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: 16b


t = 0.75s

Work Step by Step

In the previous section we found that a = $(6-8t)i $ Therefore, to find when the acceleration is 0, we plug in 0 for a and solve for the t at which this occurs. The j and k components are always 0 so we need only worry about the i component. The magnitude of this vector is given by: $magnitude = \sqrt(i^2 + j^2 +k^2) $ where i, j, and k represent the i, j, and k components respectively Therefore, we have $ magnitude = \sqrt((6-8t)^2) $ which is just 6-8t Therefore we can solve for when a = 0: $ 0 = 6-8t $ so then t = 0.75s
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.