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: 5e

Answer

12 m

Work Step by Step

To find the displacement between t = 0 and t = 4 we need to find the difference between the position at t=0 and the position at t =4. The position is given by the equation $ x = 3t - 4t^2 + t^3 $ where x is in meters and t is in seconds To find the position at t = 0 s we need to find x(0), or in other words, plug in 0 for t in the equation and teo find the position at t = 4s, we do the same for 4. This gives us : For t = 0: $ x = 3(0) - 4(0)^2 + (0)^3 = 0 -4*(0) + 0 = 0 - 0 + 0 = 0 $ Therefore at t =0, x = 0 For t = 4: $ x = 3(4) - 4(4)^2 + (4)^3 = 12 - 64 + 64 = 12 $ Therefore at t = 4, x = 12 To find the displacement we take x(4) - x(0), which is 12m - 0m = 12m
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.