# HW 003 - hanna (brh687) – hk 3 – Opyrchal – (11105) 1...

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Unformatted text preview: hanna (brh687) – hk 3 – Opyrchal – (11105) 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 10.0 points A particle moving at a velocity of 3 . 1 m / s in the positive x direction is given an accelera- tion of 0 . 7 m / s 2 in the positive y direction for 8 . 7 s. What is the final speed of the particle? Correct answer: 6 . 8336 m / s. Explanation: Let : v xf = v xi = 3 . 1 m / s . a y = 0 . 7 m / s 2 , and t = 8 . 7 s . The vertical velocity undergoes constant ac- celeration: v yf = v yi + a t = 0 + (0 . 7 m / s 2 )(8 . 7 s) = 6 . 09 m / s . Thus v f = radicalBig v 2 xf + v 2 yf = radicalBig (3 . 1 m / s) 2 + (6 . 09 m / s) 2 = 6 . 8336 m / s . 002 (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 . 5 m, y = 5 . 5 m, and has velocity vectorv o = (8 . 5 m / s) ˆ ı + ( − 6 m / s) ˆ . The acceleration is given by vectora = (7 . 5 m / s 2 ) ˆ ı + (8 . 5 m / s 2 ) ˆ . What is the x component of velocity after 3 . 5 s? Correct answer: 34 . 75 m / s. Explanation: Let : a x = 7 . 5 m / s 2 , v xo = 8 . 5 m / s , and t = 3 . 5 s . After 3 . 5 s, vectorv x = vectorv xo + vectora x t = (8 . 5 m / s) ˆ ı + (7 . 5 m / s 2 ) ˆ ı (3 . 5 s) = (34 . 75 m / s) ˆ ı . 003 (part 2 of 3) 10.0 points What is the y component of velocity after 3 . 5 s? Correct answer: 23 . 75 m / s. Explanation: Let : a y = 8 . 5 m / s 2 and v yo = − 6 m / s ....
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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 003 - hanna (brh687) – hk 3 – Opyrchal – (11105) 1...

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