Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.1 Angles and Their Measure - 6.1 Assess Your Understanding - Page 363: 111

Answer

$2291.94\text{mph}$

Work Step by Step

The distance from the Moon to the Earth is: $r=2.39\cdot10^5$ miles. The Moon takes $27.3$ days($=27.3\cdot24$ hours) to make a revolution, hence its angular velocity is: $\omega=\frac{2\pi}{t}=\frac{2\pi}{27.3\cdot24}=9.589\cdot10^{-3}\frac{\text{rad}}{\text{hr}}.$ Therefore the linear velocity at $40^\circ$: $v=\omega r\\v=9.589\cdot10^{-3}(2.39\cdot10^5)=2291.94\text{mph}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.