problem02_09 solution

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.9: a) At 0 , 0 1 1 = = x t , so Eq (2.2) gives . s m 0 . 12 s) 0 . 10 ( s) 0 . 10 )( s m 120 . 0 ( s) 0 . 10 )( s m 4 . 2 ( 3 3 2 2 2 2 av = - = = t x v b) From Eq. (2.3), the instantaneous velocity as a function of time is , ) s m 360 . 0 ( ) s m 80 . 4 ( 3 2 2 3 2 2 t t ct bt v x - = - = so i) , 0 ) 0 ( = x v ii) , s m 0 . 15 s) 0 . 5 )( s m 360 . 0 ( ) s 0 . 5 )( s m 80 . 4 ( ) s 0 . 5 ( 2 3 2 = - = x v and iii) . s m 0 . 12 s) 0 . 10 )( s m 360 . 0 ( ) s 0 . 10 )( s m 80 . 4 ( ) s 0 . 10 ( 2 3 2 = - = x v c) The car is at rest when 0 = x v . Therefore
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Unformatted text preview: ) s m 360 . ( ) s m 80 . 4 ( 2 3 2 =-t t . The only time after = t when the car is at rest is s 3 . 13 3 2 s m 360 . s m 80 . 4 = = t...
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## This document was uploaded on 02/04/2008.

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