Practice 03 Solutions - practice 03 GADHIA, TEJAS Due: Jan...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = v y t v x t = v y g t a t a t tan = v y g t a t tan + g t = v y t = v y a tan + 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) + ( 4 . 5 m / s) . The accelera- tion is given by vectora = (3 m / s 2 ) + (1 m / s 2 ) . 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) + (3 m / s 2 ) (1 . 5 s) = (11 . 5 m / s) . 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) + (1 m / s 2 ) (1 . 5 s) = ( 3 m / s) . 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 ) ....
View Full Document

Page1 / 7

Practice 03 Solutions - practice 03 GADHIA, TEJAS Due: Jan...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online