A hiker starts her trip by walking 25.0 km southeast from her car and sets up her tent for the night.
On the second day, she continues her journey by walking 40.0 km in a direction 60° north of east, where she eventually finds a forest ranger’s tower.
Determine the hiker’s final position relative to her starting point (car).
The correct answer and explanation is :
To solve this problem, we need to determine the hiker’s final position relative to her starting point (the car) after two days of walking. The key here is to break the movements into components along the east and north directions (using Cartesian coordinates), then combine them to find the final position.
Day 1:
The hiker walks 25.0 km southeast. Southeast corresponds to an angle of 45° south of east. This means that the hiker is moving both in the eastward (x) and southward (y) directions. We can break this movement into two components using trigonometry:
- Eastward component (x₁):
( x₁ = 25.0 \, \text{km} \times \cos(45^\circ) = 25.0 \, \text{km} \times 0.7071 \approx 17.68 \, \text{km} ) - Southward component (y₁):
( y₁ = 25.0 \, \text{km} \times \sin(45^\circ) = 25.0 \, \text{km} \times 0.7071 \approx 17.68 \, \text{km} )
Thus, after day 1, the hiker has moved 17.68 km east and 17.68 km south.
Day 2:
On the second day, the hiker walks 40.0 km in a direction 60° north of east. We can break this movement into eastward (x) and northward (y) components:
- Eastward component (x₂):
( x₂ = 40.0 \, \text{km} \times \cos(60^\circ) = 40.0 \, \text{km} \times 0.5 = 20.0 \, \text{km} ) - Northward component (y₂):
( y₂ = 40.0 \, \text{km} \times \sin(60^\circ) = 40.0 \, \text{km} \times 0.866 \approx 34.64 \, \text{km} )
Thus, after day 2, the hiker has moved 20.0 km east and 34.64 km north.
Final Position:
To find the final position, we add up the components from both days:
- Total eastward distance:
( x_{\text{final}} = x₁ + x₂ = 17.68 \, \text{km} + 20.0 \, \text{km} = 37.68 \, \text{km} ) - Total northward distance:
( y_{\text{final}} = y₂ – y₁ = 34.64 \, \text{km} – 17.68 \, \text{km} = 16.96 \, \text{km} )
Thus, the hiker’s final position relative to the car is 37.68 km east and 16.96 km north. The final displacement (straight-line distance from the starting point) can be calculated using the Pythagorean theorem:
[
\text{Displacement} = \sqrt{x_{\text{final}}^2 + y_{\text{final}}^2} = \sqrt{(37.68)^2 + (16.96)^2} \approx \sqrt{1417.1 + 287.4} \approx \sqrt{1704.5} \approx 41.3 \, \text{km}
]
Thus, the hiker’s final position relative to the starting point is approximately 41.3 km at an angle north of east. The angle can be calculated using:
[
\theta = \tan^{-1}\left(\frac{y_{\text{final}}}{x_{\text{final}}}\right) = \tan^{-1}\left(\frac{16.96}{37.68}\right) \approx \tan^{-1}(0.45) \approx 24.4^\circ
]
Hence, the hiker’s final position is 41.3 km at 24.4° north of east relative to her starting point.