# HW 3 - Morgan, Mena – Homework 3 – Due: Sep 26 2007,...

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Unformatted text preview: Morgan, Mena – Homework 3 – Due: Sep 26 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 9 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 3) 10 points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 9 m, y = 6 m, and has velocity ~v o = (8 . 5 m / s) ˆ ı + (- 8 m / s) ˆ . The acceleration is given by ~a = (7 . 5 m / s 2 ) ˆ ı + (8 . 5 m / s 2 ) ˆ . What is the x component of velocity after 6 s? Correct answer: 53 . 5 m / s. Explanation: Let : a x = 7 . 5 m / s 2 , v xo = 8 . 5 m / s , and t = 6 s . After 6 s, ~v x = ~v xo + ~a x t = (8 . 5 m / s) ˆ ı + (7 . 5 m / s 2 ) ˆ ı (6 s) = (53 . 5 m / s) ˆ ı . 002 (part 2 of 3) 10 points What is the y component of velocity after 6 s? Correct answer: 43 m / s. Explanation: Let : a y = 8 . 5 m / s 2 and v yo =- 8 m / s . ~v y = ~v yo + ~a y t = (- 8 m / s) ˆ + (8 . 5 m / s 2 ) ˆ (6 s) = (43 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 6 s? Correct answer: 224 . 379 m. Explanation: Let : d o = (9 m , 6 m) , v o = (8 . 5 m / s ,- 8 m / s) , and a = (7 . 5 m / s 2 , 8 . 5 m / s 2 ) . From the equation of motion, ~ d = ~ d o + ~v o t + 1 2 at 2 = h (9 m) ˆ ı + (6 m) ˆ i + [(8 . 5 m / s) ˆ ı + (- 8 m / s) ˆ ] (6 s) + 1 2 h (7 . 5 m / s 2 ) ˆ ı + (8 . 5 m / s 2 ) ˆ i (6 s) 2 = (195 m) ˆ ı + (111 m) ˆ , so | ~ d | = q d 2 x + d 2 y = q (195 m) 2 + (111 m) 2 = 224 . 379 m ....
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## This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas.

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HW 3 - Morgan, Mena – Homework 3 – Due: Sep 26 2007,...

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