ric cargo bike is travelling along a cycle lane. The speed-time graph below shows the motion of Cargo Bike A. The rider of Cargo Bike A observes a pedestrian stepping into the path at time \$t = 0.00 \\text{ s}\$, and applies the brakes at \$t = 0.80 \\text{ s}\$. The bike comes to a complete stop at \$t = 2.40 \\text{ s}\$.\n\n![Speed-time graph for Cargo Bike A](https://assets.mathsgenie.co.uk/aqa-qb/e4459613-b267-4ab6-bf83-515a9b701ac0.png)","difficulty":5,"difficultyBand":"medium","totalMarks":9,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"0491a254-d2c4-4ac9-bfbd-5fa0768388eb","label":"a","content":"Show that the thinking distance is equal to the braking distance for Cargo Bike A by calculating both values from the graph.","markscheme":"- Calculates the thinking distance for Cargo Bike A: $\\text{speed} \\times \\text{time} = 16 \\times 0.80 = 12.8\\text{ m}$ M1\n- Calculates the braking distance for Cargo Bike A using the triangular area: $\\frac{1}{2} \\times (2.40 - 0.80) \\times 16 = 12.8\\text{ m}$ M1\n- Concludes that both distances are equal to $12.8\\text{ m}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"area_under_graph","label":"Area under the graph","steps":[{"type":"text","order":0,"title":"Use graph areas","content":"On a speed–time graph, the area under the line represents the distance travelled. The horizontal section from $0.00\\,\\text{s}$ to $0.80\\,\\text{s}$ is the thinking-distance area; the triangular section after that is the braking-distance area.","highlights":[{"text":"applies the brakes at \$t = 0.80 \\text{ s}\$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"d_{\\mathrm{thinking}}=16\\,\\mathrm{m\\,s^{-1}}\\times0.80\\,\\mathrm{s}"},{"latex":"d_{\\mathrm{thinking}}=\\htmlData{anchor=thinking-distance}{12.8\\,\\mathrm{m}}"}],"highlights":[{"color":"teal","style":"box","anchor":"thinking-distance"}]},"order":1,"title":"Calculate thinking distance","description":"The bike's speed stays constant during the thinking time, so use distance = speed multiplied by time. Substituting the speed and thinking time earns the first method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"t_{\\mathrm{braking}}=2.40\\,\\mathrm{s}-0.80\\,\\mathrm{s}=1.60\\,\\mathrm{s}"},{"latex":"d_{\\mathrm{braking}}=\\frac{1}{2}\\times1.60\\,\\mathrm{s}\\times16\\,\\mathrm{m\\,s^{-1}}"},{"latex":"d_{\\mathrm{braking}}=\\htmlData{anchor=braking-distance}{12.8\\,\\mathrm{m}}"}],"highlights":[{"color":"orange","style":"box","anchor":"braking-distance"}]},"order":2,"title":"Calculate braking distance","description":"The braking section is a triangle. Its base is the braking duration: 2.40 s minus 0.80 s, so use 2.40 - 0.80. Using the triangular area with this braking time earns the second method mark."},{"type":"text","order":3,"title":"Compare the distances","content":"Both areas are $12.8\\,\\text{m}$, so the thinking distance is equal to the braking distance. This clear comparison earns the accuracy mark.","highlights":[{"text":"Show that the thinking distance is equal to the braking distance for Cargo Bike A","location":"specification"}]}]}],"generatedAt":"2026-07-22T12:02:50.127Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]},{"id":"5c68328f-6bbf-4cb1-8985-618cfd7aa31b","label":"b","content":"Another cargo bike, Cargo Bike B, is also travelling at \$16 \\text{ m/s}\$. Its rider has a longer reaction time and applies the brakes at \$t = 1.10 \\text{ s}\$. Cargo Bike B decelerates at the same rate as Cargo Bike A. Determine the total stopping distance of Cargo Bike B.","markscheme":"- Calculates the thinking distance for Cargo Bike B: $16 \\times 1.10 = 17.6\\text{ m}$ M1\n- Identifies that Cargo Bike B has the same deceleration rate and initial speed, so its braking distance is also $12.8\\text{ m}$ M1\n- Determines the total stopping distance for Cargo Bike B: $17.6 + 12.8 = 30.4\\text{ m}$ A1","mar

Physics on the move

Question 16

Physics on the move Questions

Physics on the move

Question 16

Physics on the move Questions