Fundamentals of Physics Extended (10th Edition)

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

Chapter 2 - Motion Along a Straight Line - Problems - Page 33: 19


acceleration = $ 20 m/s^2 $

Work Step by Step

Acceleration is given by $\frac{\Delta v}{\Delta t} $ or in other words, the change in velocity over the amount of time that elapsed. In the question, we are given that the time interval is 2.4 seconds so we need to find the change in velocity. The initial speed is 18m/s, so if we designate this to be in the positive x direction then the velocity is + 18m/s. We are told that 2.4s later, the speed was 30m/s in the OPPOSITE direction (the negative x) so therefore the velocity at this point is -30m/s. The $ Delta v = 18m/s - (-30m/s) = 48m/s $ Since this occurs over a period of 2.4seconds then we have 48m/s divided by 2.4s, which gives $20 m/s^2 $
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.