AQA AS Level Physics Paper 2, November 2021: Question 23

Velocity-Time Graph Distance Calculation

What this question tests

This question assesses your ability to interpret velocity-time graphs, specifically distinguishing between total distance travelled and displacement. It tests the application of area-under-the-graph techniques where velocity becomes negative (reversal of direction).

Question 23

Determining Distance from a Velocity-Time Graph

✅ Correct Answer

B (190 m)

Awarded for correctly calculating the net displacement (or accounting for the direction correctly when finding distance).

💡 Key Knowledge

The area under a velocity-time graph represents displacement.
Distance is a scalar quantity; areas below the time axis (negative velocity) add to the total distance travelled because distance cannot be negative.
Geometric shapes (trapeziums and triangles) make area calculations straightforward.

🧠 Exam Technique

Break the motion down into distinct time intervals. Calculate the positive area (forward motion) and the negative area (backward motion) separately before combining them based on whether the question asks for displacement or distance.

❌ Common Errors

Subtracting the area below the axis instead of adding it (which would give displacement instead of total distance).
Misreading the time coordinates on the grid scale, especially around the deceleration phase and the negative velocity region.

📐 Step-by-Step Calculation

To find the total distance from the start position at t = 40 s , calculate the total area enclosed between the graph line and the time axis, treating all areas as positive:

Interval 1 (Acceleration, 0 to 5 s):
Triangle with base = 5 s, height = 15 m s⁻¹.
Area = 0.5 × 5 × 15 = 37.5 m
Interval 2 (Constant velocity, 5 to 15 s):
Rectangle with width = (15 - 5) = 10 s, height = 15 m s⁻¹.
Area = 10 × 15 = 150.0 m
Interval 3 (Deceleration to zero, 15 to 20 s):
Triangle with base = (20 - 15) = 5 s, height = 15 m s⁻¹.
Area = 0.5 × 5 × 15 = 37.5 m
Interval 4 (Stationary, 20 to 30 s):
Zero velocity, so area = 0 m
Interval 5 (Reverse motion, 30 to 40 s):
Triangle with base = (40 - 30) = 10 s, height = -7 m s⁻¹.
Magnitude of area = 0.5 × 10 × 7 = 35.0 m
Total Distance:
Sum of all areas = 37.5 + 150.0 + 37.5 + 0 + 35.0 = 260 m ... Wait, let's re-verify the graph values and options!
Let's check the peak rectangle: from 5 to 15 is 10 s. Area = 150.
Let's check triangle 0-5: 0.5 × 5 × 15 = 37.5.
Let's check triangle 15-20: 0.5 × 5 × 15 = 37.5.
Total positive area = 37.5 + 150 + 37.5 = 225 m.
Negative triangle from 30 to 40 s: base = 10 s, peak height = -7 m s⁻¹? Let's check grid lines at t=40: velocity is at -7 m s⁻¹.
Area = 0.5 × 10 × 7 = 35 m.
If displacement: 225 - 35 = 190 m (Option B).
If total distance: 225 + 35 = 260 m (Option C).
Examiner note: The question asks for the distance of the vehicle from its start position (which strictly means magnitude of displacement, i.e., how far away it is linearly from the origin, or the exam board tested displacement terminology under the word 'distance'). Following the mark scheme key B, displacement is calculated as 225 - 35 = 190 m . Always watch out whether exam boards use 'distance from start' interchangeably with displacement!

AQA AS Level Physics Paper 2, November 2021: Question 23

Question

Mark scheme

How to answer it