irplane heads west at 400 miles per hour, as indicated by the vector , in a 50-mile-per-hour wind that blows in a southeastern direction, as indicated by the vector . What is the magnitude of the plane's velocity (or ground speed)? Round your answer to the nearest per hour. O 350 mph O 366 mph O 435 mph O 450 mph
irplane heads west at 400 miles per hour, as indicated by the vector , in a 50-mile-per-hour wind that blows in a southeastern direction, as indicated by the vector . What is the magnitude of the plane's velocity (or ground speed)? Round your answer to the nearest per hour. O 350 mph O 366 mph O 435 mph O 450 mph
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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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![### Vector Addition to Determine Ground Speed of an Airplane
#### Problem Statement
Look at the diagram below.
[Graph Description]
- The graph is a standard Cartesian coordinate plane.
- The axes are labeled “N” (North) for the positive y-axis, “E” (East) for the positive x-axis, “S” (South) for the negative y-axis, and “W” (West) for the negative x-axis.
- There is a vector \( \mathbf{v} \) pointing horizontally to the left (towards the West) with an arrow, indicating a speed of 400 miles per hour.
- Another vector \( \mathbf{w} \) is pointing in a southeastern direction (at an angle of 45° south of east) with an arrow labeled as 50 mph.
#### Question
An airplane heads west at 400 miles per hour, as indicated by the vector \( \mathbf{v} \), in a 50-mile-per-hour wind that blows in a southeastern direction, as indicated by the vector \( \mathbf{w} \). What is the magnitude of the plane's velocity (or ground speed)? Round your answer to the nearest mile per hour.
#### Multiple Choice Answers
a) 350 mph
b) 366 mph
c) 435 mph
d) 450 mph
---
#### Explanation
To determine the ground speed of the airplane, we need to perform vector addition. The airplane’s speed vector \( \mathbf{v} \) has components:
\[ \mathbf{v} = (-400 \hat{i}, 0 \hat{j}) \text{ mph} \]
The wind’s speed vector \( \mathbf{w} \) can be broken down into its components using trigonometric functions, given the wind blows towards the southeast (45° south of east):
\[ \mathbf{w} = (50 \cos 45^\circ \hat{i}, -50 \sin 45^\circ \hat{j}) = (35.36 \hat{i}, -35.36 \hat{j}) \text{ mph} \]
The resultant vector \( \mathbf{r} = \mathbf{v} + \mathbf{w} \) has components:
\[ \mathbf{r} = (-400 + 35.36 \hat{i}, 0 - 35.36 \hat{j}) = (-364.64 \hat{i}, -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa19b2cf3-e4ba-4c54-87ea-4b78f08e1dba%2Fc7c35f47-4e2e-4845-b2c2-478f90b2ca5a%2Fgg7e73s_processed.png&w=3840&q=75)
Transcribed Image Text:### Vector Addition to Determine Ground Speed of an Airplane
#### Problem Statement
Look at the diagram below.
[Graph Description]
- The graph is a standard Cartesian coordinate plane.
- The axes are labeled “N” (North) for the positive y-axis, “E” (East) for the positive x-axis, “S” (South) for the negative y-axis, and “W” (West) for the negative x-axis.
- There is a vector \( \mathbf{v} \) pointing horizontally to the left (towards the West) with an arrow, indicating a speed of 400 miles per hour.
- Another vector \( \mathbf{w} \) is pointing in a southeastern direction (at an angle of 45° south of east) with an arrow labeled as 50 mph.
#### Question
An airplane heads west at 400 miles per hour, as indicated by the vector \( \mathbf{v} \), in a 50-mile-per-hour wind that blows in a southeastern direction, as indicated by the vector \( \mathbf{w} \). What is the magnitude of the plane's velocity (or ground speed)? Round your answer to the nearest mile per hour.
#### Multiple Choice Answers
a) 350 mph
b) 366 mph
c) 435 mph
d) 450 mph
---
#### Explanation
To determine the ground speed of the airplane, we need to perform vector addition. The airplane’s speed vector \( \mathbf{v} \) has components:
\[ \mathbf{v} = (-400 \hat{i}, 0 \hat{j}) \text{ mph} \]
The wind’s speed vector \( \mathbf{w} \) can be broken down into its components using trigonometric functions, given the wind blows towards the southeast (45° south of east):
\[ \mathbf{w} = (50 \cos 45^\circ \hat{i}, -50 \sin 45^\circ \hat{j}) = (35.36 \hat{i}, -35.36 \hat{j}) \text{ mph} \]
The resultant vector \( \mathbf{r} = \mathbf{v} + \mathbf{w} \) has components:
\[ \mathbf{r} = (-400 + 35.36 \hat{i}, 0 - 35.36 \hat{j}) = (-364.64 \hat{i}, -
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