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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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
Transcribed Image Text:### Vector Displacement Problem
#### Problem Statement:
You walk 53 meters to the north, then turn 60° to your right (this is 60° from the South to North axis and therefore 30° from the horizontal West to East axis), and then you walk another 45 meters. Determine the direction of your net displacement vector. Express your answer as an angle relative to the east axis.
#### Multiple Choice Options:
- 69° N of E
- 57° N of E
- 63° N of E
- 50° N of E
### Explanation:
To solve this problem:
1. **Vector Representation:**
- First displacement (N): \( \vec{A} = 53 \text{ m} \) north.
- Second displacement at 30° from east axis: \( \vec{B} = 45 \text{ m} \) at 30° north of east.
2. **Component Form Representation:**
- For \( \vec{A} \) (North direction):
- \( A_x = 0 \)
- \( A_y = 53 \text{ m} \)
- For \( \vec{B} \) (30° north of east direction):
- \( B_x = 45 \cos(30^\circ) = 45 \times \frac{\sqrt{3}}{2} \approx 38.97 \text{ m} \)
- \( B_y = 45 \sin(30^\circ) = 45 \times \frac{1}{2} = 22.5 \text{ m} \)
3. **Net Displacement Components:**
- \( R_x = A_x + B_x = 0 + 38.97 = 38.97 \text{ m} \)
- \( R_y = A_y + B_y = 53 + 22.5 = 75.5 \text{ m} \)
4. **Magnitude of Net Displacement:**
- \( R = \sqrt{R_x^2 + R_y^2} = \sqrt{(38.97)^2 + (75.5)^2} \approx \sqrt{1518.5 + 5700.25} \approx \sqrt{7218.75} \approx 84.96 \text{ m} \)
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