Kinematics and Motion Graphs: Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Kinematics and Motion Graphs revision notes

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Section A

1. Distinguish between distance and displacement, and between speed and velocity. [4]

2. State what the gradient and the area under a velocity–time graph represent. [2]

Section B

3. A car accelerates uniformly from 8.0 m/s to 26.0 m/s in 6.0 s.

(a) Calculate the acceleration. [2] (b) Calculate the distance travelled. [3] (c) Sketch or describe the velocity–time graph. [2]

4. A ball is dropped from rest and falls freely for 2.5 s. Take g = 9.8 m/s².

(a) Calculate its velocity on impact. [2] (b) Calculate the height it fell from. [2] (c) Explain how the answers would change if air resistance were significant. [3]

5. A stone is thrown vertically upwards at 15 m/s. Take g = 9.8 m/s².

(a) Calculate the maximum height reached. [3] (b) Calculate the total time before it returns to the thrower’s hand. [2] (c) State its acceleration at the highest point and explain. [2]

6. Explain what terminal velocity is and why a skydiver reaches it. [4]

7. A cyclist accelerates uniformly from rest to 12 m/s in 4.0 s, then travels at a constant 12 m/s for a further 10 s. Using the area under a speed–time graph, find the total distance travelled. [4]

8. State what each of the following looks like on a distance–time graph: (a) an object at rest, (b) an object accelerating. [2]

9. A car decelerates uniformly from 20 m/s to 5 m/s in 3.0 s. Calculate its acceleration, and explain why deceleration does not need a different equation. [3]

10. A walker travels 400 m east, then 400 m west back to the start. State the total distance travelled and the total displacement. [2]

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Answers

1. Distance is the total path length travelled — a scalar [1]; displacement is the straight-line distance in a stated direction from the start — a vector [1]. Speed is the rate of change of distance, a scalar [1]; velocity is the rate of change of displacement, a vector [1].

2. Gradient = acceleration [1]. Area under the graph = displacement [1] — calculated by splitting the graph into simple shapes (rectangles/triangles), this technique is used for motion with constant speed or constant acceleration; a graph with changing acceleration is not calculated this way at this level.

3. (a) a = (v − u) ÷ t = (26.0 − 8.0) ÷ 6.0 [1] = 3.0 m/s² [1]. (b) The speed–time graph is a trapezium: split it into a rectangle of height 8.0 m/s and a triangle of height (26.0 − 8.0) = 18.0 m/s, both of base 6.0 s [1]. Area = (8.0 × 6.0) + ½ × 6.0 × 18.0 = 48 + 54 [1] = 102 m [1]. (c) A straight line of positive gradient [1] starting at 8.0 m/s and finishing at 26.0 m/s at t = 6.0 s [1].

4. (a) Acceleration = change in velocity ÷ time taken, so change in velocity = acceleration × time = 9.8 × 2.5 [1]; the ball starts from rest, so its velocity on impact is 24.5 m/s [1]. (b) The speed–time graph is a triangle (the speed rises uniformly from 0 to 24.5 m/s over 2.5 s), so height fallen = area under the graph = ½ × 2.5 × 24.5 [1] ≈ 30.6 m [1]. (c) Air resistance acts upwards, opposing the motion [1], so the resultant force and therefore the acceleration are less than g [1]; the ball would reach a lower final velocity and would have fallen a shorter distance in the same time [1].

5. (a) Time to reach maximum height (v = 0): a = (v − u)/t → t = (0 − 15) ÷ (−9.8) ≈ 1.53 s [1]. The speed–time graph from launch to maximum height is a triangle (speed falling uniformly from 15 m/s to 0), so maximum height = area under the graph = ½ × 1.53 × 15 [1] ≈ 11.5 m [1]. (b) Time up = 15 ÷ 9.8 = 1.53 s [1]; total time = 3.1 s [1]. (c) 9.8 m/s² downwards [1]; gravity still acts on the stone even though its velocity is momentarily zero [1].

6. Terminal velocity is the constant maximum velocity reached when the resultant force is zero [1]. Initially the skydiver’s weight is greater than the air resistance, so there is a downward resultant force and acceleration [1]. As speed increases the air resistance increases [1]; when air resistance equals weight the resultant force is zero, so the acceleration is zero and the velocity stays constant [1].

7. Stage 1 (triangle) = ½ × base × height = ½ × 4.0 × 12 [1] = 24 m. Stage 2 (rectangle) = base × height = 10 × 12 [1] = 120 m [1]. Total distance = 24 + 120 = 144 m [1].

8. (a) A flat (horizontal) line [1]. (b) A curve getting steeper [1].

9. a = (v − u) ÷ t = (5 − 20) ÷ 3.0 [1] = −5.0 m/s² [1]. It is the same acceleration equation, simply giving a negative value because the velocity is decreasing — deceleration is not a separate formula [1].

10. Distance = 800 m [1] (the full path length); displacement = 0 [1] (the walker ends up back where they started, so there is no net change in position).

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Where marks are usually lost

• Using the area under a distance–time graph.
• Forgetting the negative sign for g when motion is upwards.
• Saying acceleration is zero at the top of a projectile’s flight.
• Claiming air resistance is constant.
• Splitting a speed–time graph into the wrong shapes, or forgetting to add the areas of each stage together.
• Confusing distance–time graph shapes with speed–time graph shapes — the same-looking line means something different on each.
• Thinking deceleration needs a different formula — it’s the same a = (v − u) ÷ t, simply giving a negative answer.