A helicopter flies for 80 miles. then turns and flies at a heading of 60°for 40 miles. What is the magnitude and direction of the net displacement of the helicopter. Include a drawing

Please find attached the motion of the helicopter (drawn to scale) showing the magnitude and the direction of the net displacement of the helicopter obtained by measurement to be 105 miles and 19° respectively

By calculation, the magnitude of the net displacement is approximately 105.83 miles and the direction of the net displacement is approximately 19.1066°

Explanation:

The parameters of the motion of the helicopter are;

The distance the helicopter flies before turning = 80 miles

The distance the helicopter then turns and flies = 40 miles

The direction the helicopter turns = 60°

Therefore, we have;

The total length the helicopter flies in the original direction = The component of the length of the 40 mile flight that is in the original direction + The distance the helicopter flies before turning

∴ The total length the helicopter flies in the original direction = 40 × cos(60°) + 80 = 20 + 80 = 100

The total length the helicopter flies in the original direction = 100 miles

The component of the 40 mile flight that is perpendicular to the original direction = 40 × sin(60°) = 40 × √3/2

The magnitude of the net displacement = √(100² + (40×√3/2)²) ≈ 105.83

The magnitude of the net displacement ≈ 105.83 miles

The direction of the net displacement = tan⁻¹((40 ×√3/2)/(100)) ≈ 19.1066°

The direction of the net displacement ≈ 19.1066°.

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