Question
What is the net displacement of a party of hikers who walked 8 km from camp on a 30-degree course and then 6 km on a 160-degree course?
Answers
GPT-4o mini
To calculate the net displacement of the hikers, we can break down their movements into vector components, using trigonometric functions.
1. **First Leg: 8 km at 30 degrees**
- We need to find the x (east-west) and y (north-south) components.
- The angle of 30 degrees is measured from the positive x-axis (east).
- \( x_1 = 8 \cos(30^\circ) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \approx 6.93 \) km (east)
- \( y_1 = 8 \sin(30^\circ) = 8 \cdot \frac{1}{2} = 4 \) km (north)
2. **Second Leg: 6 km at 160 degrees**
- The angle of 160 degrees is measured from the positive x-axis (east), so we again find the x and y components.
- \( x_2 = 6 \cos(160^\circ) = 6 \cdot \cos(160^\circ) \)
- \( x_2 = 6 \cdot (-\cos(20^\circ)) = 6 \cdot (-0.9397) \approx -5.64 \) km (west)
- \( y_2 = 6 \sin(160^\circ) = 6 \cdot \sin(160^\circ) \)
- \( y_2 = 6 \cdot \sin(20^\circ) \approx 6 \cdot 0.3420 \approx 2.05 \) km (north)
3. **Total Displacement Components:**
- Sum the x-components:
\[ x_{\text{total}} = x_1 + x_2 = 6.93 - 5.64 \approx 1.29 \text{ km (east)} \]
- Sum the y-components:
\[ y_{\text{total}} = y_1 + y_2 = 4 + 2.05 \approx 6.05 \text{ km (north)} \]
4. **Magnitude of Net Displacement:**
- Use the Pythagorean theorem to find the distance:
\[
d = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = \sqrt{(1.29)^2 + (6.05)^2}
\]
\[
d = \sqrt{1.6641 + 36.6025} = \sqrt{38.2666} \approx 6.18 \text{ km}
\]
5. **Direction of Net Displacement:**
- The angle \( \theta \) can be found using the tangent function:
\[
\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = \tan^{-1}\left(\frac{6.05}{1.29}\right) \approx \tan^{-1}(4.69) \approx 78.56^\circ
\]
- This angle is measured from the positive x-axis (east).
**Final Result:**
The net displacement of the hikers is approximately **6.18 km** at an angle of about **78.56 degrees** from the east (or towards the north-east).
1. **First Leg: 8 km at 30 degrees**
- We need to find the x (east-west) and y (north-south) components.
- The angle of 30 degrees is measured from the positive x-axis (east).
- \( x_1 = 8 \cos(30^\circ) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \approx 6.93 \) km (east)
- \( y_1 = 8 \sin(30^\circ) = 8 \cdot \frac{1}{2} = 4 \) km (north)
2. **Second Leg: 6 km at 160 degrees**
- The angle of 160 degrees is measured from the positive x-axis (east), so we again find the x and y components.
- \( x_2 = 6 \cos(160^\circ) = 6 \cdot \cos(160^\circ) \)
- \( x_2 = 6 \cdot (-\cos(20^\circ)) = 6 \cdot (-0.9397) \approx -5.64 \) km (west)
- \( y_2 = 6 \sin(160^\circ) = 6 \cdot \sin(160^\circ) \)
- \( y_2 = 6 \cdot \sin(20^\circ) \approx 6 \cdot 0.3420 \approx 2.05 \) km (north)
3. **Total Displacement Components:**
- Sum the x-components:
\[ x_{\text{total}} = x_1 + x_2 = 6.93 - 5.64 \approx 1.29 \text{ km (east)} \]
- Sum the y-components:
\[ y_{\text{total}} = y_1 + y_2 = 4 + 2.05 \approx 6.05 \text{ km (north)} \]
4. **Magnitude of Net Displacement:**
- Use the Pythagorean theorem to find the distance:
\[
d = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = \sqrt{(1.29)^2 + (6.05)^2}
\]
\[
d = \sqrt{1.6641 + 36.6025} = \sqrt{38.2666} \approx 6.18 \text{ km}
\]
5. **Direction of Net Displacement:**
- The angle \( \theta \) can be found using the tangent function:
\[
\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = \tan^{-1}\left(\frac{6.05}{1.29}\right) \approx \tan^{-1}(4.69) \approx 78.56^\circ
\]
- This angle is measured from the positive x-axis (east).
**Final Result:**
The net displacement of the hikers is approximately **6.18 km** at an angle of about **78.56 degrees** from the east (or towards the north-east).