m/s².2. stage 1: ½ × 8 × 16 = 64 m. Stage 2: 16 × 12 = 192 m. Stage 3: ½ × 6 × 16 = 48 m. Total = 304 m.
3. 304 ÷ 26 = 11.7 m/s.
Problem 17 Spot the errors
1. Firstly, a flat line means the object is stationary.2. Then a steeper line on a speed-time graph means a larger acceleration, not a larger speed.
3. Finally, negative acceleration means the velocity is decreasing; the object may still move forward while slowing down.
Problem 18 A falling ball
1. Firstly, v = 9.8 × t: 9.8 m/s, 19.6 m/s, 29.4 m/s.2. Then a straight line from the origin with a gradient of 9.8 m/s².
3. area of a triangle = ½ × 3.0 × 29.4 = 44.1 m.
Problem 19 Chasing cars
1. Firstly, t = 20 ÷ 4.0 = 5.0 s.2. Then car A: 20 × 5.0 = 100 m. Car B: ½ × 5.0 × 20 = 50 m.
3. Finally, a is 50 m ahead. After that both cars travel at 20 m/s, so the gap does not change.
Problem 20 A tram journey
1. Firstly, a = 15 ÷ 10 = 1.5 m/s². Deceleration = 15 ÷ 5 = 3.0 m/s².2. distances: ½ × 10 × 15 = 75 m; 15 × 40 = 600 m; ½ × 5 × 15 = 37.5 m. Total = 712.5 m.
3. Finally, 712.5 ÷ 55 = 13 m/s (12.95…).
