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Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 26.0 \(\mathrm{m}\) in a direction \(60.0^{\circ}\) west of north. Jane walks 16.0 \(\mathrm{m}\) in a direction \(30.0^{\circ}\) south of west. They then stop and turn to face each other. (a) What is the distance between them? (b) In what direction should Ricardo walk to go directly toward Jane?

Short Answer

Expert verified
Ricardo and Jane are 9.9 m apart. Ricardo should walk 30.1° north of west to reach Jane.

Step by step solution

01

Break Down Directions

To solve this problem, we'll use components of vectors in two directions: North-South (vertical) and East-West (horizontal). Ricardo walks at an angle, so does Jane. Let's decompose their walks into these components.
02

Calculate Ricardo's Components

Ricardo's walk vector is 26.0 m at 60.0° west of north. For the northward (vertical) component: \[ R_y = 26.0 \cos(60.0°) \]For the westward (horizontal) component:\[ R_x = 26.0 \sin(60.0°) \]
03

Evaluate Calculations for Ricardo

Calculate the components using trigonometric functions:\[ R_y = 26.0 \times 0.5 = 13.0 \text{ m North} \]\[ R_x = 26.0 \times \frac{\sqrt{3}}{2} \approx 22.5 \text{ m West} \]
04

Calculate Jane's Components

Jane's walk vector is 16.0 m at 30.0° south of west. For the southward (vertical) component: \[ J_y = 16.0 \sin(30.0°) \] For the westward (horizontal) component:\[ J_x = 16.0 \cos(30.0°) \]
05

Evaluate Calculations for Jane

Calculate the components using trigonometric functions:\[ J_y = 16.0 \times 0.5 = 8.0 \text{ m South} \]\[ J_x = 16.0 \times \frac{\sqrt{3}}{2} \approx 13.9 \text{ m West} \]
06

Determine Relative Position

Ricardo's position relative to the starting point is at 13.0 m north, 22.5 m west.Jane's position relative to the starting point is 8.0 m south, 13.9 m west. We'll subtract Jane's position from Ricardo's for components:Vertical component:\[ 13.0 \text{ m North} - 8.0 \text{ m South} = 5.0 \text{ m} \] Horizontal component:\[ 22.5 \text{ m West} - 13.9 \text{ m West} = 8.6 \text{ m West} \]
07

Calculate Distance Between Them

Use the Pythagorean theorem to find the distance between them:\[ d = \sqrt{(8.6)^2 + (5.0)^2} \]Calculate:\[ d = \sqrt{73.96 + 25} = \sqrt{98.96} \approx 9.9 \text{ m} \]
08

Determine Direction Ricardo Should Walk

Use trigonometry to find the angle relative to due west:\[ \tan(\theta) = \frac{5.0}{8.6} \]Find the angle:\[ \theta = \tan^{-1}\left(\frac{5.0}{8.6}\right) \approx 30.1° \]So, Ricardo should walk \(30.1^{\circ}\) north of west to reach Jane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry
Trigonometry is incredibly useful in breaking down vectors into components. When you have a movement or force that is not straight along an axis, trigonometric functions like sine and cosine help to find the components along the axes.

For example, in our problem, Ricardo and Jane move in directions that form angles with the standard North-South and East-West directions. To find out how far they move north or west, we apply trigonometric functions:
  • The cosine function gives us the component along the adjacent side of the angle (northward for Ricardo).
  • The sine function helps find the component along the opposite side of the angle (westward for Ricardo).
By breaking the movements into components, we simplify calculations and gain a clearer view of each person's movement in the pasture.
Pythagorean theorem
The Pythagorean theorem is a powerhouse for determining distances. In a right triangle, it relates the lengths of the two legs to the hypotenuse. The formula is: \[c^2 = a^2 + b^2 \]

In the context of vector problems, we use this theorem to find the overall distance between two points after we calculate their respective horizontal and vertical components. For Ricardo and Jane, we determined their relative horizontal position (8.6 m) and vertical position (5.0 m).

By plugging these into the formula, we compute the straight-line distance between them:
  • Square the horizontal and vertical components.
  • Add these squares together.
  • Take the square root of the sum to find the hypotenuse or the direct distance.
This simple principle forms the basis of vector analysis, enabling easy calculation of resultant movements or forces in various fields.
Angles and direction
Understanding angles and direction is essential in vector analysis, especially in navigation and physics. Ricardo and Jane's movements are initially specified in terms of angles relative to cardinal directions (north, west, etc.). These angles help define the trajectory and allow us to convert their movements into standard component vectors.

Here's a quick guide to understanding their specified directions:
  • "West of North" means starting from North and turning towards West.
  • "South of West" means starting from West and turning towards South.
To solve problems like these, it's pivotal to clearly define directions and reference angles. Once the angles and directions are understood, they guide the use of trigonometric functions to resolve vectors into components, significantly easing the calculation process.
Problem-solving strategy
Successfully solving vector problems often comes down to a clear strategy. Here is a strategic approach used in the exercise, applicable to similar problems:
  • **Break Down Movements:** Start by clearly understanding and breaking each movement into its directional components.
  • **Calculate Components:** Use trigonometric functions to compute both horizontal and vertical components of each vector.
  • **Find Relative Position:** Determine the relative position by comparing components and calculating differences.
  • **Compute Distances and Directions:** Use mathematical tools like the Pythagorean theorem to find distances, and trigonometry (e.g., tangent function) to find direction angles.
Following this structured approach can efficiently tackle complex vector scenarios, ensuring you accurately find distances and directional paths even in multifaceted problems.

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