Homework 03 Solutions - homework 03 GADHIA, TEJAS Due: Feb...

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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 =...
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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 at Austin.

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Homework 03 Solutions - homework 03 GADHIA, TEJAS Due: Feb...

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