College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Margaret walks to the store using the following path: 0.710 mi west, 0.350 mi north, 0.270 mi east. Assume north to be along the +y-axis and west to be along the –x-axis.

What is the magnitude of her total displacement?

[Text box for the answer]

This problem involves calculating total displacement by considering movements along the x and y coordinate axes. The movement involves three segments:

1. 0.710 miles to the west (–x direction)
2. 0.350 miles to the north (+y direction)
3. 0.270 miles to the east (+x direction)

The displacement is a vector quantity, representing the shortest distance from the starting point to the final position. To find the magnitude of this displacement, one can use the Pythagorean theorem to sum vector components in the respective directions.

There are no graphs or diagrams accompanying the text.
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Transcribed Image Text:Margaret walks to the store using the following path: 0.710 mi west, 0.350 mi north, 0.270 mi east. Assume north to be along the +y-axis and west to be along the –x-axis. What is the magnitude of her total displacement? [Text box for the answer] This problem involves calculating total displacement by considering movements along the x and y coordinate axes. The movement involves three segments: 1. 0.710 miles to the west (–x direction) 2. 0.350 miles to the north (+y direction) 3. 0.270 miles to the east (+x direction) The displacement is a vector quantity, representing the shortest distance from the starting point to the final position. To find the magnitude of this displacement, one can use the Pythagorean theorem to sum vector components in the respective directions. There are no graphs or diagrams accompanying the text.
Certainly! Below is the transcription and explanation for the educational website:

---

**Text:**

Margaret walks to the store using the following path: 0.710 mi west, 0.350 mi north, 0.270 mi east. Assume north to be along the +y-axis and west to be along the -x-axis.

Find the direction of the displacement vector. Enter the answer as an angle north of west.

____° north of west

**Explanation:**

In this problem, Margaret's path involves three distinct movements:

1. **0.710 miles west**: She moves in the negative x-direction.
2. **0.350 miles north**: She moves in the positive y-direction.
3. **0.270 miles east**: She moves in the positive x-direction.

The task is to calculate the resultant displacement vector's direction, defined as the angle north of west. The process involves summing the total movements in the x-direction (west and east) and the y-direction (north), then using trigonometry to find the vector's angle relative to the west (negative x-axis).

To find the resultant displacement:
- Total x-displacement = -0.710 mi + 0.270 mi = -0.440 mi
- Total y-displacement = 0.350 mi

Using these components, the angle θ (north of west) can be found using the arctangent function:
\[ \theta = \arctan\left(\frac{\text{y-displacement}}{\text{x-displacement}}\right) \]

This angle provides the direction of the net displacement vector relative to the west.

--- 

Note: Ensure students understand vector addition and the application of trigonometric functions to calculate angles in navigation problems.
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Transcribed Image Text:Certainly! Below is the transcription and explanation for the educational website: --- **Text:** Margaret walks to the store using the following path: 0.710 mi west, 0.350 mi north, 0.270 mi east. Assume north to be along the +y-axis and west to be along the -x-axis. Find the direction of the displacement vector. Enter the answer as an angle north of west. ____° north of west **Explanation:** In this problem, Margaret's path involves three distinct movements: 1. **0.710 miles west**: She moves in the negative x-direction. 2. **0.350 miles north**: She moves in the positive y-direction. 3. **0.270 miles east**: She moves in the positive x-direction. The task is to calculate the resultant displacement vector's direction, defined as the angle north of west. The process involves summing the total movements in the x-direction (west and east) and the y-direction (north), then using trigonometry to find the vector's angle relative to the west (negative x-axis). To find the resultant displacement: - Total x-displacement = -0.710 mi + 0.270 mi = -0.440 mi - Total y-displacement = 0.350 mi Using these components, the angle θ (north of west) can be found using the arctangent function: \[ \theta = \arctan\left(\frac{\text{y-displacement}}{\text{x-displacement}}\right) \] This angle provides the direction of the net displacement vector relative to the west. --- Note: Ensure students understand vector addition and the application of trigonometric functions to calculate angles in navigation problems.
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