Chapter 1: Problem 75
A dog in an open field runs 12.0 m east and then 28.0 m in a direction 50.0\(^{\circ}\) west of north. In what direction and how far must the dog then run to end up 10.0 m south of her original starting point?
Short Answer
Expert verified
The dog must run approximately 31.6 m at an angle 140.4° south of east.
Step by step solution
01
Understand the Problem
First, break down the problem into its components. The dog runs 12.0 m east and then 28.0 m at an angle of 50.0\(^{\circ}\) west of north. We need to find the third displacement such that the final position is 10.0 m south of the starting point.
02
Coordinate System Setup
Establish a coordinate system with the origin as the dog's starting point. Let east be the positive x-direction and north be the positive y-direction.
03
Calculate First Displacement Components
The first displacement is 12.0 m east, which translates to \((12.0, 0)\) in our coordinate system.
04
Calculate Second Displacement Components
The second displacement is 28.0 m at an angle 50.0\(^{\circ}\) west of north.- Northward component: \(28.0 \cos(50.0^{\circ})\)- Westward component: \(-28.0 \sin(50.0^{\circ})\)Convert this to components in our coordinate system: \((-28.0 \sin(50.0^{\circ}), 28.0 \cos(50.0^{\circ}))\).
05
Calculate Final Position from Displacement
Add the first and second displacement components:\[ (12.0 + (-28.0 \sin(50.0^{\circ})), 0 + 28.0 \cos(50.0^{\circ}))\]Calculate to find the resultant position after these displacements.
06
Determine Required Displacement
The final position should be \((0, -10.0)\). Set up the equation from the resultant of steps 4 and 5 to solve for the third displacement \((x, y)\): \[ (x_{ ext{final}}, y_{ ext{final}}) = (0, -10.0)\]Substitute for \(x_{ ext{resultant}}\) and \(y_{ ext{resultant}}\) from previous step to find \((x, y)\).
07
Solve for Magnitude and Direction
Use \\[d = \sqrt{x^2 + y^2}\\]to find the magnitude of the displacement, and \\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\\]to find the direction south of east.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate System
When solving problems related to vector addition and displacements, establishing a coordinate system is essential. This provides a reference frame for analyzing the movements and positions involved in the problem.
In a typical coordinate system:
By establishing this coordinate system, we can easily track the dog's position after each movement in terms of its x and y coordinates.
In a typical coordinate system:
- The x-axis often represents horizontal movement. In many problems, right or east is taken as the positive x-direction.
- The y-axis often represents vertical movement, where up or north is the positive y-direction.
By establishing this coordinate system, we can easily track the dog's position after each movement in terms of its x and y coordinates.
Displacement Components
Displacement components help break down a movement or shift into manageable parts along the coordinate axes. Think of them as the building blocks of a vector that help us understand complex motions simply.
For instance, when the dog runs 12.0 meters east, its displacement in terms of components is straightforward:
For instance, when the dog runs 12.0 meters east, its displacement in terms of components is straightforward:
- Eastward (x-direction): 12.0 meters or (12.0, 0)
- Northward (y-direction): 0 meters as there is no northward displacement.
- Northward component: 28.0 meters multiplied by cos(50°)
- Westward component: 28.0 meters multiplied by sin(50°) but in the negative x-direction because it’s towards the west.
Resultant Vector
The resultant vector is the net displacement resulting from multiple movements or actions and is a critical concept in vector addition.
To find this, we add the respective components of each vector involved. Let’s consider what happens after the dog’s two movements:
To find this, we add the respective components of each vector involved. Let’s consider what happens after the dog’s two movements:
- The first movement, 12.0 meters east, gives a component of (12.0, 0).
- The second movement, 28.0 meters at 50° west of north, results in components: \(-28.0 \sin(50.0^{\circ})\) meters in x (west) and \(28.0 \cos(50.0^{\circ})\) meters in y (north).
Trigonometric Functions
Trigonometric functions play a pivotal role in decomposing vectors into components when the movements are at an angle. They relate the angles in triangle geometry to the ratios of a right triangle’s sides.
Consider the function involved:
Consider the function involved:
- Cosine (\( \cos \)): Relates the adjacent side of an angle to the hypotenuse; used to find the northward component: \(28.0 \cos(50^{\circ})\).
- Sine (\( \sin \)): Relates the opposite side of an angle to the hypotenuse; useful for finding the westward component: \( -28.0 \sin(50^{\circ}) \).