ction time) at \$15 \\text{ m/s}\$. Use the equation:\n\\[\\text{distance travelled} = \\text{speed} \\times \\text{time}\\]\nGive your answer to 2 decimal places.","markscheme":"- Rearranges equation to $\\text{time} = \\frac{\\text{distance}}{\\text{speed}}$ and substitutes values: $\\frac{9}{15}$ M1\n- Evaluates to $0.60\\text{ s}$ (must be to 2 d.p.) A1","marks":2,"order":2,"markingRubric":null,"explanation":{"methods":[{"id":"rearrange_formula","label":"Rearrange the formula","steps":[{"type":"text","order":0,"title":"Read the thinking distance","content":"At a speed of $15\\text{ m/s}$, read from the straight **thinking distance** line on the supplied graph: the distance is $9\\text{ m}$.\n\nUse the thinking distance, not the curved braking-distance line.","highlights":[{"text":"**thinking time**","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{distance travelled}=\\text{speed}\\times\\text{time}"},{"latex":"\\text{time}=\\frac{\\text{distance travelled}}{\\text{speed}}"},{"latex":"\\text{time}=\\htmlData{anchor=substituted}{\\frac{9}{15}}"}]},"order":1,"title":"Rearrange and substitute","description":"Divide both sides of the distance equation by speed. Using 9 m and 15 m/s in this form earns the first mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{time}=\\htmlData{anchor=fraction}{\\frac{9}{15}}"},{"latex":"\\text{time}=\\htmlData{anchor=answer}{0.60\\text{ s}}"}],"connections":[{"to":"answer","from":"fraction","color":"green","label":"calculate"}]},"order":2,"title":"Evaluate the fraction","description":"Now calculate the fraction and give the time to 2 decimal places. This earns the second mark."}]}],"generatedAt":"2026-07-22T09:27:40.637Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]},{"id":"13b25126-0951-4c1a-81b0-d9fdbd094924","label":"d","content":"State **one** factor that could increase thinking distance.","markscheme":"- Identifies a factor increasing reaction time/thinking distance (e.g., driver tiredness, using a phone, consumption of alcohol/drugs, distraction) B1","marks":1,"order":3,"markingRubric":null,"explanation":{"methods":[{"id":"reaction_time","label":"Link reaction time to distance","steps":[{"type":"text","order":0,"title":"Identify the relevant time","content":"Thinking distance is the distance travelled **before** the driver begins to brake. It increases when the driver's **reaction time** is longer: they continue moving for longer before responding to a hazard.","highlights":[{"text":"State **one** factor that could increase thinking distance.","location":"specification"}]},{"type":"text","order":1,"title":"State a valid factor","content":"Tiredness makes a driver take longer to react to a hazard. The quadricycle therefore travels further before the brakes are applied.\n\n**Answer:** Driver tiredness.\n\n**B1:** tiredness increases reaction time, so it increases thinking distance.","highlights":[{"text":"State **one** factor that could increase thinking distance.","location":"specification"}]}]}],"generatedAt":"2026-07-22T03:01:34.412Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]},{"id":"8b3c2e1e-9265-4b1e-bbc1-27f7742c98fe","label":"e","content":"Calculate the **stopping distance** at \$20 \\text{ m/s}\$. Use the graph to help you.","markscheme":"- Identifies thinking distance as $12\\text{ m}$ and braking distance as $20\\text{ m}$ at $20\\text{ m/s}$ from the graph M1\n- Sums values to get stopping distance of $32\\text{ m}$ A1","marks":2,"order":4,"markingRubric":null,"explanation":{"methods":[{"id":"read_and_add","label":"Read the graph and add","steps":[{"type":"graph","order":0,"title":"Read both distances","viewport":{"top":50,"left":0,"right":30,"bottom":0},"axisLabels":{"x":"Speed (m/s)","y":"Distance (m)"},"description":"Stopping distance is the thinking distance plus the braking distance. At $20\\text{ m/s}$, read $12\\text{ m}$ for thinking distance and $20\\text{ m}$

Physics on the move

Question 6

Physics on the move Questions

Physics on the move

Question 6

Physics on the move Questions