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solution_pdf3 - billing (cab4763) HW 3 Opyrchal (11109) 1...

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Unformatted text preview: billing (cab4763) HW 3 Opyrchal (11109) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 4 m, y = 9 m, and has velocity vectorv o = (9 m / s) + (- 5 . 5 m / s) . The acceleration is given by vectora = (1 m / s 2 ) + (3 . 5 m / s 2 ) . What is the x component of velocity after 1 . 5 s? Correct answer: 10 . 5 m / s. Explanation: Let : a x = 1 m / s 2 , v xo = 9 m / s , and t = 1 . 5 s . After 1 . 5 s, vectorv x = vectorv xo + vectora x t = (9 m / s) + (1 m / s 2 ) (1 . 5 s) = (10 . 5 m / s) . 002 (part 2 of 3) 10.0 points What is the y component of velocity after 1 . 5 s? Correct answer:- . 25 m / s. Explanation: Let : a y = 3 . 5 m / s 2 and v yo =- 5 . 5 m / s . vectorv y = vectorv yo + vectora y t = (- 5 . 5 m / s) + (3 . 5 m / s 2 ) (1 . 5 s) = (- . 25 m / s) . 003 (part 3 of 3) 10.0 points What is the magnitude of the displacement from the origin ( x = 0 m, y = 0 m) after 1 . 5 s? Correct answer: 19 . 2058 m. Explanation: Let : d o = (4 m , 9 m) , v o = (9 m / s ,- 5 . 5 m / s) , and a = (1 m / s 2 , 3 . 5 m / s 2 ) . From the equation of motion, vector d = vector d o + vectorv o t + 1 2 a t 2 = bracketleftBig (4 m) + (9 m) bracketrightBig + [(9 m / s) + (- 5 . 5 m / s) ] (1 . 5 s) + 1 2 bracketleftBig (1 m / s 2 ) + (3 . 5 m / s 2 ) bracketrightBig...
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solution_pdf3 - billing (cab4763) HW 3 Opyrchal (11109) 1...

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