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. a. Find the component form of the velocity of the airplane. b. Find the actual ground speed and direction of the airplane.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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
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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