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Practice 03 Solutions - practice 03 GADHIA TEJAS Due 10:00...

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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 0 g t . The horizontal velocity is v x t = v y 0 + 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 0 g t a t a t tan θ = v y 0 g t a t tan θ + g t = v y 0 t = v y 0 a tan θ + g = (7 . 9 m / s) (2 . 1 m / s 2 ) tan(54 ) + (9 . 8 m / s 2 ) = 0 . 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:
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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 ) . From the equation of motion, vector d = vector d o + vectorv o t + 1 2 a t 2 = bracketleftBig (2 m) ˆ ı + (5 m) ˆ bracketrightBig + [(7 m / s) ˆ ı + ( 4 . 5 m / s) ˆ ] (1 . 5 s) + 1 2 bracketleftBig (3 m / s 2 ) ˆ ı + (1 m / s 2 ) ˆ bracketrightBig (1 . 5 s) 2 = (15 . 875 m) ˆ ı + ( 0 . 625 m) ˆ  , so | vector d | = radicalBig d 2 x + d 2 y = radicalBig (15 . 875 m) 2 + ( 0 . 625 m) 2 = 15 . 8873 m . Question 6 part 1 of 1 10 points A cannon fires a 0 . 278 kg shell with initial velocity v i = 11 m / s in the direction θ = 56 above the horizontal.
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