Question

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hikes 2.7 km in a direction of 16° east of south to the scenic overlook.

What is the magnitude of the hiker’s resultant displacement? Round your answer to the nearest tenth.

5.6 km

What is the direction of the hiker’s resultant displacement? Round your answer to nearest whole degree.

77 ° south of west

Answer 1

The hiker's resultant displacement is approximately 6.1 km, at a direction of 116.5° south of west. The provided rounded answers (5.6 km and 77° south of west) are not accurate.

To find the resultant displacement, we can use vector addition. The hiker's displacement consists of two vectors, one going 3.5 km at 55° south of west, and the other going 2.7 km at 16° east of south.

1. Convert the vectors into their horizontal (x) and vertical (y) components:

For the first vector (3.5 km at 55° south of west):

`\[ x_1 = 3.5 \, \text{km} \cdot \cos(55^\circ) \] \[ y_1 = -3.5 \, \text{km} \cdot \sin(55^\circ) \]` negative because it's south

For the second vector (2.7 km at 16° east of south):

`\[ x_2 = 2.7 \, \text{km} \cdot \sin(16^\circ) \] \\ y_2 = -2.7 \, \text{km} \cdot \cos(16^\circ)` (negative because it's south)

2. Add the horizontal and vertical components separately:

`\[ x_{\text{total}} = x_1 + x_2 \] \[ y_{\text{total}} = y_1 + y_2 \]`

3. Find the magnitude of the resultant displacement:

`\[ \text{Resultant Displacement} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} \]`

4. Find the direction:

`\[ \text{Direction} = \tan^(-1)\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) \]`

Now, let's calculate:

`\[ x_1 = 3.5 \, \text{km} \cdot \cos(55^\circ) \approx 1.92 \, \text{km} \]\[ y_1 = -3.5 \, \text{km} \cdot \sin(55^\circ) \approx -2.83 \, \text{km} \]`

`\[ x_2 = 2.7 \, \text{km} \cdot \sin(16^\circ) \approx 0.74 \, \text{km} \]\[ y_2 = -2.7 \, \text{km} \cdot \cos(16^\circ) \approx -2.59 \, \text{km} \]\[ x_{\text{total}} = 1.92 \, \text{km} + 0.74 \, \text{km} \approx 2.66 \, \text{km} \]\[ y_{\text{total}} = -2.83 \, \text{km} - 2.59 \, \text{km} \approx -5.42 \, \text{km} \]`

`\[ \text{Resultant Displacement} = √(2.66^2 + (-5.42)^2) \approx 6.1 \, \text{km} \]\[ \text{Direction} = \tan^(-1)\left((-5.42)/(2.66)\right) \approx -63.5^\circ \]`

However, since the hiker is south of the starting point, we need to add 180° to the direction:

`\[ \text{Direction} = -63.5^\circ + 180^\circ \approx 116.5^\circ \]`

So, the resultant displacement is approximately 6.1 km at a direction of 116.5° south of west. The given answers are rounded values that closely match these calculations.