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Unformatted text preview: sela (ds39632) – HW03 – janow – (11196) 1 This printout should have 10 questions. Multiplechoice 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 = 6 m, y = 3 . 5 m, and has velocity vectorv o = (8 m / s) ˆ ı +( 3 m / s) ˆ . The acceleration is given by vectora = (7 m / s 2 ) ˆ ı + (1 . 5 m / s 2 ) ˆ . What is the x component of velocity after 4 . 5 s? Correct answer: 39 . 5 m / s. Explanation: Let : a x = 7 m / s 2 , v xo = 8 m / s , and t = 4 . 5 s . After 4 . 5 s, vectorv x = vectorv xo + vectora x t = (8 m / s) ˆ ı + (7 m / s 2 ) ˆ ı (4 . 5 s) = (39 . 5 m / s) ˆ ı . 002 (part 2 of 3) 10.0 points What is the y component of velocity after 4 . 5 s? Correct answer: 3 . 75 m / s. Explanation: Let : a y = 1 . 5 m / s 2 and v yo = 3 m / s . vectorv y = vectorv yo + vectora y t = ( 3 m / s) ˆ + (1 . 5 m / s 2 ) ˆ (4 . 5 s) = (3 . 75 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 4 . 5 s? Correct answer: 112 . 994 m. Explanation: Let : d o = (6 m , 3 . 5 m) , v o = (8 m / s , 3 m / s) , and a = (7 m / s 2 , 1 . 5 m / s 2 ) . From the equation of motion, vector d = vector d o + vectorv o t + 1 2 a t 2 = bracketleftBig (6 m) ˆ ı + (3 . 5 m) ˆ bracketrightBig + [(8 m / s) ˆ ı + ( 3 m / s) ˆ ] (4 . 5 s) + 1 2 bracketleftBig (7 m / s 2 ) ˆ ı + (1 . 5 m / s 2 ) ˆ bracketrightBig (4 . 5 s)...
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This note was uploaded on 03/09/2012 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro

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