solutions1B

solutions1B - Magnitude = 2 2 125 4 . 91 = 155 N (B)...

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PHYS 152 EXAM 1B SOLUTIONS Fall Semester 2008 PROBLEM 1 (A) 3 2 1 . 0 2 . 1 ) 3 . 0 2 . 1 ( t t dt t v ; s v s v a av 4 ) 0 ( ) 4 ( = –0.4 m/s 2 (B) Velocity has extrema when 2 3 . 0 2 . 1 ) ( t t a = 0. Solve for t : t = ±2 sec. Maximum at t = +2 sec. Plug into v ( t ) to get v max = 1.6 m/s (C) 4 0 4 2 4 0 4 1 . 0 6 . 0 ) ( t t dt t v x = 3.2 m PROBLEM 2 (A) Let + x be north, then x police = x speeder 2 2 1 0 at t v = v speeder t . 0 + ½(2.5) t 2 = 30 t Solve: t = 24 sec (B) x = ½(2.5)(24) 2 = 720 m (C) v = v 0 + at = 0 + (2.5)(24) = 60 m/s PROBLEM 3 (A) Σ F x = 130sin20° – 80cos60° – 120cos37° = –91.4 N Σ F y = 80sin60° – 120sin37° + 130cos20° = –125 N
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Unformatted text preview: Magnitude = 2 2 125 4 . 91 = 155 N (B) Direction 4 . 91 125 tan 1 = 54 from x axis or 234 from + x axis (C) m = w / g = 5 kg, so a = F / m = 31 m/s 2 PROBLEM 4 v = 20 m/s; v x = 6 m/s; v y = 2 2 6 20 = 19.1 m/s (A) 1 cos v v x = 72.5 (B) 2 2 1 gt t v y y y 0 = 45 + 19.1 t 4.9 t 2 use quadratic formula to get t = 5.55 sec (C) x = v x t = (6 m/s)(5.55 s) = 33.3 m...
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This note was uploaded on 01/13/2011 for the course BUS A202 taught by Professor Tindall during the Spring '10 term at IUPUI.

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