Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 49: 20

Answer

(a) My average speed is 25.0 m/s. Equal time is spent at 20.0 m/s and 30.0 m/s. (b) $Average$ $speed=25.0m/s$

Work Step by Step

$10.0 min = (10.0min)(\frac{60s}{1min})=600 s$ $Distance_{1}=(average speed_{1})(time elapsed_{1})=(20.0m/)(600s)=12000m$ $Distance_{2}=(average speed_{2})(time elapsed_{2})=(30.0m/)(600s)=18000m$ $Total$ $time$ $elapsed=600s+600s=1200s$ $Total$ $distance = 12000m+18000m=30000m$ $Average$ $speed=\frac{total distance}{total time elapsed}=\frac{3000m}{1200s}=25.0m/s$
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.