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problem02_64 solution

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

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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
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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....
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