Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems: 14b

Answer

a avg = 1.58 $m/s^2$

Work Step by Step

The magnitude of the acceleration is the magnitude of the acceleration vector. Since the vector was $ a = -(6/4)i + 0j + (2/4)k $ we can find the magnitude as $\sqrt (i^2 + j^2 +k ^2) $ This gives us $ \sqrt ((6/4)^2 + 0^2 + (2/4)^2) = \sqrt(2.25 + 0 + 0.25) = 1.58 $
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.