HS3 - Adams Russell Homework 3 Due midnight Inst Opyrchal H...

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Adams, Russell – Homework 3 – Due: Sep 27 2006, midnight – Inst: Opyrchal, H 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A particle moving at a velocity of 2 . 6 m / s in the positive x direction is given an accelera- tion of 2 . 8 m / s 2 in the positive y direction for 3 . 2 s. What is the final speed of the particle? Correct answer: 9 . 32961 m / s. Explanation: Let : v xf = v xi = 2 . 6 m / s . a y = 2 . 8 m / s 2 , and t = 3 . 2 s . The vertical velocity undergoes constant ac- celeration: v yf = v yi + a t = 0 + (2 . 8 m / s 2 )(3 . 2 s) = 8 . 96 m / s . Thus v f = q v 2 xf + v 2 yf = q (2 . 6 m / s) 2 + (8 . 96 m / s) 2 = 9 . 32961 m / s . keywords: 002 (part 1 of 3) 10 points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 4 . 5 m, y = 1 . 5 m, and has velocity ~v o = (5 m / s) ˆ ı + ( - 6 . 5 m / s) ˆ  . The acceleration is given by ~a = (9 m / s 2 ) ˆ ı + (1 . 5 m / s 2 ) ˆ  . What is the x component of velocity after 6 s? Correct answer: 59 m / s. Explanation: Let : a x = 9 m / s 2 , v xo = 5 m / s , and t = 6 s . After 6 s, ~v x = ~v xo + ~a x t = (5 m / s) ˆ ı + (9 m / s 2 ) ˆ ı (6 s) = (59 m / s) ˆ ı . 003 (part 2 of 3) 10 points What is the y component of velocity after 6 s? Correct answer: 2 . 5 m / s. Explanation: Let : a y = 1 . 5 m / s 2 and v yo = - 6 . 5 m / s . ~v y = ~v yo + ~a y t = ( - 6 . 5 m / s) ˆ + (1 . 5 m / s 2 ) ˆ (6 s) = (2 . 5 m / s) ˆ .
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