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solution-hw4

# solution-hw4 - lai(yl8859 HW04 Ross(15200 This print-out...

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lai (yl8859) – HW04 – Ross – (15200) 1 This print-out should have 8 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 = 5 . 5 m, y = 8 . 5 m, and has velocity vectorv o = (1 . 5 m / s) ˆ ı + ( - 9 m / s) ˆ  . The acceleration is given by vectora = (7 m / s 2 ) ˆ ı + (4 . 5 m / s 2 ) ˆ  . What is the x component of velocity after 5 s? Correct answer: 36 . 5 m / s. Explanation: Let : a x = 7 m / s 2 , v xo = 1 . 5 m / s , and t = 5 s . After 5 s, vectorv x = vectorv xo + vectora x t = (1 . 5 m / s) ˆ ı + (7 m / s 2 ) ˆ ı (5 s) = (36 . 5 m / s) ˆ ı . 002 (part 2 of 3) 10.0 points What is the y component of velocity after 5 s? Correct answer: 13 . 5 m / s. Explanation: Let : a y = 4 . 5 m / s 2 and v yo = - 9 m / s . vectorv y = vectorv yo + vectora y t = ( - 9 m / s) ˆ + (4 . 5 m / s 2 ) ˆ (5 s) = (13 . 5 m / s) ˆ . 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 5 s? Correct answer: 102 . 422 m. Explanation: Let : d o = (5 . 5 m , 8 . 5 m) , v o = (1 . 5 m / s , - 9 m / s) , and a = (7 m / s 2 , 4 . 5 m / s 2 ) . From the equation of motion, vector d = vector d o + vectorv o t + 1 2 a t 2 = bracketleftBig (5 . 5 m) ˆ ı + (8 . 5 m) ˆ bracketrightBig + [(1 . 5 m / s) ˆ ı + ( - 9 m / s) ˆ ] (5 s) + 1 2 bracketleftBig (7 m / s 2 ) ˆ ı + (4 . 5 m / s 2 ) ˆ bracketrightBig (5 s) 2 = (100 . 5 m) ˆ ı + (19 . 75 m) ˆ  , so | vector d | = radicalBig d 2 x + d 2 y = radicalBig (100 . 5 m) 2 + (19 . 75 m) 2 = 102 . 422 m .

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solution-hw4 - lai(yl8859 HW04 Ross(15200 This print-out...

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