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HW-03-solutions

# HW-03-solutions - kuruvila(lk5992 HW-03 gokce(P111S11 This...

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kuruvila (lk5992) – HW-03 – gokce – (P111S11) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points A particle has an initial horizontal velocity of 2 . 8 m / s and an initial upward velocity of 3 . 8 m / s. It is then given a horizontal accelera- tion of 2 . 5 m / s 2 and a downward acceleration of 2 m / s 2 . What is its speed after 3 s? Correct answer: 10 . 5323 m / s. Explanation: BasicConcepts: The direction of the motion depends only on the horizontal and vertical components of the velocity at any moment. Solution: For the horizontal motion, v x = v 0 x + a x t = 2 . 8 m / s + ( 2 . 5 m / s 2 ) (3 s) = 10 . 3 m / s For the vertical motion, v y = v 0 y - a y t = 3 . 8 m / s - ( 2 m / s 2 ) (3 s) = - 2 . 2 m / s A resultant negative velocity v y would indi- cate downward motion. A right triangle is formed by the compo- nents, so v = radicalBig v 2 x + v 2 y = radicalBig (10 . 3 m / s) 2 + ( - 2 . 2 m / s) 2 = 10 . 5323 m / s 002(part2of2)10.0points What is the direction of its velocity at this time with respect to the horizontal? Answer between - 180 and +180 . Correct answer: - 12 . 0568 . Explanation: The vertical component v y is the side oppo- site the angle θ and the horizontal component v x is the side adjacent to the angle, so tan θ = v y v x θ = arctan v y v x = arctan - 2 . 2 m / s 10 . 3 m / s = 10 . 5323 m / s The angle must be in degrees and a positive angle indicates upward motion while a nega- tive angle indicates downward motion. 003(part1of3)10.0points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 2 . 5 m, y = 5 . 5 m, and has velocity vectorv o = (1 m / s) ˆ ı + ( - 5 . 5 m / s) ˆ  . The acceleration is given by vectora = (7 m / s 2 ) ˆ ı + (5 m / s 2 ) ˆ  . What is the x component of velocity after 8 . 5 s? Correct answer: 60 . 5 m / s. Explanation: Let : a x = 7 m / s 2 , v xo = 1 m / s , and t = 8 . 5 s . After 8 . 5 s, vectorv x = vectorv xo + vectora x t = (1 m / s) ˆ ı + (7 m / s 2 ) ˆ ı (8 . 5 s) = (60 . 5 m / s) ˆ ı . 004(part2of3)10.0points What is the y component of velocity after 8 . 5 s? Correct answer: 37 m / s.

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kuruvila (lk5992) – HW-03 – gokce – (P111S11) 2 Explanation: Let : a y = 5 m / s 2 and v yo = - 5 . 5 m / s . vectorv y = vectorv yo + vectora y t = ( - 5 . 5 m / s) ˆ + (5 m / s 2 ) ˆ (8 . 5 s) = (37 m / s) ˆ . 005(part3of3)10.0points What is the magnitude of the displacement from the origin ( x = 0 m, y = 0 m) after 8 . 5 s? Correct answer: 298 . 422 m. Explanation: Let : d o = (2 . 5 m , 5 . 5 m) , v o = (1 m / s , - 5 . 5 m / s) , and a = (7 m / s 2 , 5 m / s 2 ) . From the equation of motion, vector d = vector d o + vectorv o t + 1 2 a t 2 = bracketleftBig (2 . 5 m) ˆ ı + (5 . 5 m) ˆ bracketrightBig + [(1 m / s) ˆ ı + ( - 5 . 5 m / s) ˆ ] (8 . 5 s) + 1 2 bracketleftBig (7 m / s 2 ) ˆ ı + (5 m / s 2 ) ˆ bracketrightBig (8 . 5 s) 2 = (263 . 875 m) ˆ ı + (139 . 375 m) ˆ  , so | vector d | = radicalBig d 2 x + d 2 y = radicalBig (263 . 875 m) 2 + (139 . 375 m) 2 = 298 . 422 m .
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HW-03-solutions - kuruvila(lk5992 HW-03 gokce(P111S11 This...

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