Shakir Mugasha · GCSE Combined Science (Trilogy) · AQA · Higher Tier
AQA 8464 · Spec 6.5.4 · Stopping distance & acceleration (a = Δv / t)
Graduated · definitions → table look-ups → single-formula → multi-step
Sd = Td + Bd · Td = v × t · a = Δv / t · v–t & d–t graphs · show all working · keep the negative sign
This homework covers two linked topics from AQA GCSE Combined Science: Trilogy (Higher Tier, code 8464) — Section 6.5.4 Forces and Motion: stopping distance (Thinking distance + Braking distance, the Highway Code table, and the thinking-distance formula Td = v × t) and acceleration (a = Δv / t, including negative / deceleration answers). The questions start with straightforward definitions and table look-ups, then ramp up to multi-step calculations using both formulas.
Show every step of your working: write down what you are given, the equation you are using, the substitution, and the answer with its unit. Do not round until the end.
Remember: when the velocity is decreasing, Δv = final − initial will be negative and so will the acceleration — the negative sign must appear in the answer. Also remember the speed conversions given on the exam equation sheet (30 mph = 13.4 m/s, 20 mph ≈ 8.94 m/s, 50 mph ≈ 22.35 m/s).
Questions 16 to 20 introduce velocity–time and distance–time graphs — each comes with its own labelled graph drawn to scale. For these, the gradient of a velocity–time graph is the acceleration, and the area under a velocity–time graph is the distance travelled. For distance–time graphs, the gradient gives the speed — a horizontal line means the object is stationary.
Use this Highway Code table for Questions 4, 5, 9, 13, 14 and 15. Values are taken from the official Highway Code and are the numbers you should know for the exam.
| Speed (mph) | Thinking Distance (m) | Braking Distance (m) | Stopping Distance (m) |
|---|---|---|---|
| 20 | 6 | 6 | 12 |
| 30 | 9 | 14 | 23 |
| 40 | 12 | 24 | 36 |
| 50 | 15 | 38 | 53 |
| 60 | 18 | 55 | 73 |
| 70 | 21 | 75 | 96 |
Pattern to spot: Thinking distance rises by exactly 3 m for every 10 mph increase. Braking distance rises by much more — at 70 mph it is already 75 m, roughly the length of a football pitch.
Define stopping distance. Write the equation that links stopping distance, thinking distance and braking distance, using the abbreviations Sd, Td and Bd.
Show your working fully on a separate piece of paper.
Define thinking distance in a single sentence. Your answer must include the words distance, reaction time and hazard.
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Define braking distance in a single sentence. Your answer must include the words distance and braking force.
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A car is travelling at 30 mph. Use the Highway Code table to state its thinking distance, braking distance and stopping distance, and then write the one-line addition that gives the total stopping distance.
Show your working fully on a separate piece of paper.
A lorry is travelling at 70 mph. Use the Highway Code table to state its thinking distance, braking distance and stopping distance. Identify which of these three values is described as "about the length of a football pitch" and state its value.
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A driver is travelling at 30 mph (13.4 m/s) with a reaction time of 0.6 s. Calculate the thinking distance. Use the formula Td = v × t.
Show your working fully on a separate piece of paper.
A car accelerates from rest (0 m/s) to 20 m/s in 5 seconds. Calculate its acceleration. Use the formula a = Δv / t.
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A car decelerates from 20 m/s to 8 m/s in 4 seconds. Calculate its acceleration. Your final answer must include the negative sign — write down why the sign is negative in one sentence alongside the calculation.
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Using the Highway Code values for 40 mph, write out the calculation that adds the thinking distance and the braking distance to give the total stopping distance. State the final answer.
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A van travels at 22 m/s with a reaction time of 0.7 s. Calculate the thinking distance using Td = v × t. Give your answer to one decimal place.
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A rocket accelerates from rest at 12 m/s² for 6 seconds. By rearranging a = Δv / t (so that Δv = a × t), find the rocket's final velocity.
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A bicycle travels at a constant velocity of 8 m/s for 5 seconds. What is its acceleration? Add one sentence explaining why the answer is what it is.
Show your working fully on a separate piece of paper.
A driver travelling at 20 mph (≈ 8.94 m/s) has a thinking distance of 6 m.
(a) Calculate the driver's reaction time using a rearrangement of Td = v × t.
(b) Use the Highway Code table to find the total stopping distance at 20 mph.
Show your working fully on a separate piece of paper.
Use the Highway Code table to calculate:
(i) the ratio of braking distance at 60 mph to braking distance at 30 mph,
(ii) the ratio of thinking distance at 60 mph to thinking distance at 30 mph.
Then write two sentences commenting on what these ratios tell you about how each type of distance depends on speed. (Reference the ½mv² kinetic-energy idea if it helps your explanation.)
Show your working fully on a separate piece of paper.
A car is travelling at 50 mph (≈ 22.35 m/s) when the driver sees a hazard. The driver's reaction time is 0.7 s.
(a) Calculate the thinking distance in metres using Td = v × t.
(b) Use the Highway Code table to find the total stopping distance at 50 mph.
(c) By what factor does the total stopping distance exceed the thinking distance alone? (Divide and round to one decimal place.)
Show your working fully on a separate piece of paper.
A car's velocity–time graph is shown below. From t = 0 s to t = 5 s the velocity rises in a straight diagonal from 0 m/s to 25 m/s. From t = 5 s to t = 8 s the velocity is constant at 25 m/s.
Calculate the car's acceleration during the first 5 seconds of the journey using a = Δv / t.
Show your working fully on a separate piece of paper.
A van's velocity–time graph is shown below. From t = 0 s to t = 4 s the velocity falls in a straight diagonal from 30 m/s down to 10 m/s. From t = 4 s to t = 7 s the velocity is constant at 10 m/s.
Calculate the van's acceleration during the first 4 seconds. Your answer must include the negative sign — write one sentence explaining why the sign is negative alongside the calculation.
Show your working fully on a separate piece of paper.
A cyclist's distance–time graph is shown below, with the following three phases:
(0 s, 0 m) to (6 s, 60 m).60 m from t = 6 s to t = 10 s.(10 s, 60 m) to (16 s, 120 m).(a) Calculate the cyclist's speed in Phase A in m/s.
(b) Calculate the cyclist's speed in Phase C in m/s.
(c) Explain in one sentence what is happening in Phase B, and explain why a horizontal line on a distance–time graph means the object is stationary.
Show your working fully on a separate piece of paper.
A bus's velocity–time graph is shown below. The graph has three phases:
(0 s, 0 m/s) to (4 s, 20 m/s) (acceleration).20 m/s from t = 4 s to t = 10 s (constant velocity).(10 s, 20 m/s) to (14 s, 0 m/s) (deceleration).Calculate the total distance travelled over the 14 seconds, by finding the area under the graph. Split the area into three shapes — a triangle for Phase A, a rectangle for Phase B, and a triangle for Phase C — and state the area of each shape and the total.
Show your working fully on a separate piece of paper.
A lorry's velocity–time graph is shown below:
(0 s, 0 m/s) to (5 s, 30 m/s).30 m/s from t = 5 s to t = 8 s.(8 s, 30 m/s) to (13 s, 0 m/s).(a) Calculate the lorry's deceleration in Phase C using a = Δv / t. The answer must include the negative sign.
(b) Calculate the total distance travelled over the 13 seconds by finding the area under the graph. Split it into a triangle (Phase A), a rectangle (Phase B) and a triangle (Phase C) and state the area of each.
(c) State the braking distance in Phase C alone (the area of the Phase C triangle), and comment in one sentence on whether this deceleration looks realistic for a typical road lorry.
Show your working fully on a separate piece of paper.
Homework prepared for Shakir Mugasha — GCSE Combined Science tuition, 7 July 2026. Twenty questions on stopping distance (Sd = Td + Bd, Highway Code table, Td = v × t), acceleration (a = Δv / t, including deceleration), and velocity–time & distance–time graph interpretation (gradient = acceleration or speed; area under a v–t graph = distance). Questions start with definitions and table look-ups and increase in difficulty to multi-step graph calculations. Show all working on a separate piece of paper.
Prepared by Dynamic Brain Hub · Human Experts - Cutting Edge AI Systems.