two curves starting from the origin (0,0):\n1. A thick straight line representing thinking distance: it passes through (0,0), (10, 6) and (20, 12) and (30, 18).\n2. A curved line representing braking distance: it curves upwards from (0,0), passing through (10, 5) and (20, 20) and (30, 45).](https://assets.mathsgenie.co.uk/aqa-qb/40ea240d-2d3e-4012-adbc-db417e1c103c.png)","difficulty":5,"difficultyBand":"medium","totalMarks":11,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"26d146dc-887c-4d40-895c-b08bcc4e7351","label":"a","content":"Describe the relationship between the speed of the car and its thinking distance. Use data from the graph to support your answer.","markscheme":"- States that thinking distance is directly proportional to speed (or increases linearly and starts at the origin) M1\n- Supports with data from the graph: e.g., at $10\\text{ m/s}$ the thinking distance is $6\\text{ m}$, and when speed doubles to $20\\text{ m/s}$, the thinking distance doubles to $12\\text{ m}$ A1","marks":2,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"read_the_graph","label":"Read the graph","steps":[{"type":"text","order":0,"title":"Identify the line","content":"Use the straight line labelled **thinking distance**, not the curved braking-distance line. It starts at the origin and rises by equal amounts for equal increases in speed, which is the graph shape for a directly proportional relationship.","highlights":[{"text":"Describe the relationship between the speed of the car and its thinking distance.","location":"specification"}]},{"type":"text","order":2,"title":"State and support","content":"Write a relationship **and** a numerical comparison:\n\n> Thinking distance is directly proportional to speed. For example, when speed doubles from $10\\text{ m/s}$ to $20\\text{ m/s}$, thinking distance doubles from $6\\text{ m}$ to $12\\text{ m}$.\n\nStating “directly proportional” earns **M1**. The two values that demonstrate the doubling earn **A1**.","highlights":[{"text":"Use data from the graph to support your answer.","location":"specification"}]}]}],"generatedAt":"2026-07-22T09:38:32.123Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]},{"id":"da066976-43df-4e5b-8c47-bdb770f6ee78","label":"b","content":"Use the graph to find the thinking distance of the car at a speed of \$25\\text{ m/s}\$.","markscheme":"- Identifies thinking distance as $15\\text{ m}$ at a speed of $25\\text{ m/s}$ B1","marks":1,"order":1,"markingRubric":null,"explanation":{"methods":[{"id":"read_thinking_distance","label":"Read the thinking-distance line","steps":[{"type":"text","order":0,"title":"Read the graph","content":"At $25\\text{ m/s}$ on the horizontal axis, trace vertically up to the straight **thinking distance** line. Then trace horizontally to the distance axis and read $15\\text{ m}$.\n\nWrite **$15\\text{ m}$**. This exact value earns the B1 mark.","highlights":[{"text":"thinking distance of the car at a speed of \$25\\text{ m/s}\$","location":"specification"}]}]}],"generatedAt":"2026-07-22T12:16:07.470Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]},{"id":"43011212-8b68-406c-8384-2ab3464f41bf","label":"c","content":"Calculate the thinking time (reaction time) of the driver at \$25\\text{ m/s}\$. Use the formula:\n\\[\\text{distance} = \\text{speed} \\times \\text{time}\\]\nGive your answer to 2 decimal places.","markscheme":"- Rearranges formula to $\\text{time} = \\frac{\\text{distance}}{\\text{speed}}$ and substitutes values: $\\text{time} = \\frac{15}{25}$ M1\n- Obtains $\\text{time} = 0.60\\text{ s}$ (to 2 decimal places) A1","marks":2,"order":2,"markingRubric":null,"explanation":{"methods":[{"id":"formula_rearrangement","label":"Rearrange the formula","steps":[{"type":"text","order":0,"title":"Read the thinking distance","content":"Use the **straight line labelled thinking distance**, not the curved braking-distance line. At a speed of $25\\text{ m/s}$, read

Physics on the move

Question 2

Physics on the move Questions

Physics on the move

Question 2

Physics on the move Questions