Travel 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: Travel Graphs revision notes

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Section A

1. State what the gradient of a distance–time graph represents, and what a horizontal section indicates. [2]

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

Section B

3. A cyclist’s journey is described as follows: accelerates uniformly from rest to 8 m s⁻¹ in 10 s; travels at 8 m s⁻¹ for 40 s; decelerates uniformly to rest in 20 s.

(a) Sketch or describe the velocity–time graph. [3] (b) Calculate the acceleration in the first stage. [2] (c) Calculate the total distance travelled. [4] (d) Calculate the average speed for the whole journey. [2]

4. A distance–time graph shows a curve of increasing gradient.

(a) Describe the motion. [2] (b) Explain how you would find the speed at a particular instant from such a graph. [3]

5. Two runners start together. Runner A’s distance–time graph is a straight line; runner B’s is a curve that starts shallow and steepens, crossing A’s line at 40 s.

(a) Describe each runner’s motion. [3] (b) Explain what is happening at the crossing point. [2] (c) State which runner is ahead at 20 s and justify your answer. [2]

6. A velocity–time graph shows a line starting at +6 m s⁻¹ and falling steadily to −4 m s⁻¹ over 10 s.

(a) Calculate the acceleration. [2] (b) Explain the significance of the graph crossing the time axis. [2] (c) Explain how the total displacement differs from the total distance travelled. [3]

7. A conversion graph for miles to kilometres is a straight line through the origin. When x = 50 miles, y = 80 km.

(a) State what the gradient of this graph represents. [1]

(b) Calculate the gradient, and use it to convert 120 miles to kilometres. [3]

8. A graph shows the total cost of a taxi journey, y (in $), against the distance travelled, x (in km). The line passes through (0, 5) and (10, 25).

(a) Calculate the gradient of the graph and interpret it in context, with units. [3]

(b) Explain what the y-intercept represents. [2]

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Answers

1. The gradient represents the speed [1]; a horizontal section means the object is stationary [1].

2. The gradient represents the acceleration [1]; the area under the graph represents the displacement (equal to the distance travelled only if the velocity does not change sign) [1].

3. (a) A straight line rising from the origin to (10, 8) [1]; a horizontal line from (10, 8) to (50, 8) [1]; a straight line falling from (50, 8) to (70, 0) [1]. (b) a = (8 − 0) ÷ 10 [1] = 0.8 m s⁻² [1]. (c) Stage 1: ½ × 10 × 8 = 40 m [1]. Stage 2: 40 × 8 = 320 m [1]. Stage 3: ½ × 20 × 8 = 80 m [1]. Total = 440 m [1]. (d) 440 ÷ 70 [1] = 6.3 m s⁻¹ [1].

4. (a) The object is speeding up [1], because the gradient — and therefore the speed — is increasing [1]. (b) Draw a tangent to the curve at that point [1]; construct a large right-angled triangle on the tangent [1]; the speed is the change in distance divided by the change in time for that triangle [1].

5. (a) Runner A moves at a constant speed throughout, since the gradient is uniform [1]. Runner B starts slowly and accelerates, as the gradient increases [1]. Runner B’s gradient starts below A’s, equals it partway through, and exceeds it in the period just before the 40 s crossing — this is why B is catching up and draws level at that point [1]. (b) At 40 s the two runners have travelled the same distance from the start — they are level [1]; after that point B, being faster, moves ahead [1]. (c) Runner A [1], because at 20 s A’s line is above B’s curve, meaning A has covered a greater distance [1].

6. (a) a = (−4 − 6) ÷ 10 [1] = −1.0 m s⁻² [1]. (b) The object is momentarily at rest [1], and afterwards it is moving in the opposite direction [1]. (c) The area above the axis is positive displacement and the area below is negative [1]. Displacement is the difference between the two areas, while distance is the sum of their magnitudes [1]. Here the areas are ½ × 6 × 6 = 18 m and ½ × 4 × 4 = 8 m, so the displacement is 10 m but the distance travelled is 26 m [1].

7. (a) The gradient represents the conversion factor between miles and kilometres [1].

(b) Gradient = 80 ÷ 50 = 1.6 [1]. 120 miles = 120 × 1.6 [1] = 192 km [1].

8. (a) Gradient = (25 − 5) ÷ (10 − 0) = 20 ÷ 10 = 2 [1]. This represents a cost of $2 per kilometre travelled [2].

(b) The y-intercept represents a fixed charge of $5, payable regardless of distance travelled — for example a call-out fee [2].

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Where marks are usually lost

• Reading a horizontal line on a velocity–time graph as “stationary”.
• Finding an instantaneous speed from a curve without drawing a tangent.
• Using a small triangle on the tangent, which magnifies the error.
• Adding the areas below the axis when displacement is asked for.
• Giving a gradient’s value without interpreting it in context and stating its units.
• Forgetting that a non-zero y-intercept represents a fixed starting value, not part of the rate itself.