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.
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.
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
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![### 2. Part 2 - Find the error(s) and solve the problem correctly.
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.
#### Questions:
a. Find the component form of the velocity of the airplane.
b. Find the actual ground speed and direction of the airplane.
#### Answer:
a. The velocity vector \( v \) of the airplane is calculated using its speed and direction:
\[
v = 460 \times \langle \cos 170^\circ, \sin 170^\circ \rangle = \langle -453.01, 79.88 \rangle
\]
b. Next, determine the wind vector \( w \) using its speed and direction:
\[
w = 80 \times \langle \cos 200^\circ, \sin 200^\circ \rangle = \langle -75.18, -27.36 \rangle
\]
Add the airplane's velocity vector \( v \) and the wind vector \( w \) to find the resultant velocity vector \( v + w \):
\[
\text{Velocity vector} = v + w = \langle -528.19, 52.52 \rangle
\]
Calculate the actual speed (ground speed) by finding the magnitude of the resultant velocity vector:
\[
\text{Actual speed} = \| v + w \| = \sqrt{(-528.19)^2 + (52.52)^2} \approx 530.79 \text{ mph}
\]
Determine the actual direction \( \theta \), noting that an angle larger than 180° is used:
\[
\text{Actual direction:} \ \theta = 180^\circ + \tan^{-1} \left( \frac{-528.19}{52.52} \right) = 95.68^\circ
\]
These steps outline the procedure to calculate the component form of the airplane's velocity, and subsequently, the actual ground speed and direction considering wind influence. The process involves vector addition and trigonometric functions to resolve the movement vectors accurately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feeea96f9-d9e2-4e41-9df9-d54882930b66%2F29b4b561-0835-4a14-8229-218562c9fccd%2Fl99osr8_processed.png&w=3840&q=75)
Transcribed Image Text:### 2. Part 2 - Find the error(s) and solve the problem correctly.
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.
#### Questions:
a. Find the component form of the velocity of the airplane.
b. Find the actual ground speed and direction of the airplane.
#### Answer:
a. The velocity vector \( v \) of the airplane is calculated using its speed and direction:
\[
v = 460 \times \langle \cos 170^\circ, \sin 170^\circ \rangle = \langle -453.01, 79.88 \rangle
\]
b. Next, determine the wind vector \( w \) using its speed and direction:
\[
w = 80 \times \langle \cos 200^\circ, \sin 200^\circ \rangle = \langle -75.18, -27.36 \rangle
\]
Add the airplane's velocity vector \( v \) and the wind vector \( w \) to find the resultant velocity vector \( v + w \):
\[
\text{Velocity vector} = v + w = \langle -528.19, 52.52 \rangle
\]
Calculate the actual speed (ground speed) by finding the magnitude of the resultant velocity vector:
\[
\text{Actual speed} = \| v + w \| = \sqrt{(-528.19)^2 + (52.52)^2} \approx 530.79 \text{ mph}
\]
Determine the actual direction \( \theta \), noting that an angle larger than 180° is used:
\[
\text{Actual direction:} \ \theta = 180^\circ + \tan^{-1} \left( \frac{-528.19}{52.52} \right) = 95.68^\circ
\]
These steps outline the procedure to calculate the component form of the airplane's velocity, and subsequently, the actual ground speed and direction considering wind influence. The process involves vector addition and trigonometric functions to resolve the movement vectors accurately.
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