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Unformatted text preview: kuruvila (lk5992) HW03 gokce (P111S11) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points 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: Basic Concepts: 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 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 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 (part 2 of 2) 10.0 points 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 and the horizontal component v x is the side adjacent to the angle, so tan = v y v x = 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 (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 = 2 . 5 m, y = 5 . 5 m, and has velocity vectorv o = (1 m / s) + ( 5 . 5 m / s) . The acceleration is given by vectora = (7 m / s 2 ) + (5 m / s 2 ) . 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) + (7 m / s 2 ) (8 . 5 s) = (60 . 5 m / s) . 004 (part 2 of 3) 10.0 points What is the y component of velocity after 8 . 5 s? Correct answer: 37 m / s. kuruvila (lk5992) HW03 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) + (5 m / s 2 ) (8 . 5 s) = (37 m / s) . 005 (part 3 of 3) 10.0 points 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 ) ....
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This note was uploaded on 02/12/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro

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