Question

U s Help System Announcements PRINTER VERSION BACK NEXT Chapter 02, Problem 22 Consult Interactive Solution 2.22 before beginning this problem. A car is traveling along a straight road at a velocity of +28.8 m/s when its engine cuts out. For the next 2.89 seconds, the car slows down, and its average acceleration is a,. For the next 6.46 seconds, the car slows down further, and its average acceleration is an. The velocity of the car at the end of the 9.35-second period is +15.9 m/s. The ratio of the average acceleration values is a/a, = 1.77. Find the velocity of the car at the end of the initial 2.89-second interval. Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work

Answer 1

Page 20.8m V215- 9m 2.8gree for period o 2.89 see - = W'= u-caut 9x2.89=20.8-V .. Now for 2.89 9.35 time period Valsug = V1 - 9286.96 - Q2 X6-46 = v' 15.9 by 0 40 divide both equ C4 x 2.84, 288-v1 a 646 . v'- 15.9 s 1.77 2.09 20.8-11 N'-189 6.46 -- =) 6.9918 XVI – 12.59 = 20.2 -U! 1:7910_xvla 91.19 * Ve 23.09 m/s