UpStudy Free Solution:

To determine who is closer to the visitor center, we need to calculate the final distances of Mikayla and Paula from the visitor center using vector addition and trigonometry.

Mikayla's Path

1. First Leg:

- Hikes 7 miles east.

- Coordinates after this leg: \(( 7, 0) \).

2. Second Leg:

- Turns 40° north of east and hikes 5.5 miles.

- We need to break this into x (east) and y (north) components:

\[x_ 2 = 5.5 \cos ( 40^ \circ ) \]

\[y_ 2 = 5.5 \sin ( 40^ \circ ) \]

- Calculations:

\[x_ 2 \approx 5.5 \times 0.766 \approx 4.213\]

\[y_ 2 \approx 5.5 \times 0.643 \approx 3.537\]

- Coordinates after the second leg:

\[( 7 + 4.213, 0 + 3.537) = ( 11.213, 3.537) \]

3. Distance from the visitor center:

\[d_ M = \sqrt { ( 11.213) ^ 2 + ( 3.537) ^ 2} \]

\[d_ M \approx \sqrt { 125.735 + 12.508} \approx \sqrt { 138.243} \approx 11.76 \text { miles} \]

Paula's Path

1. First Leg:

- Hikes 7 miles west.

- Coordinates after this leg: \(( - 7, 0) \).

2. Second Leg:

- Turns 35° south of west and hikes 5.5 miles.

- We need to break this into x (west) and y (south) components:

\[x_ 2 = 5.5 \cos ( 35^ \circ ) \]

\[y_ 2 = - 5.5 \sin ( 35^ \circ ) \]

- Calculations:

\[x_ 2 \approx 5.5 \times 0.819 \approx 4.505\]

\[y_ 2 \approx - 5.5 \times 0.574 \approx - 3.157\]

- Coordinates after the second leg:

\[( - 7 - 4.505, 0 - 3.157) = ( - 11.505, - 3.157) \]

3. Distance from the visitor center:

\[d_ P = \sqrt { ( - 11.505) ^ 2 + ( - 3.157) ^ 2} \]

\[d_ P \approx \sqrt { 132.538 + 9.967} \approx \sqrt { 142.505} \approx 11.93 \text { miles} \]

Conclusion

Mikayla's distance from the visitor center is approximately 11.76 miles, while Paula's distance is approximately 11.93 miles. Therefore, Mikayla is closer to the visitor center.

Justification:

Using vector addition and trigonometry, the calculated distances show that Mikayla's final position is closer to the visitor center compared to Paula's.

Supplemental Knowledge

Understanding vector addition and trigonometry is essential for solving problems involving distances and directions. Here are some key concepts:

1. Vector Components:

- Any vector can be broken down into its horizontal (x) and vertical (y) components using trigonometric functions.

- For a vector with magnitude \(r\) making an angle \(\theta \) with the horizontal axis:

\[x = r \cos ( \theta ) \]

\[y = r \sin ( \theta ) \]

2. Pythagorean Theorem:

- To find the resultant distance from the origin (or any point), use the Pythagorean theorem:

\[d = \sqrt { x^ 2 + y^ 2} \]

3. Example Problem:

- Suppose a person walks 8 miles north and then 6 miles east. To find their final distance from the starting point:

- First, break down each leg into components:

- North leg: \(x_ 1 = 0, y_ 1 = 8\)

- East leg: \(x_ 2 = 6, y_ 2 = 0\)

- Sum the components to get total displacement:

- Total \(x = x_ 1 + x_ 2 = 0 + 6 = 6\)

- Total \(y = y_ 1 + y_ 2 = 8 + 0 = 8\)

- Use Pythagorean theorem to find distance:

\[d = \sqrt { 6^ 2 + 8^ 2} = \sqrt { 36 + 64} = \sqrt { 100} = 10\]

This method can be applied to more complex paths involving angles.

Mastering vector addition and trigonometry will drastically expand your problem-solving skills in mathematics and physics. At UpStudy, our detailed explanations make even complex topics easy to comprehend.
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