Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.1 Angles and Their Measures - 5.1 Assess Your Understanding - Page 388: 111

Answer

$2292\text{ mi/h}$

Work Step by Step

We have: $r=2.39\times10^{5}\,miles$ $\omega=\dfrac{1\text{ revolution}}{27.3\,day}=\dfrac{1\text{ revolution}}{27.3\,day}\times\dfrac{2\pi\,rad}{1\text{revolution}}\times\dfrac{1\text{day}}{24\text{ h}}$ $\omega=0.0095897\text{ rad/h}$ The formula for linear speed $v$ is $v=r\omega$. Using this formula gives: $v=2.39\times10^{5}\text{miles}\times0.0095897\text{ rad/h}$ $v=2292\text{ miles/h}$
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