37. Go The components of vector A are A, and A, (both positive), and the angle that it makes with respect to the positive x axis is 0. Find the angle 0 if the components of the displacement vector A are (a) A, = 12 m and A, 12 m, (b) A, = 17 m and Ay 12 m, and (c) A, A, = 17 m. = 12 m and =
37. Go The components of vector A are A, and A, (both positive), and the angle that it makes with respect to the positive x axis is 0. Find the angle 0 if the components of the displacement vector A are (a) A, = 12 m and A, 12 m, (b) A, = 17 m and Ay 12 m, and (c) A, A, = 17 m. = 12 m and =
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Question 37: Vector Components and Angles**
**Problem Statement:**
The components of vector **A** are \( A_x \) and \( A_y \) (both positive), and the angle that it makes with respect to the positive x axis is \( \theta \). Find the angle \( \theta \) if the components of the displacement vector **A** are:
**(a)** \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \)
**(b)** \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \)
**(c)** \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \)
**Explanation:**
### Solution Overview:
To find the angle \( \theta \) made with the positive x-axis, you can use the trigonometric relation involving the arctangent function:
\[ \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \]
### Detailed Solutions:
1. **Case (a):**
Given \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \),
\[
\theta = \tan^{-1} \left( \frac{12}{12} \right) = \tan^{-1}(1) = 45^\circ
\]
2. **Case (b):**
Given \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \),
\[
\theta = \tan^{-1} \left( \frac{12}{17} \right)
\]
Using a calculator or trigonometric tables to find the inverse tangent,
\[
\theta \approx 35.5^\circ
\]
3. **Case (c):**
Given \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \),
\[
\theta = \tan^{-1} \left( \frac{17}{12}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2c8c9d8-cb7d-4bb3-b241-ba372e4ba26e%2Fc7d58235-02e7-42f9-8cd0-b08c7571527b%2Fcifvhl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 37: Vector Components and Angles**
**Problem Statement:**
The components of vector **A** are \( A_x \) and \( A_y \) (both positive), and the angle that it makes with respect to the positive x axis is \( \theta \). Find the angle \( \theta \) if the components of the displacement vector **A** are:
**(a)** \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \)
**(b)** \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \)
**(c)** \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \)
**Explanation:**
### Solution Overview:
To find the angle \( \theta \) made with the positive x-axis, you can use the trigonometric relation involving the arctangent function:
\[ \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \]
### Detailed Solutions:
1. **Case (a):**
Given \( A_x = 12 \, \text{m} \) and \( A_y = 12 \, \text{m} \),
\[
\theta = \tan^{-1} \left( \frac{12}{12} \right) = \tan^{-1}(1) = 45^\circ
\]
2. **Case (b):**
Given \( A_x = 17 \, \text{m} \) and \( A_y = 12 \, \text{m} \),
\[
\theta = \tan^{-1} \left( \frac{12}{17} \right)
\]
Using a calculator or trigonometric tables to find the inverse tangent,
\[
\theta \approx 35.5^\circ
\]
3. **Case (c):**
Given \( A_x = 12 \, \text{m} \) and \( A_y = 17 \, \text{m} \),
\[
\theta = \tan^{-1} \left( \frac{17}{12}
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