AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2023: Question 7

Key facts you need

• Gradient of a velocity–time graph = acceleration.
• For deceleration, the gradient is negative, but the deceleration value is usually given as a positive number.
• Thinking distance = speed × reaction time.
• Braking distance can be found from the area under the graph during braking.
• Stopping distance = thinking distance + braking distance.

Examiner insight

• Use values read carefully from the graph. The mark scheme allowed small variation in the final time reading.
• Show your working. Even if one number is slightly off, method marks can still be awarded.
• For explanation questions, make sure you give both the idea and the reason, not just a short statement.

✅ Correct answer

You can use two points from the sloping section of the graph:

a = (26 − 0) ÷ (5.25 − 2.0)

a = 26 ÷ 3.25 = 8

Deceleration = 8 m/s²

📐 Calculation steps

1. Pick two points on the straight braking line, for example (2.0 s, 26 m/s) and (5.25 s, 0 m/s) .
2. Use acceleration = change in velocity ÷ time taken .
3. a = (0 − 26) ÷ (5.25 − 2.0)
4. This gives −8 m/s² , so the deceleration is 8 m/s² .

💡 Key knowledge

• A straight line on a velocity–time graph means constant acceleration or constant deceleration.
• The steeper the slope, the larger the acceleration or deceleration.
• Because the car is slowing down, the gradient is negative.
• In GCSE answers, “deceleration” is normally written as a positive value unless the question specifically asks for acceleration.

❌ Common errors

• Using the flat section from 0 to 2.0 s. That section shows constant velocity, so acceleration there is 0.
• Forgetting the unit m/s² .
• Reading the stopping time inaccurately from the graph.
• Leaving the answer as −8 when the question asks for deceleration.

✅ Correct answer

Thinking distance: s = v × t = 26 × 0.75 = 19.5 m

Braking distance: s = ((5.25 − 2.0) × 26) ÷ 2 = 42.25 m

Stopping distance: 19.5 + 42.25 = 61.75 m

Stopping distance = 61.75 m

Acceptable rounded answers: 61.8 m or 61.7 m depending on rounding.

📐 Step-by-step method

1. Find thinking distance
   The car is still moving at 26 m/s during the reaction time.
   thinking distance = speed × reaction time
   = 26 × 0.75 = 19.5 m
2. Find braking distance
   On a velocity–time graph, distance travelled = area under the graph.
   The braking section is a triangle with:
   • base = 5.25 − 2.0 = 3.25 s
   • height = 26 m/s
   braking distance = ½ × base × height
   = ½ × 3.25 × 26 = 42.25 m
3. Add them
   stopping distance = 19.5 + 42.25 = 61.75 m

💡 Key knowledge

• Thinking distance is the distance travelled while the driver is reacting.
• Braking distance is the distance travelled after the brakes start working.
• Stopping distance is the total of both distances.
• Area under a velocity–time graph gives distance travelled.
• You could also find braking distance using v² − u² = 2as , rearranged to s = (u² − v²) ÷ 2a .

🧠 Exam technique

• Use the 0.75 s reaction time exactly. The mark scheme specifically required this to get the thinking distance marks.
• Clearly separate thinking distance and braking distance in your working.
• If you use the graph area method, label the shape as a triangle. This makes your reasoning clear.
• If your value from part (a) was slightly wrong, you could still get follow-through marks in part (b) if your method was correct.

❌ Common errors

• Using the whole graph area and forgetting to add a separate thinking distance.
• Using 5.25 s as the braking time instead of 5.25 − 2.0 = 3.25 s .
• Forgetting that the braking section is a triangle, so you must divide by 2.
• Mixing up stopping distance and braking distance.
• Not giving the final unit in m .

✅ Correct answers

Any one of these full 2-mark explanations would score:

• The brakes can overheat, so the brakes will not work properly.
• Large deceleration can lead to loss of control because the tyres lose traction / grip.
• The greater the deceleration, the greater the force, and large forces can cause injury.

💡 Key knowledge

• A larger deceleration means the speed changes more quickly.
• A faster change in velocity means a larger force is involved.
• If tyres lose grip, the car may skid and the driver can lose control.
• Very heavy braking can also heat the brakes too much.

🧠 Exam technique

• Give a because statement. Example: “Large deceleration can cause loss of control because the tyres lose grip.”
• For a 2-mark explanation, write two linked points.
• Top answers were specific: they mentioned traction, overheating, or larger force.

❌ Common errors

• Saying only “it could cause a crash” or “it is dangerous” without any physics explanation.
• Giving one short point only, such as “the car may skid”, without explaining that this happens because the tyres lose grip.
• The mark scheme says to ignore vague references to accidents and crashes throughout.

07.1 Model answer

a = (26 − 0) ÷ (5.25 − 2.0) = 8

Deceleration = 8 m/s²

07.2 Model answer

Thinking distance = 26 × 0.75 = 19.5 m

Braking distance = area of triangle = ((5.25 − 2.0) × 26) ÷ 2 = 42.25 m

Stopping distance = 19.5 + 42.25 = 61.75 m

07.3 Model answer

Large decelerations can be dangerous because the tyres may lose grip, which can cause the driver to lose control of the car.