To find the total displacement and direction of the plane, we’ll approach this as a vector problem.

1. Calculate the Total Displacement

The displacement is the straight-line distance from City A to City B. Since the plane flies 1540 km west and 1160 km south, we can treat this as a right triangle problem where:

• The westward component, $x = 1540$ km
• The southward component, $y = 1160$ km

Using the Pythagorean theorem, the total displacement $d$ is: $d = \sqrt{x^2 + y^2} = \sqrt{(1540)^2 + (1160)^2}$

Calculating, $d = \sqrt{2371600 + 1345600} = \sqrt{3717200} \approx 1928 \text{ km}$

So, the total displacement is approximately 1928 km.

2. Calculate the Direction

The direction can be found by calculating the angle $\theta$ south of west using the tangent function: $\tan(\theta) = \frac{y}{x} = \frac{1160}{1540}$

Calculating, $\theta = \arctan\left(\frac{1160}{1540}\right) \approx 37.45^\circ$

Thus, the direction of the plane's displacement is approximately 37.45° south of west.

Summary

• Total Displacement: 1928 km
• Direction: 37.45° south of west

Would you like more details on the steps or have any questions?

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1. How is the Pythagorean theorem applied in displacement problems?
2. What are the differences between displacement and distance?
3. How do you determine direction when dealing with vectors?
4. What is the importance of trigonometric functions in direction calculation?
5. How can vectors be resolved into components for navigation problems?

Tip: In vector problems, always sketch a diagram—it helps visualize components and angle relationships.