31 ms august 31 2006 physics for scientistsengineers

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Unformatted text preview: tarts at x = 0 and swims to x = 50 m. It takes her 38.2 s • Average velocity x2 ! x1 50 m ! 0 m 50 = = m/s = 1.31 m/s "t 38.2 s 38.2 • Average speed vleg 1 = vleg 1 = 1.31 m/s August 31, 2006 Physics for Scientists&Engineers 1 15 August 31, 2006 Physics for Scientists&Engineers 1 16 E x a m p le : S w im m in g la p s ( 3 ) ! Second Leg: • Swimmer starts at x = 50 m and swims to x = 0 m. It takes her 42.5 s • Average velocity vleg 2 = x2 ! x1 0 m ! 50 m !50 = = m/s = !1.18 m/s "t 42.5 s 42.5 E x a m p le : S w im m in g la p s ( 4 ) ! Entire lap: • Swimmer starts at x = 0 m, swims to x = 50 m, and then back to 0. It takes her 38.2 s + 42.5 s = 80.7 s • Average velocity: 0 • D is p la c e m e n t is 0 • C a n a ls o s h o w th is b y ta k in g th e tim e - w e ig h te d a v e r a g e • Average speed vleg 2 = 1.18 m/s v= vleg 1 ! "t1 + vleg 2 ! "t2 "t1 + "t2 = (1.31 m/s)(38.2 s)+( # 1.18 m/s)(42.5 s) =0 38.2 s+42.5 s ! Physics for Scientists&Engineers 1 17 • Average speed: use total distance = 100 m and total time v= !x1lap + !x2lap !t = 100 m = 1.24 m/s 80.7 s • (again same result obtained from weighted average) August 31, 2006 August 31, 2006 Physics for Scientists&Engineers 1 18...
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This note was uploaded on 03/01/2010 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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