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: 16a

Answer

a = -18i $m/s^2$

Work Step by Step

The velocity is given by $ v = (6t-4t^2)i + 8j $ The acceleration is given by the derivative of the velocity so we need to find: $\frac{d}{dt}((6t-4t^2)i + 8j)$ We find this to be $a = (6 -4(2)t)i + 0j = (6-8t)i $ To find the acceleration at t = 3 we just need to plug in 3 for t in the expression above which gives $a = (6-8(3))i = (6-24)i = -18i $
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.