Motion graphs worksheet answers for GCSE physics (free PDF download)

Problem 1 Graph facts

1. speed.
2. Then, acceleration.
3. Next, distance travelled.
4. Then, stationary.

Problem 2 Reading a straight line

1. Moreover, the line passes through 25 m at 10 s.

2. speed = 50 ÷ 20 = 2.5 m/s.

3. d = 2.5 × 16 = 40 m.

Problem 3 Gradient from two points

1. 30 − 6 = 24 m.

2. 8 − 2 = 6 s.

3. gradient = 24 ÷ 6 = 4.0 m/s, the speed.

Problem 4 Describe the sections

1. Next, a: the object is stationary. B: it moves at a high constant speed. Then, C: it moves at a low constant speed.

2. Section B has the steepest gradient, so it is the fastest.

Problem 5 Reading a velocity graph

1. In addition, at 3 s the line is at 6 m/s.

2. a = (12 − 0) ÷ 6 = 2.0 m/s².

3. The line is horizontal, so the bike moves at a constant 12 m/s with zero acceleration.

Problem 6 Rectangle area

1. A horizontal line at 5.0 m/s from 0 to 12 s.

2. area = 5.0 × 12 = 60 m.

3. Also, it represents the distance travelled.

Problem 7 Triangle area

1. In addition, a straight line from the origin sloping up to 9.0 m/s at 6.0 s.

2. a = 9.0 ÷ 6.0 = 1.5 m/s².

3. area = ½ × 6.0 × 9.0 = 27 m.

Problem 8 A walk in stages

1. Section 1: 30 ÷ 20 = 1.5 m/s. Then, section 2: 0 m/s. Section 3: (90 − 30) ÷ (90 − 50) = 60 ÷ 40 = 1.5 m/s.

2. average speed = 90 ÷ 90 = 1.0 m/s.

Problem 9 Out and back

1. 120 ÷ 30 = 4.0 m/s.

2. Also, the return is 120 m in 90 − 50 = 40 s: 3.0 m/s.

3. Distance = 120 + 120 = 240 m. He is back home, so the displacement is 0 m.

Problem 10 Trapezoid journey

1. a = 6.0 ÷ 3.0 = 2.0 m/s².

2. a = (0 − 6.0) ÷ 5.0 = −1.2 m/s².

3. areas: ½ × 3 × 6 = 9 m; 10 × 6 = 60 m; ½ × 5 × 6 = 15 m. Total = 84 m.

Problem 11 From a table to a graph

1. Moreover, the points lie on a straight line through the origin.

2. a = 3 ÷ 2 = 1.5 m/s².

3. area = ½ × 8 × 12 = 48 m.

Problem 12 Two lines compared

1. A: 40 ÷ 10 = 4.0 m/s. B: 25 ÷ 10 = 2.5 m/s.

2. At 20 s A is at 80 m and B at 50 m, so the gap is 30 m.

Problem 13 Slowing to rest

1. Gradient = (0 − 18) ÷ 9.0 = −2.0 m/s².

2. The negative sign shows the velocity is decreasing, so the van decelerates.

3. area = ½ × 9.0 × 18 = 81 m.

Problem 14 Tangent to a curve

1. Then, (11.0 − 3.0) ÷ (3.0 − 1.0) = 8.0 ÷ 2.0 = 4.0 m/s.

2. In addition, it is the speed of the coconut at 2 s.

3. Yes: the curve gets steeper, so the gradient (speed) increases.

Problem 15 From distance to speed steps

1. 0–10 s: 50 ÷ 10 = 5.0 m/s. 10–30 s: 80 ÷ 20 = 4.0 m/s. 30–40 s: 0 m/s.

2. Three flat steps: 5.0 m/s, then 4.0 m/s, then 0 m/s.

3. 130 ÷ 40 = 3.25, so 3.3 m/s.

Problem 16 Robot on a rail

1. Then, stage 1: 4.0 × 5.0 = +20 m. Stage 2: 0 m. So, stage 3: −2.0 × 6.0 = −12 m.

2. displacement = 20 + 0 − 12 = +8 m. Distance = 20 + 12 = 32 m.

3. total time = 14 s. average speed = 32 ÷ 14 = 2.3 m/s. Average velocity = 8 ÷ 14 = 0.57 m/s.

Problem 17 Counting squares

1. Also, one square is 1.0 m/s × 1.0 s = 1.0 m.

2. area ≈ 38 + 14 ÷ 2 = 38 + 7 = 45 squares.

3. Moreover, the distance is about 45 m.

Problem 18 Find the top speed

1. triangle = ½ × 5.0 × V = 2.5 V. Rectangle = 15 × V = 15 V.

2. 2.5 V + 15 V = 17.5 V = 210, so V = 12 m/s.

3. a = 12 ÷ 5.0 = 2.4 m/s².

Problem 19 Spot the graph errors

1. The distance from the start is decreasing, so the object is moving back towards the start. Then, a straight line means constant speed, not slowing.

2. Then, a horizontal line means constant velocity; the object is stationary only if the line is at 0 m/s.

3. Next, a curve that gets flatter (decreasing gradient).

Problem 20 Overtaking scooter

1. In addition, lorry: area of a rectangle = 12 t. Also, scooter: area of a triangle = ½ × t × (2.0 t) = t².

2. t² = 12 t, so t = 12 s (both are then 144 m from the start).

3. v = a t = 2.0 × 12 = 24 m/s, which is twice the lorry’s speed.