Calculus: Early Transcendentals
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ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Problem Statement:**
A plane is flying with an airspeed of 200 mph in the direction of N 50° E and a 40 mi/hr wind is blowing directly from the west. Approximate the true course (direction) and ground speed of the airplane using vectors.
**Solution Explanation:**
To solve this problem, we can use vector addition to combine the airplane's airspeed vector and the wind speed vector to find the resulting ground speed vector.
1. **Airspeed Vector:**
- The airspeed vector has a magnitude of 200 mph.
- The direction is N 50° E. In vector notation, this can be broken into components:
- North component: \( 200 \cos(50°) \)
- East component: \( 200 \sin(50°) \)
2. **Wind Speed Vector:**
- The wind speed vector has a magnitude of 40 mph.
- The wind is blowing directly from the west, which means its direction is to the east.
- Therefore, the wind vector has only an eastward component of 40 mph.
3. **Resultant Ground Speed Vector:**
- Add the north components and the east components of the airspeed and wind vectors to find the resultant vector:
- North component: only affected by the airspeed vector \( 200 \cos(50°) \)
- East component: \( 200 \sin(50°) + 40 \)
4. **Magnitude and Direction of the Resultant Vector:**
- To find the ground speed (magnitude of the resultant vector):
\[
\text{Ground Speed} = \sqrt{(200 \cos(50°))^2 + (200 \sin(50°) + 40)^2}
\]
- To find the direction (angle θ from the north):
\[
\tan(θ) = \frac{\text{East Component}}{\text{North Component}} = \frac{200 \sin(50°) + 40}{200 \cos(50°)}
\]
\[
θ = \tan^{-1}\left(\frac{200 \sin(50°) + 40}{200 \cos(50°)}\right)
\]
Using these calculations, students can find the true course (direction) and ground speed of the airplane. To complete the calculations](https://content.bartleby.com/qna-images/question/5c941f29-ddef-4885-9d8b-99e8246804cb/dd392cae-dc30-4762-b50d-e5eede3ad8b9/cm6mnz.jpeg)
Transcribed Image Text:**Problem Statement:**
A plane is flying with an airspeed of 200 mph in the direction of N 50° E and a 40 mi/hr wind is blowing directly from the west. Approximate the true course (direction) and ground speed of the airplane using vectors.
**Solution Explanation:**
To solve this problem, we can use vector addition to combine the airplane's airspeed vector and the wind speed vector to find the resulting ground speed vector.
1. **Airspeed Vector:**
- The airspeed vector has a magnitude of 200 mph.
- The direction is N 50° E. In vector notation, this can be broken into components:
- North component: \( 200 \cos(50°) \)
- East component: \( 200 \sin(50°) \)
2. **Wind Speed Vector:**
- The wind speed vector has a magnitude of 40 mph.
- The wind is blowing directly from the west, which means its direction is to the east.
- Therefore, the wind vector has only an eastward component of 40 mph.
3. **Resultant Ground Speed Vector:**
- Add the north components and the east components of the airspeed and wind vectors to find the resultant vector:
- North component: only affected by the airspeed vector \( 200 \cos(50°) \)
- East component: \( 200 \sin(50°) + 40 \)
4. **Magnitude and Direction of the Resultant Vector:**
- To find the ground speed (magnitude of the resultant vector):
\[
\text{Ground Speed} = \sqrt{(200 \cos(50°))^2 + (200 \sin(50°) + 40)^2}
\]
- To find the direction (angle θ from the north):
\[
\tan(θ) = \frac{\text{East Component}}{\text{North Component}} = \frac{200 \sin(50°) + 40}{200 \cos(50°)}
\]
\[
θ = \tan^{-1}\left(\frac{200 \sin(50°) + 40}{200 \cos(50°)}\right)
\]
Using these calculations, students can find the true course (direction) and ground speed of the airplane. To complete the calculations
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