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Unformatted text preview: practice 03 – GADHIA, TEJAS – Due: Jan 31 2007, 10:00 pm 1 Question 1 part 1 of 1 10 points Initially (at time t = 0) a particle is moving vertically at 7 . 9 m / s and and horizontally at 0 m / s. The particle accelerates horizontally at 2 . 1 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . At what time will the particle be traveling at 54 ◦ with respect to the horizontal? Correct answer: 0 . 622518 s (tolerance ± 1 %). Explanation: Basic Concept Kinematic Equation v = v o + a t . Solution The vertical velocity is v y t = v y − g t . The horizontal velocity is v x t = v y + a t = a t . v x t v y t v t 54 ◦ The vertical component is the opposite side and the horizontal component is the adjacent side to the angle, so tan θ = v y t v x t = v y − g t a t a t tan θ = v y − g t a t tan θ + g t = v y t = v y a tan θ + g = (7 . 9 m / s) (2 . 1 m / s 2 ) tan(54 ◦ ) + (9 . 8 m / s 2 ) = . 622518 s . Question 2 part 1 of 1 10 points A particle moving at a velocity of 8 . 1 m / s in the positive x direction is given an accel eration of 7 . 8 m / s 2 in the positive y direction for 8 s. What is the final speed of the particle? Correct answer: 62 . 9235 m / s (tolerance ± 1 %). Explanation: Let : v xf = v xi = 8 . 1 m / s . a y = 7 . 8 m / s 2 , and t = 8 s . The vertical velocity undergoes constant ac celeration: v yf = v yi + a t = 0 + (7 . 8 m / s 2 )(8 s) = 62 . 4 m / s . Thus v f = radicalBig v 2 xf + v 2 yf = radicalBig (8 . 1 m / s) 2 + (62 . 4 m / s) 2 = 62 . 9235 m / s . Question 3 part 1 of 3 10 points A particle moves in the xy plane with con stant acceleration. At time zero, the particle is at x = 2 m, y = 5 m, and has velocity vectorv o = (7 m / s) ˆ ı + ( − 4 . 5 m / s) ˆ . The accelera tion is given by vectora = (3 m / s 2 ) ˆ ı + (1 m / s 2 ) ˆ . What is the x component of velocity after 1 . 5 s? Correct answer: 11 . 5 m / s (tolerance ± 1 %). Explanation: practice 03 – GADHIA, TEJAS – Due: Jan 31 2007, 10:00 pm 2 Let : a x = 3 m / s 2 , v xo = 7 m / s , and t = 1 . 5 s . After 1 . 5 s, vectorv x = vectorv xo + vectora x t = (7 m / s) ˆ ı + (3 m / s 2 ) ˆ ı (1 . 5 s) = (11 . 5 m / s) ˆ ı . Question 4 part 2 of 3 10 points What is the y component of velocity after 1 . 5 s? Correct answer: − 3 m / s (tolerance ± 1 %). Explanation: Let : a y = 1 m / s 2 and v yo = − 4 . 5 m / s . vectorv y = vectorv yo + vectora y t = ( − 4 . 5 m / s) ˆ + (1 m / s 2 ) ˆ (1 . 5 s) = ( − 3 m / s) ˆ . Question 5 part 3 of 3 10 points What is the magnitude of the displacement from the origin ( x = 0 m, y = 0 m) after 1 . 5 s? Correct answer: 15 . 8873 m (tolerance ± 1 %). Explanation: Let : d o = (2 m , 5 m) , v o = (7 m / s , − 4 . 5 m / s) , and a = (3 m / s 2 , 1 m / s 2 ) ....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Acceleration

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