AQA A-Level Physics Paper 1, June 2022: Question 22

1 mark · Medium difficulty · Multiple Choice

Calculate the northward distance travelled by an aeroplane flying at a speed of 150 m s^-1 on a bearing of 60° east of north after one hour.

Practise this question

Question

A multiple choice question numbered 2.2 asking for the northward displacement of an aeroplane flying at 150 m s^-1 at 60° east of north after 1 hour. An accompanying diagram shows a vector pointing up and right at 60° to a vertical North line. Four options are provided: A 270 km, B 470 km, C 510 km, and D 540 km.
Question text

22 An aeroplane flies horizontally at 150 m s−1 along a bearing 60° east of north.

How far north from its starting position is the aeroplane after one hour?

[1 mark]

A 270 km

B 470 km

C 510 km

D 540 km

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option A (270 km).

22 A (AO1) 270 km

How to answer it

Question 22: Aeroplane Vector Resolution

📌 What this question tests

This question assesses your ability to resolve velocity vectors into perpendicular components using trigonometry, convert units of time into seconds, calculate linear displacement (distance = velocity × time), and correctly interpret navigational bearings relative to cardinal directions.

Question 22 (Multiple Choice)

Calculate the northward displacement of an aeroplane flying at 150 m s⁻¹ on a bearing of 60° east of north for one hour.

✅ Correct Answer

A: 270 km

Mark Awarded: 1 / 1 (AO1)

💡 Key Knowledge

  • Bearings: "60° east of north" means the angle is measured clockwise from the North line towards the East.
  • Vector Components: The northward component of velocity uses the cosine function because the angle is given relative to the north axis: v_north = v cos(θ) .
  • Time Conversion: 1 hour = 60 minutes × 60 seconds = 3600 s.

🧠 Exam Technique

  • Always sketch or visualise the right-angled triangle formed by the velocity vector and the coordinate axes.
  • Check whether the angle provided is with respect to the horizontal or vertical axis before choosing between sin and cos .

❌ Common Errors

  • Using sine instead of cosine: Because the 60° angle is measured from the North axis, the adjacent side corresponds to the northward direction ( cos ). Using sin(60°) calculates the eastward component instead.
  • Unit conversion failures: Forgetting to convert hours into seconds, leading to severe undercalculations.

📐 Step-by-Step Calculation

  1. Find the northward velocity component:
    v_north = 150 × cos(60°) = 150 × 0.5 = 75 m s⁻¹
  2. Convert time into seconds:
    t = 1 hour = 3600 s
  3. Calculate northward displacement:
    Distance = velocity × time = 75 × 3600 = 270,000 m
  4. Convert metres to kilometres:
    270,000 m ÷ 1000 = 270 km (matching Option A)

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.