## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 107: 29

#### Answer

The plane must fly at a speed of 1670 km/h (which is equal to 1040 mph) from east to west.

#### Work Step by Step

For the sun to stand still relative to the passengers, the plane must fly at the same speed as the earth's tangential speed at the equator, but in the opposite direction as the earth's rotation. Therefore, the plane must fly from east to west. We can find the angular speed of the earth as it rotates once each day. $\omega = \frac{2\pi~rad}{(24)(3600~s)}$ $\omega = 7.27\times 10^{-5}~rad/s$ $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.4\times 10^6~m)$ $v = 465~m/s$ We can convert the speed to km/h. $v = (465~m/s)(\frac{1~km}{1000~m})(\frac{3600~s}{1~h})$ $v = 1670~km/h$ We can convert the speed to mph. $v = (465~m/s)(\frac{1~mi}{1609~m})(\frac{3600~s}{1~h})$ $v = 1040~mph$ The plane must fly at a speed of 1670 km/h (which is equal to 1040 mph) from east to west.

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