This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 03 GADHIA, TEJAS Due: Feb 7 2007, 4:00 am 1 Question 1 part 1 of 2 10 points A particle travels to the right at a constant rate of 5 . 1 m / s. It suddenly is given a vertical acceleration of 2 . 8 m / s 2 for 4 . 7 s. What is its direction of travel after the acceleration with respect to the horizontal? Answer between 180 and +180 . Correct answer: 68 . 8168 (tolerance 1 %). Explanation: Basic Concepts The direction of the motion depends only on the horizontal and vertical components of the velocity at any moment. For the horizontal motion, a x = 0, so the ve locity remains the same throughout the mo tion, and v x = v x = v Solution For the vertical motion, v y = 0, so v y = v y + a y t = a y t v x v y v opposite the angle and the horizontal com ponent v x is the side adjacent to the angle, so tan = v y v x = arctan v y v x Question 2 part 2 of 2 10 points What is the speed at this time? Correct answer: 14 . 1137 m / s (tolerance 1 %). Explanation: so v = radicalBig v 2 x + v 2 y Question 3 part 1 of 2 10 points A particle at rest undergoes an acceleration of 1 . 5 m / s 2 to the right and 4 m / s 2 up. a) What is its speed after 7 . 7 s? Correct answer: 32 . 8944 m / s (tolerance 1 %). Explanation: Basic Concepts The direction of the motion depends only on the horizontal and vertical com ponents of the velocity at any moment. v x v y v Solution: For the horizontal motion, v x = 0, so v x = v x + a x t = a x t For the vertical motion, v y = 0, so v y = v y + a y t = a y t The resultant velocity is the hypotenuse of the triangle formed by the components, so v = radicalBig v 2 x + v 2 y Question 4 part 2 of 2 10 points b) What is its direction with respect to the horizontal at this time? Answer between 180 and +180 . Correct answer: 69 . 444 (tolerance 1 %). 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 The angle must be expressed in degrees. homework 03 GADHIA, TEJAS Due: Feb 7 2007, 4:00 am 2 Question 5 part 1 of 2 10 points A particle has constant acceleration vector vector a with components a x = 2 . 2 m / s 2 and a y = 3 . 1 m / s 2 . Its initial velocity vector vector v (at time t = 0) has components v x = 3 m / s and v y = 2 . 5 m / s. What is the particles speed at time t = 3 . 3 s? Correct answer: 12 . 846 m / s (tolerance 1 %). Explanation: Basic Concepts: For a motion with a constant acceleration vector, the particles velocity vector changes with time according to vector v ( t ) = t vector a + vector v . (1) And speed is the magnitude of the velocity vector, v = bardbl vector v bardbl = radicalBig ( v x ) 2 + ( v y ) 2 . (2) Solution: In vector notations, the velocity of the par ticle at time t is given by eq. (1). To evaluate this equation, it is easiest to switch to compo nent notations, thus: v x =...
View
Full
Document
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 at Austin.
 Spring '08
 Turner
 Acceleration, Work

Click to edit the document details