problem02_64 solution

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
2.64: Taking the start of the race as the origin, runner A's speed at the end of 30 m can be found from: s m 80 . 9 s m 96 s m 96 ) m 30 )( s m 6 . 1 ( 2 0 ) ( 2 2 2 A 2 2 2 0 A 2 0A 2 A = = = + = - + = v x x a v v A’s time to cover the first 30 m is thus: s 13 . 6 s m 6 . 1 s m 80 . 9 2 A 0A A = = - = a v v t and A’s total time for the race is: s 8 . 38 s m 80 . 9 m ) 30 350 ( s 13 6 = - + . B’s speed at the end of 30 m is found from: s m 95 . 10 s m 120 s m 120 ) m 30 )( s m 0 . 2 ( 2 0 ) ( 2 2 2 B 2 2 2 0 B 2 0B 2 B = = = + = - + = v x x a v v B’s time for the first 30 m is thus
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s 48 . 5 s m . 2 s m 95 . 10 2 B 0B B = =-= a v v t and B's total time for the race is: s 7 . 34 s m 95 . 10 m ) 30 350 ( s 48 5 =-+ . B can thus nap for s 1 . 4 7 . 34 8 . 38 =-and still finish at the same time as A....
View Full Document

This document was uploaded on 02/04/2008.

Ask a homework question - tutors are online