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WileyPLUS × Chegg.com -> C fi Dedugen.wileyplus.com/edugen/student/mainfr.uni Apps Sign in to Office 365 CS2113 - Programm... myLSU - Single Sign....Q Anatomy Ch. 3 flash....Uber WileyPLUS: MyWiley PLUS| Help | Contact Us | Log Out WileyPLUS Halliday, Fundamentals of physics, 10e Halliday, Fundamentals of Physics, 10e Physics (1202/2110/2112/2113) Home Read, Study & Practice Assignment GradebookORION AssignmentOpen Assignment FULL SCREEN PRINTER VERSION BACK NEXT ASSIGNMENT RESOURCES Question 18 Hw 03 (S16) A cat rides a merry-go-round turning with uniform circular motion. At time t1 = 2.20 s, the cat's velocity is (2.90 m/s) + (2.70 m/s) J , measured on a horizontal xy coordinate system. At t2 = 5.60 s, its velocity is (-2.90 m/s) i + (-2.70) j . what are (a) the magnitude of the cat's centripetal acceleration and (b) the magnitude of the cat's average acceleration during the time interval t2 - t1, which is less than a period of the motion? (a) Number (b) Number Question 5 Question 8 Unit Unit SHOW HINT Question Attempts: 0 of 8 used SAVE FOR LATER SUBMIT ANSWER uestion 18 Copyright © 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved JENG 11:09 PM 2/14/2016

Explanation / Answer

As we see form the Final & Initial Velcocity coordinates, It makes 1/2 rev in 2 seconds. The magnitude of the tangential velocity remains the same, so there is no angular acceleration.

= (1/2 rev)/(3.4 sec) * (2 rad/rev)
= /3.4 rad/sec
r = v/
a_centripetal = r *^2 = (v/)***2 = v*
a_centripetal = sqrt(2.9^2 + 2.7^2) */3.4 = 3.96*/3.4
a_centripetal = 3.66 m/s**2

(b)
a_avg = v/t
a_avg = (3.96- (-3.96))/(5.6 - 2.2)
a_avg = 2.33 m/s^2

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