solution-hw2

solution-hw2 - lai (yl8859) – HW02 – Ross – (15200) 1...

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Unformatted text preview: lai (yl8859) – HW02 – Ross – (15200) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A man jogs at a speed of 1 . 8 m / s. His dog waits 1 . 9 s and then takes off running at a speed of 2 . 7 m / s to catch the man. How far will they have each traveled when the dog catches up with the man? Correct answer: 10 . 26 m. Explanation: Let : v man = 1 . 8 m / s , Δ t = 1 . 9 s , and v dog = 2 . 7 m / s . The distance that separates the man and the dog when the dog takes off is d = v man Δ t = (1 . 8 m / s) (1 . 9 s) = 3 . 42 m . The dog catches up to the man when x man = x dog d + v man t = v dog t t = d v dog- v man t = 3 . 42 m 2 . 7 m / s- 1 . 8 m / s = 3 . 8 s , so the distance is d = v dog t = (2 . 7 m / s)(3 . 8 s) = 10 . 26 m . 002 (part 1 of 3) 10.0 points The position of a softball tossed vertically upward is described by the equation y = c 1 t- c 2 t 2 , where y...
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This note was uploaded on 02/24/2011 for the course PHYS 152 taught by Professor Button during the Winter '08 term at IUPUI.

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solution-hw2 - lai (yl8859) – HW02 – Ross – (15200) 1...

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