Math Problem Statement

A person leaves home and walks 2 miles west, then 5 miles southwest.

How far from home is she?

miles

In what direction must she walk to head directly home?

degrees North of East

Solution

To solve this problem, we can break the person's movement into two segments: the first segment (2 miles west) and the second segment (5 miles southwest). We can then use the Pythagorean theorem to find the straight-line distance from her final position to the starting point (home), and trigonometry to determine the direction she must head to return home.

Step 1: Break down the movements

  • The first movement is 2 miles west. This is simply a horizontal displacement of 2 miles to the left (west).
  • The second movement is 5 miles southwest. Southwest is at a 45-degree angle south of west, which means the movement forms a 45-degree angle with the horizontal axis. Using trigonometry, we can break this movement into two components:
    • West component: 5×cos(45)5 \times \cos(45^\circ)
    • South component: 5×sin(45)5 \times \sin(45^\circ)

Step 2: Calculate total displacement

  • West component of the second movement: 5×cos(45)=5×22=5×0.7071=3.535 miles west5 \times \cos(45^\circ) = 5 \times \frac{\sqrt{2}}{2} = 5 \times 0.7071 = 3.535 \text{ miles west}
  • South component of the second movement: 5×sin(45)=5×22=5×0.7071=3.535 miles south5 \times \sin(45^\circ) = 5 \times \frac{\sqrt{2}}{2} = 5 \times 0.7071 = 3.535 \text{ miles south}

Step 3: Total displacement in the west and south directions

  • Total westward displacement: 2+3.535=5.5352 + 3.535 = 5.535 miles to the west
  • Total southward displacement: 3.535 miles to the south

Step 4: Find the straight-line distance from home

To find the straight-line distance from her home (her total displacement), we use the Pythagorean theorem: d=(5.535)2+(3.535)2d = \sqrt{(5.535)^2 + (3.535)^2} d=30.646+12.5=43.1466.57 milesd = \sqrt{30.646 + 12.5} = \sqrt{43.146} \approx 6.57 \text{ miles}

Step 5: Find the direction to head home

To find the direction, we use the inverse tangent function (tan1\tan^{-1}) to calculate the angle north of east: θ=tan1(southward displacementwestward displacement)\theta = \tan^{-1}\left(\frac{\text{southward displacement}}{\text{westward displacement}}\right) θ=tan1(3.5355.535)tan1(0.638)32.5\theta = \tan^{-1}\left(\frac{3.535}{5.535}\right) \approx \tan^{-1}(0.638) \approx 32.5^\circ This means she must walk 32.5° east of north to head directly home.

Final Answer:

  • Distance from home: 6.57\approx 6.57 miles
  • Direction: 32.532.5^\circ North of East

Would you like a more detailed explanation of any step?

Related questions:

  1. How would the displacement change if she had walked only 3 miles southwest instead of 5?
  2. What is the formula for calculating the straight-line distance using the Pythagorean theorem?
  3. How can we adjust this solution if the person had walked 5 miles southeast instead of southwest?
  4. How can vector addition help in understanding the displacement?
  5. What would the result be if she had walked 2 miles north first instead of west?

Tip: Always break down the motion into components based on the direction (like north-south or east-west) before applying Pythagoras or trigonometric functions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Pythagorean Theorem
Vector Addition

Formulas

Pythagorean Theorem: d = √(x² + y²)
Cosine and Sine for vector components: x = r * cos(θ), y = r * sin(θ)
Inverse Tangent: θ = tan⁻¹(y / x)

Theorems

Pythagorean Theorem
Trigonometric Functions
Vector Addition

Suitable Grade Level

Grades 9-12