Problem 75 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The dog must run approximately 31.6 m at an angle 140.4° south of east.

Step 1: 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.

Step 2: 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.

Step 3: Calculate First Displacement Components

The first displacement is 12.0 m east, which translates to \((12.0, 0)\) in our coordinate system.

Step 4: 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: \((-28.0 \sin(50.0^{\circ}), 28.0 \cos(50.0^{\circ}))\).

Step 5: 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.

Step 6: 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 \((x, y)\).

Step 7: 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.

Key Concepts

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:

• 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.

For the dog's movement scenario, the starting point is chosen as the origin (0,0). This means any displacement can be expressed as a change in x (east-west) and y (north-south) components from this origin.

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:

• Eastward (x-direction): 12.0 meters or (12.0, 0)
• Northward (y-direction): 0 meters as there is no northward displacement.

Displacements at angles require the use of trigonometric functions to break them into components. For example, the dog’s 28.0 meters run at an angle of 50° west of north involves:

• 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.

The resulting vector in terms of components could be written as: \((-28.0 \sin(50.0^{\circ}), 28.0 \cos(50.0^{\circ}))\). Breaking vectors this way allows combining them easily to find the overall effect of multiple movements.

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:

• 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).

We simply add these components:\[x_{resultant} = 12.0 + (-28.0 \sin(50.0^{\circ}))\]\[y_{resultant} = 0 + 28.0 \cos(50.0^{\circ})\]Thus, the resultant vector is a combined effect of all individual displacements. By calculating the resultant, we can find the dog’s new position relative to the starting point. Knowing this is vital as it allows us to determine the further steps or corrections required, like finding a new displacement.

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:

• 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: \( -28.0 \sin(50^{\circ}) \).

These trigonometric relations allow precisely calculating the direction and magnitude of movement components. Beyond decomposition, once both components are known for several displacements, trigonometry can again be applied to find angles and directions from the resultant vector. This involves using arctangent (\( \tan^{-1} \)) to calculate direction angles from the ratio of y to x components, crucial for determining paths relative to standard directions such as east or south, as in the dog's third run.