Stopping distance, reaction time and braking – GCSE Physics exam questions with mark scheme

1Grade 4–55 marks

A car is travelling at 20 m/s. The driver sees a hazard and has a reaction time of 0.70 s. The braking distance of the car is 24 m.

(a) Calculate the thinking distance. [2]

(b) Calculate the stopping distance of the car. [1]

(a) s = 20 × 0.70(1)
(a) 14 m(1)
(b) 14 + 24 = 38 m (allow ecf from (a))(1)
(c) reaction time increases (allow: driver reacts more slowly)(1)
(c) so the thinking distance increases (allow: the car travels further before the brakes are applied), so the stopping distance increases (braking distance is unchanged, but stating this is not needed for the mark)(1)
Answer: (a) 14 m (b) 38 m (c) longer reaction time → longer thinking distance → longer stopping distance
Examiner tip: Tiredness affects the DRIVER, so it changes thinking distance — not braking distance.

2Grade 1–3Foundation4 marks

A driver sees a hazard in the road ahead and makes an emergency stop.

(a) Complete the equation.

stopping distance = thinking distance + ____________________ [1]

(b) What is meant by the thinking distance? [1]

(d) Which of the following is a typical reaction time for a driver? Tick (✓) one box. [1]

0.006 s	☐
0.6 s	☐
6 s	☐
60 s	☐

(a) braking distance(1)
(b) the distance the vehicle travels during the driver’s reaction time
allow the distance travelled between the driver seeing the hazard and applying the brakes
do not accept ‘the time the driver takes to react’ (a time, not a distance)(1)
(c) the distance the vehicle travels after the brakes are applied (until it stops)
allow the distance travelled under the braking force(1)
(d) 0.6 s
more than one box ticked scores 0(1)
Answer: (a) braking distance (b) distance travelled during the driver’s reaction time (c) distance travelled while the brakes act, until the car stops (d) 0.6 s
Examiner tip: Thinking distance is a distance, not a time: ‘how long the driver takes to react’ scores nothing.

3Grade 4–56 marks

The table shows the thinking distance and braking distance for a car at different speeds on a dry road. The driver has a reaction time of 0.50 s.

Speed in m/s	Thinking distance in m	Braking distance in m
10	5.0	8
15	7.5	18
20	10.0	32
25	X	50

(a) Calculate value X. [2]

(b) Calculate the stopping distance of the car at 25 m/s. [1]

(c) Compare how the thinking distance and the braking distance change as the speed of the car increases. Use data from the table in your answer. [2]

(d) Give the range of typical reaction times for a driver. [1]

(a) X = 25 × 0.50
allow s = v t with correct substitution(1)
(a) 12.5 (m)
allow 13 (m)(1)
(b) 62.5 (m)
allow 62–63 (m)
allow ecf: their X + 50(1)
(c) thinking distance is (directly) proportional to speed / thinking distance doubles when speed doubles (e.g. 5.0 m at 10 m/s, 10.0 m at 20 m/s)
allow thinking distance increases by the same amount (2.5 m) for each 5 m/s(1)
(c) braking distance increases more rapidly than thinking distance / more than doubles when speed doubles (e.g. 8 m at 10 m/s but 32 m at 20 m/s, four times greater)
allow braking distance increases by a larger amount each time
a comparison using data is needed for both marks; ‘both increase’ alone scores 0(1)
(d) 0.2 s to 0.9 s
allow any range within 0.2–1.0 s that has a width of at least 0.5 s(1)
Answer: (a) 12.5 m (b) 62.5 m (c) thinking distance ∝ speed (doubles); braking distance increases faster (×4 when speed doubles) (d) 0.2–0.9 s
Examiner tip: ‘Compare’ needs both distances and some numbers: say what thinking distance does AND how braking distance is different.

4Grade 6–7Triple only8 marks

The graph shows how the thinking distance and the stopping distance of a car depend on its speed on a dry road and on an icy road. The solid lines show the stopping distance and the dashed line shows the thinking distance. The same driver drives the car on both roads.

(a) Use the graph to determine the stopping distance on the dry road when the car is travelling at 15 m/s. [1]

(b) Use the graph to determine the reaction time of the driver. [2]

(d) Explain why the thinking distance is the same on both roads. [1]

(e) The stopping distance does not increase in direct proportion to the speed.

Explain why. [2]

(a) 24 (m)
allow 23–25 (m)(1)
(b) correct pair of values read from the thinking distance line, e.g. 15 m at 25 m/s, 6 m at 10 m/s, 12 m at 20 m/s
allow reaction time = thinking distance ÷ speed(1)
(b) 0.6 (s)
allow 0.58–0.62 (s) from a correct reading(1)
(c) thinking distance at 15 m/s = 9 m and stopping distance (icy) at 15 m/s = 54 m
allow 8–10 m for the thinking distance and 53–55 m for the stopping distance
allow thinking distance calculated as 15 × their reaction time from (b)(1)
(c) 45 (m)
allow 43–47 (m)
allow ecf from their readings: stopping distance − thinking distance(1)
(d) the road surface does not affect the driver’s reaction time
allow thinking distance depends only on speed and reaction time (which are the same)
allow the brakes have not yet been applied during the thinking distance(1)
(e) the braking distance increases faster than (in proportion to) the speed / the braking distance is not proportional to speed
allow braking distance is proportional to speed²
allow thinking distance is proportional to speed but braking distance is not(1)
(e) because the kinetic energy (that the braking force must transfer) depends on speed squared / doubling speed gives four times the kinetic energy
allow more work must be done by the brakes (so a greater distance is needed for the same braking force)(1)
Answer: (a) 24 m (b) 0.6 s (c) 45 m (d) road surface does not affect reaction time (e) braking distance ∝ v² because KE ∝ v²
Examiner tip: In (c) the graph line is the stopping distance — subtract the thinking distance to get the braking distance.

5Grade 6–7Higher5 marks

A family car is travelling along a motorway at 31 m/s. The driver brakes and the car stops. The braking distance is 75 m.

(a) Estimate the average braking force on the car. [4]

(b) Suggest why the braking force from the brakes would actually be slightly smaller than your estimate. [1]

(a) a sensible estimate of the mass of a car: 800–2000 kg(1)
(a) E_k = ½ × m × 31²
e.g. ½ × 1000 × 31² = 480 500 J(1)
(a) work done = kinetic energy and W = F × s with s = 75 m: F = E_k ÷ 75
allow use of v² − u² = 2as and F = ma as an alternative method(1)
(a) braking force consistent with their mass, e.g. 6400 N for 1000 kg
allow answers that round correctly, e.g. 6407 N, 6000 N
range 5100–12 800 N for masses 800–2000 kg(1)
(b) air resistance / drag (and friction in other parts of the car) also does work on the car / also acts against the motion
allow other resistive forces help to slow the car(1)
Answer: (a) about 6000 N (e.g. 6400 N for a 1000 kg car) (b) air resistance also does work slowing the car
Examiner tip: An ‘estimate’ question expects you to choose a sensible value yourself — write your mass down clearly so the examiner can follow it.

6Grade 8–9Higher8 marks

A car of mass 1500 kg is travelling at 24 m/s. The driver sees a fallen tree lying across the road 55 m ahead.

The driver’s reaction time is 0.45 s. When the brakes are applied, the car decelerates with a constant braking force of 9000 N.

(a) Calculate the thinking distance. [2]

(b) Calculate the deceleration of the car. [2]

(d) Determine whether the car stops before it reaches the tree. Give a reason for your answer. [1]

(a) s = 24 × 0.45(1)
(a) 10.8 (m)
allow 11 (m)(1)
(b) a = 9000 ÷ 1500
allow 9000 = 1500 × a(1)
(b) 6 (m/s²)
allow −6 (m/s²)(1)
(c) 0² − 24² = 2 × (−6) × s
allow 24² = 2 × 6 × s
allow ecf from (b)(1)
(c) s = 576 ÷ 12(1)
(c) 48 (m)
allow ecf from (b) correctly calculated
alternative: E_k = ½ × 1500 × 24² = 432 000 J (1); s = 432 000 ÷ 9000 (1); 48 m (1)(1)
(d) no, stopping distance is 58.8 m, which is greater than 55 m
allow ecf: compare their (a) + (c) with 55 m
a conclusion with no comparison of distances scores 0(1)
Answer: (a) 10.8 m (b) 6 m/s² (c) 48 m (d) no — stopping distance 58.8 m > 55 m
Examiner tip: In (d) the braking distance alone (48 m) is less than 55 m — you must add the thinking distance to compare.

7Grade 1–3Foundation4 marks

Different factors can affect the stopping distance of a car.

(a) Complete the table. Tick (✓) one box in each row to show whether the factor increases the thinking distance or the braking distance. [2]

Factor	Increases thinking distance	Increases braking distance
driver is tired	☐	☐
road is icy	☐	☐
tyres are worn	☐	☐
driver is sending a text message	☐	☐

(b) Explain why the braking distance of a car is greater on an icy road than on a dry road. [2]

(a) two or three rows correct
tired → thinking; icy → braking; worn tyres → braking; texting → thinking(1)
(a) all four rows correct
a row with both boxes ticked is wrong(1)
(b) there is less friction (force) between the tyres and the road
allow less grip
do not accept ‘no friction’(1)
(b) so the car decelerates less / slows down more slowly (so it travels further before stopping)
allow smaller braking force (that can be applied without skidding)(1)
Answer: (a) tired, texting → thinking distance; icy road, worn tyres → braking distance (b) less friction between tyres and road, so smaller deceleration and the car travels further
Examiner tip: Driver factors (tiredness, drink, drugs, distractions) change the thinking distance; road and vehicle factors change the braking distance.

8Grade 8–9Higher7 marks

A car brakes with the same constant braking force each time it stops.

(a) Explain why doubling the speed of the car makes the braking distance four times greater. Refer to work done and kinetic energy in your answer. [3]

(b) At a speed of 12 m/s the braking distance of the car is 10 m.

Calculate the braking distance at a speed of 30 m/s. [2]

(a) kinetic energy = ½mv², so doubling the speed gives four times the kinetic energy(1)
(a) to stop the car the work done by the braking (friction) force must equal the kinetic energy / all the kinetic energy must be transferred(1)
(a) work done = force × distance, so with the same force the distance must be four times greater
allow braking distance ∝ kinetic energy (for a constant force)(1)
(b) (30 ÷ 12)² = 6.25 / braking distance ∝ speed²
allow ½ m 12² = F × 10 used to find F/m = 7.2 then 30² ÷ 2 ÷ 7.2(1)
(b) 62.5 (m)
allow 62 or 63 (m)
an answer of 25 m (direct proportion) scores 0(1)
(c) the brakes may overheat (so they become less effective / fail)
allow a large amount of energy transferred to the brakes in a short time(1)
(c) the driver may lose control of the car / the car may skid
allow the tyres lose grip; allow passengers thrown forward / injured by the large force(1)
Answer: (a) KE ∝ v², W = Fs with F constant so d ∝ KE → ×4 (b) 62.5 m (c) brakes overheat; loss of control / skidding
Examiner tip: Braking distance scales with speed squared — scaling 10 m by 30/12 (giving 25 m) is the most common wrong answer.

9Grade 6–76 marks

Two cars are travelling at the same speed along a road in heavy rain.

Driver P is alert. Driver P’s car has new tyres and new brake pads.
Driver Q is having an argument with a passenger. Driver Q’s car has worn tyres and worn brake pads.

Both drivers see the same hazard at the same moment and make an emergency stop.

Explain why the stopping distance of Driver Q’s car is greater than the stopping distance of Driver P’s car.

Your answer should refer to thinking distance, braking distance and the forces involved. [6]

Stopping distance: stopping distance = thinking distance + braking distance; both parts are greater for Driver Q(1)
Thinking distance: Driver Q is distracted (by the argument), so Driver Q’s reaction time is longer; the car travels at the same (constant) speed during the reaction time, so it travels further before the brakes are applied; thinking distance = speed × reaction time(1)
Braking distance – tyres: worn tyres have less tread so they cannot push water away as well (in heavy rain); less grip / less friction between the tyres and the wet road; the car may skid(1)
Braking distance – brakes: worn brake pads produce a smaller friction force on the discs / wheels; so a smaller braking force(1)
Forces and energy: a smaller braking force gives a smaller deceleration (F = ma); the kinetic energy of the car is the same for both cars (same speed, similar mass), so with a smaller force the work done (F × s) needs a greater distance; the car travels further before stopping(1)
The rain: the wet road reduces friction for both cars, but the effect is greater for Driver Q’s worn tyres(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically linked explanation of why BOTH the thinking distance and the braking distance are greater for Driver Q. The longer reaction time is linked to the extra thinking distance, and the worn tyres and brakes are linked through reduced friction / braking force to a smaller deceleration and a longer braking distance.
Level 2	3–4	Explains why either the thinking distance or the braking distance is greater, with a linked reason, and identifies relevant factors for the other part. Some links between steps may be missing.
Level 1	1–2	Identifies one or more relevant factors (e.g. distraction, worn tyres, worn brakes) or states that thinking or braking distance increases, with little or no explanation. Links are not made.
No relevant content	0	Nothing worthy of credit.

Answer: Q’s longer reaction time (distraction) → larger thinking distance; worn tyres (less grip on wet road) and worn brakes → smaller friction/braking force → smaller deceleration → larger braking distance; so larger stopping distance.
Examiner tip: Keep driver factors and car/road factors separate: distraction changes thinking distance only, worn tyres and brakes change braking distance only.

10Grade 4–57 marks

A student measures her reaction time using a ruler.

Her partner holds a 30 cm ruler vertically, with the zero mark level with the top of the student’s open thumb and finger. The partner drops the ruler without warning and the student catches it as quickly as possible. The drop distance is the reading on the ruler where it is caught.

The table shows her results.

Trial	1	2	3	4	5
Drop distance in cm	15.8	17.1	29.5	16.4	16.2

(a) Identify the anomalous result. [1]

(b) Calculate the mean drop distance. Do not include the anomalous result. [2]

Use the graph and your answer to (b) to determine the student’s reaction time. [1]

(d) Give one variable the student should keep the same in each trial. [1]

(e) The student repeats the test while she is having a conversation on her mobile phone.

Predict how the drop distance will change. Give a reason for your answer. [2]

(a) 29.5 (cm)
allow trial 3(1)
(b) (15.8 + 17.1 + 16.4 + 16.2) ÷ 4(1)
(b) 16.4 (cm)
allow 16.375 / 16.38 / 16
an answer of 19.0 (cm) (anomaly included) scores 1 mark(1)
(c) 0.18 (s)
allow 0.18–0.19 s
allow ecf from (b) if read correctly from the graph(1)
(d) any one from: same hand / same (dominant) hand catching; same gap between thumb and finger at the start; same starting position of the ruler (zero level with top of thumb); same ruler; same person dropping the ruler; no warning given before dropping
ignore ‘same ruler length’ on its own if ‘same ruler’ not stated
do not accept ‘same drop distance’(1)
(e) the drop distance increases(1)
(e) because her reaction time increases (as she is distracted)
allow she takes longer to react so the ruler falls for longer / accelerates for longer
reason only scores if the change is ‘increases’(1)
Answer: (a) 29.5 cm (b) 16.4 cm (c) 0.18 s (d) e.g. same catching hand (e) increases — distraction increases reaction time
Examiner tip: In (b) leave the anomaly out before working out the mean — including 29.5 and dividing by 5 gives 19.0 cm and loses the accuracy mark.

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