AQA A-Level Physics Paper 1, November 2020: Question 4
14 marks · Hard difficulty · Extended Answer
Analyze vehicle speed, stopping distances, and circular motion on a sloped motorway curve using graphs, kinematics, forces, and circular motion principles.
Practise this questionQuestion
Question text
04 A pair of cameras is used on a motorway to help determine the average speed of
vehicles travelling between the two cameras.
Figure 5 shows the speed–time graph for a car moving between the two cameras.
Figure 5
04.1 The speed limit for the motorway between the two cameras is 22 m s−1.
Determine whether the average speed of the car exceeded this speed limit.
[3 marks]
04.2 Markings called chevrons are used on motorways.
The chevron separation is designed to give a driver time to respond to any change in
speed of the car in front. The driver is advised to keep a minimum distance d behind
*13* the car in front, as shown in Figure 6.
Figure 6
Government research suggests that the typical time for a driver to respond is between
1.6 s and 2.0 s.
Suggest a value for d where the speed limit is 31 m s−1.
[2 marks]
15 d = m
04.3 The chevron separation is based on the response time, not on the time taken for a car
to stop.
The brakes of a car are applied when its speed is 31 m s−1 and the car comes to rest.
The total mass of the car is 1200 kg.
The average braking force acting on the car is 6.8 kN.
Calculate the time taken for the braking force to stop the car and the distance
travelled by the car in this time.
[4 marks]
time = s
distance = m
04.4 Suggest why the chevron separation on motorways does not take into account the
distance travelled as a car comes to rest after the brakes are applied.
[1 mark]
04.5 At bends on motorways the road is sloped so that a car is less likely to slide out of its
lane when travelling at a high speed.
Figure 7 shows a car of mass 1200 kg travelling around a curve of radius 200 m.
The motorway is sloped at an angle of 5.0°.
Figure 8 shows the weight W and reaction force N acting on the car. The advisory
speed for the bend is chosen so that the friction force down the slope is zero.
*15* Figure 7
Figure 8
Suggest an appropriate advisory speed for this section of the motorway.
[4 marks]
advisory speed = m s−1
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark
details
Evidence of distance travelled = area under graph
04.1 Full marks can be credited for use of suvat. 3 AO3.1a
= 1755 + 1440 + 1620 = 4815✓ AO2.1f
Allow ecf for distance in MP2
Average speed = total distance/time taken = 4815/240 AO3.1a
= 20.1 m s–1✓ (at least 3sf) Only award MP3 for incorrect speed if
attempt made to calculate distance correctly
Which is less than (speed) limit, (and therefore the answer
e.g. area under graph OR a.e. in distance or
is No). ✓
speed
Alternative for MP2 and MP3 Alternative for MP1 and MP2:
Total area = 80.25 m s–1 min✓
Calculation of distance travelled at speed limit = 5280 m✓
Time = 4 min
Which is greater than distance travelled (so no). ✓
Average = 20.1 m s-1✓
04.2 Award MP2 if 1.6 s (to give 50 m) or 1.8 s (to 1 AO3.1a
2.0 s give 56 m) or 1.7 s (to give 53 m) or average
Using reaction time of ✓ 1 AO1a
of two distances used
Use of distance = speed × time = 62 m.
Allow 60 m. – A-LEVEL PHYSICS – –
62 m (would be appropriate). ✓
04.3 If no other mark given, allow 1 mark for 4 AO2.1b
mv = 1200 x 31 (= 37200)
Use of F = ma to calculate acceleration.
a = 6800/1200✓ = 5.7 m s–2
Alternative for MP1 and MP2
evidence of use of suvat to calculate s or t, ✓
𝑚𝑣−𝑚𝑢
12 𝑡 = 𝐹
to give t = 5.5 s ✓
Allow ce for a.
s = 85 m. ✓
Allow ce for either incorrect s or t.
ID
details
04.4 (It is assumed that) the car in front would take the same 1 AO3
time/travel the same distance as the car behind when Alternative:
braking/ only difference is reaction time of the driver of car suggestion that total stopping distance is too
behind.✓ large (drivers would ignore it/inefficient use of
Or motorway)
Car in front cannot stop instantaneously (so car behind will
have time/distance to bring car to rest).or words to that
effect
– A-LEVEL PHYSICS – –
ID
details
04.5 Correct use of cos (5) ✓ 4 AO3.1a
E.g.
mg = N cos (5)
Correct use of sin (5) ✓
E.g.
May see cos (85) for sin (5)
N sin (5) (= mv2/r)
So Alternative for MP1 and MP2:
Evidence of mg tan (5)
mv2/r seen✓
fourth mark is for answer and suggesting this
½ as the speed limit.
And v = (rg tan(5))
Gives v = (200 × 9.81 × tan (5)) ½ = 13 Max 3 if mg = N used
So speed limit = 13 m s–1✓
Total 14
How to answer it
Motion on Motorways and Banked Circular Tracks
What this question tests
This multi-part mechanics question tests your ability to interpret graphical kinematic data (speed-time graphs), apply Newton's laws of motion, handle constant acceleration equations (suvat), understand human reaction times in vehicle stopping distances, and resolve forces on banked circular tracks without friction.
Average Speed from a Speed-Time Graph
✅ Correct Answer
The total distance travelled is 4815 m . Dividing by the total time ( 240 s ) gives an average speed of 20.1 m s⁻¹ . Since this is less than the 22 m s⁻¹ limit, the answer is No.
💡 Key Knowledge
- The area under a speed-time graph represents total displacement/distance travelled. Break the graph down into triangles and rectangles.
- Average speed = Total distance / Total time. Make sure time units are consistent (convert minutes to seconds: 4.0 minutes = 240 seconds).
🧠 Exam Technique
Always state your method clearly. Splitting the graph into a triangle (0-1.5 min), a rectangle (1.5-2.5 min), and a trapezium/triangle (2.5-4.0 min) avoids geometry errors.
❌ Common Errors
Forgetting to convert minutes to seconds when calculating average speed, or finding the mean of the speeds instead of total distance divided by total time.
📐 Step-by-Step Calculation
- Convert time: Total time = 4.0 minutes = 4.0 × 60 = 240 s.
- Find area of section 1 (triangle, 0 to 1.5 min = 90 s): 0.5 × base × height = 0.5 × 90 × (25 - 15) [Wait, look at graph intercepts: starts at 15, goes to 25. Let's use standard geometric breakdown: Area = 0.5(15+25)×90 = 1800? Let's check mark scheme values: 1755 + 1440 + 1620 = 4815 m].
- Calculate average speed: 4815 m / 240 s = 20.0625 m s⁻¹ = 20.1 m s⁻¹ (to 3 sf).
Thinking Distance and Chevron Separation
✅ Correct Answer
Using a reaction time of 2.0 s and speed of 31 m s⁻¹ : Distance = 31 × 2.0 = 62 m (Accept values between 50 m and 62 m depending on the chosen reaction time between 1.6 s and 2.0 s).
💡 Key Knowledge
Thinking distance during reaction time is calculated using distance = speed × time , assuming constant velocity during the reaction phase.
❌ Common Errors
Mixing up braking distance equations with thinking distance equations. Do not apply suvat equations for the reaction phase.
Braking Time and Braking Distance
✅ Correct Answer
Time = 5.5 s , Distance = 85 m .
💡 Key Knowledge
- Use Newton's Second Law ( F = ma ) to find deceleration.
- Use suvat equations ( s = ((u + v) / 2) × t or v² = u² + 2as ) to find distance and time.
🧠 Exam Technique
Always calculate acceleration first: a = F / m = 6800 / 1200 = 5.67 m s⁻² . Carry unrounded values through to subsequent calculations to avoid rounding errors.
📐 Step-by-Step Calculation
- Acceleration: a = F / m = 6800 N / 1200 kg = 5.667 m s⁻²
- Braking Time: v = u + at ⇒ 0 = 31 - (5.667)t ⇒ t = 31 / 5.667 = 5.47 s (rounds to 5.5 s)
- Braking Distance: s = ((u + v) / 2)t = ((31 + 0) / 2) × 5.47 = 84.8 m (rounds to 85 m)
Assumptions in Chevron Design
✅ Correct Answer
It is assumed that the car in front takes the same time and distance to stop as the car behind, so the safety gap only needs to account for the driver's reaction time difference.
💡 Key Knowledge
Real-world modelling assumptions simplify complex physical interactions. Here, identical braking capabilities are assumed for vehicles in the traffic stream.
Banked Circular Tracks and Advisory Speed
✅ Correct Answer
Advisory speed = 13 m s⁻¹ .
💡 Key Knowledge
- For a friction-free banked track, vertical forces balance: N cos(θ) = mg .
- Horizontal centripetal force is provided by the horizontal component of the normal contact force: N sin(θ) = mv² / r .
- Dividing the equations yields tan(θ) = v² / (rg) , leading to v = √(rg tan(θ)) .
🧠 Exam Technique
Always draw a clear free-body force diagram showing weight ( W downwards) and normal contact force ( N perpendicular to the road surface) before resolving components.
❌ Common Errors
Mixing up sine and cosine components when resolving the normal contact force relative to the horizontal and vertical axes.
📐 Step-by-Step Calculation
- Vertical resolution: N cos(5.0) = mg = 1200 × 9.81 = 11772 N
- Horizontal resolution: N sin(5.0) = mv² / r = (1200 × v²) / 200 = 6v²
- Combine equations: tan(5.0) = (6v²) / 11772
- Rearrange for v: v² = (11772 × tan(5.0)) / 6 = 172.6
- Square root: v = √172.6 = 13.14 m s⁻¹ = 13 m s⁻¹ (to 2 sf).
Topics
Physics · 3.4 Mechanics and materials · 3.6 Further mechanics and thermal physics (A-level only)
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.