Online Homework Answers 4 (Oct. 5, 2006)

Online Homework Answers 4 (Oct. 5, 2006) - Hood Charles...

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Hood, Charles – Homework 4 – Due: Oct 5 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 3) 10 points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 3 . 5 m, y = 3 m, and has velocity ~v o = (1 m / s) ˆ ı +( - 3 m / s) ˆ  . The acceleration is given by ~a = (1 . 5 m / s 2 ) ˆ ı + (2 m / s 2 ) ˆ  . What is the x component oF velocity aFter 8 . 5 s? Correct answer: 13 . 75 m / s. Explanation: Let : a x = 1 . 5 m / s 2 , v xo = 1 m / s , and t = 8 . 5 s . AFter 8 . 5 s, ~v x = ~v xo + ~a x t = (1 m / s) ˆ ı + (1 . 5 m / s 2 ) ˆ ı (8 . 5 s) = (13 . 75 m / s) ˆ ı . 002 (part 2 oF 3) 10 points What is the y component oF velocity aFter 8 . 5 s? Correct answer: 14 m / s. Explanation: Let : a y = 2 m / s 2 and v yo = - 3 m / s . ~v y = ~v yo + ~a y t = ( - 3 m / s) ˆ + (2 m / s 2 ) ˆ (8 . 5 s) = (14 m / s) ˆ . 003 (part 3 oF 3) 10 points What is the magnitude oF the displacement From the origin ( x = 0 m, y = 0 m) aFter 8 . 5 s? Correct answer: 82 . 8 m. Explanation: Let : d o = (3 . 5 m , 3 m) , v o = (1 m / s , - 3 m / s) , and a = (1 . 5 m / s 2 , 2 m / s 2 ) . ±rom the equation oF motion, ~ d = ~ d o + ~v o t + 1 2 a t 2 = h (3 . 5 m) ˆ ı + (3 m) ˆ i + [(1 m / s) ˆ ı + ( - 3 m / s) ˆ ] (8 . 5 s) + 1 2 h (1 . 5 m /
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Online Homework Answers 4 (Oct. 5, 2006) - Hood Charles...

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