Phys111-3sol

Phys111-3sol - Anderson, Matthew – Homework 3 – Due:...

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Unformatted text preview: Anderson, Matthew – 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 = 1 m, y = 2 . 5 m, and has velocity ~v o = (6 m / s) ˆ ı + (- 4 . 5 m / s) ˆ . The acceleration is given by ~a = (2 m / s 2 ) ˆ ı + (5 m / s 2 ) ˆ . What is the x component of velocity after 3 . 5 s? Correct answer: 13 m / s. Explanation: Let : a x = 2 m / s 2 , v xo = 6 m / s , and t = 3 . 5 s . After 3 . 5 s, ~v x = ~v xo + ~a x t = (6 m / s) ˆ ı + (2 m / s 2 ) ˆ ı (3 . 5 s) = (13 m / s) ˆ ı . 002 (part 2 of 3) 10 points What is the y component of velocity after 3 . 5 s? Correct answer: 13 m / s. Explanation: Let : a y = 5 m / s 2 and v yo =- 4 . 5 m / s . ~v y = ~v yo + ~a y t = (- 4 . 5 m / s) ˆ + (5 m / s 2 ) ˆ (3 . 5 s) = (13 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 3 . 5 s? Correct answer: 38 . 4051 m. Explanation: Let : d o = (1 m , 2 . 5 m) , v o = (6 m / s ,- 4 . 5 m / s) , and a = (2 m / s 2 , 5 m / s 2 ) . From the equation of motion, ~ d = ~ d o + ~v o t + 1 2 at 2 = h (1 m) ˆ ı + (2 . 5 m) ˆ i + [(6 m / s) ˆ ı + (- 4 . 5 m / s) ˆ ] (3 . 5 s) + 1 2 h (2 m / s 2 ) ˆ ı + (5 m / s 2 ) ˆ i (3 . 5 s) 2 = (34 . 25 m) ˆ ı + (17 . 375 m) ˆ , so | ~ d | = q d 2 x + d 2 y = q (34 . 25 m) 2 + (17 . 375 m) 2 = 38 . 4051 m ....
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This note was uploaded on 10/21/2010 for the course PHYS 003 taught by Professor Janow during the Fall '10 term at NJIT.

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Phys111-3sol - Anderson, Matthew – Homework 3 – Due:...

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