Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Find the error(s) and find the correct solution.
![### Problem:
An airplane is flying on a compass heading (bearing) of 170° at 460 mph. A wind is blowing with the bearing 200° at 80 mph.
#### Tasks:
a. Find the component form of the velocity of the airplane.
b. Find the actual ground speed and direction of the airplane.
### Solution:
#### Part (a):
To find the component form of the velocity of the airplane:
Given the airplane's speed and direction:
- Velocity \( v = 460 \) mph with a bearing of \( 170° \)
We convert this to component form using trigonometric functions:
\[ v = 460 \cdot \langle \cos 170°, \sin 170° \rangle \approx \langle -453.01, 79.88 \rangle \]
#### Part (b):
To find the ground speed and direction:
First, find the wind vector \( w \):
- Wind speed \( w = 80 \) mph with a bearing of \( 200° \)
We convert this to component form:
\[ w = 80 \cdot \langle \cos 200°, \sin 200° \rangle \approx \langle -75.18, -27.36 \rangle \]
Now, add the velocity vectors \( v \) and \( w \):
\[ \text{Velocity vector} = v + w = \langle -453.01, 79.88 \rangle + \langle -75.18, -27.36 \rangle \approx \langle -528.19, 52.52 \rangle \]
Calculate the actual ground speed:
\[ \text{Actual speed} = ||v + w|| = \sqrt{(-528.19)^2 + (52.52)^2} \approx 530.79 \text{ mph} \]
Calculate the actual direction \( \theta \):
\[ \theta = 180° + \tan^{-1} \left( \frac{-528.19}{52.52} \right) \approx 95.68° \]
### Summary:
- The component form of the velocity of the airplane is \(\langle -453.01, 79.88 \rangle\).
- The actual ground speed of the airplane is approximately 530.79 mph.
- The actual direction of the airplane is approximately 95.68](https://content.bartleby.com/qna-images/question/361784a8-d964-492d-9687-5d5c5e192aef/43a44f7a-dbb4-4d33-a153-915ba750d35c/pzuj3vh_processed.png)
Transcribed Image Text:### Problem:
An airplane is flying on a compass heading (bearing) of 170° at 460 mph. A wind is blowing with the bearing 200° at 80 mph.
#### Tasks:
a. Find the component form of the velocity of the airplane.
b. Find the actual ground speed and direction of the airplane.
### Solution:
#### Part (a):
To find the component form of the velocity of the airplane:
Given the airplane's speed and direction:
- Velocity \( v = 460 \) mph with a bearing of \( 170° \)
We convert this to component form using trigonometric functions:
\[ v = 460 \cdot \langle \cos 170°, \sin 170° \rangle \approx \langle -453.01, 79.88 \rangle \]
#### Part (b):
To find the ground speed and direction:
First, find the wind vector \( w \):
- Wind speed \( w = 80 \) mph with a bearing of \( 200° \)
We convert this to component form:
\[ w = 80 \cdot \langle \cos 200°, \sin 200° \rangle \approx \langle -75.18, -27.36 \rangle \]
Now, add the velocity vectors \( v \) and \( w \):
\[ \text{Velocity vector} = v + w = \langle -453.01, 79.88 \rangle + \langle -75.18, -27.36 \rangle \approx \langle -528.19, 52.52 \rangle \]
Calculate the actual ground speed:
\[ \text{Actual speed} = ||v + w|| = \sqrt{(-528.19)^2 + (52.52)^2} \approx 530.79 \text{ mph} \]
Calculate the actual direction \( \theta \):
\[ \theta = 180° + \tan^{-1} \left( \frac{-528.19}{52.52} \right) \approx 95.68° \]
### Summary:
- The component form of the velocity of the airplane is \(\langle -453.01, 79.88 \rangle\).
- The actual ground speed of the airplane is approximately 530.79 mph.
- The actual direction of the airplane is approximately 95.68
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