College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 96: 7


$a). 70m$ $b). 0.57min, 0.43min$

Work Step by Step

a). Base of the triangle = $50 \times cos37^{\circ}=50\times 0.8=40m$ Thus, the other side = $\sqrt (50^{2}+40^{2})=30m$ So, total distance =$40+30=70m$ b). The student walked around the route for 1 min, she spent on 40m side for $=\frac{40}{70}\times1=0.57min$ she spent on 30m side for $=\frac{30}{70}\times1=0.43min$
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.