problem03_45

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

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3.45: a) The 0 = x a and β a y 2 - = , so the velocity and the acceleration will be perpendicular only when 0 = y v , which occurs at 0 = t . b) The speed is 0 / , ) 4 ( 2 / 1 2 2 2 = + α = dt dv t β v at 0 = t . (See part d below.) c) r and v are perpendicular when their dot product is 0: 0 2 m) 0 . 30 ( ) 2 ( ) m 0 . 15 ( ) )( ( 3 2 2 2 = + - α = - × - + α α t β βt t βt βt t . Solve this for t : s 208 . 5 2 2 2 2 ) m/s 500 . 0 ( 2 m/s) 2 . 1 ( ) m/s m)(0.500 0 . 30 ( + = ± = - t , and 0 s, at which times the student is at (6.25 m,
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Unformatted text preview: 1.44 m) and (0 m, 15.0 m), respectively. d) At s 208 . 5 = t , the student is 6.41 m from the origin, at an angle of ° 13 from the x-axis. A plot of 2 / 1 2 2 ) ) ( ) ( ( ) ( t y t x t d + = shows the minimum distance of 6.41 m at 5.208 s: e) In the x- y plane the student’s path is:...
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This document was uploaded on 02/04/2008.

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