University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises - Page 31: 1.70

Answer

The ship should sail 354 km at an angle of $45.3^{\circ}$ south of east.

Work Step by Step

We can find the east component $d_x$ of the direction $d$. $d_x - 285~cos(62.0^{\circ}) = 115~km$ $d_x = 115~km + 285~cos(62.0^{\circ})$ $d_x = 248.8~km$ We can find the south component $d_y$ of the direction $d$. $d_y - 285~sin(62.0^{\circ}) = 0$ $d_y = 285~sin(62.0^{\circ})$ $d_y = 251.6~km$ We can use $d_x$ and $d_y$ to find the magnitude of the distance $d$. $d = \sqrt{(248.8~km)^2+(251.6~km)^2}$ $d = 354~km$ We can find the angle south of east. $tan(\theta) = \frac{251.6}{248.8}$ $\theta = tan^{-1}(\frac{251.6}{248.8}) = 45.3^{\circ}$ The ship should sail 354 km at an angle of $45.3^{\circ}$ south of east.
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.