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Unformatted text preview: yindeemark (rry82) – Homework 3 – chelikowsky – (59005) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A particle at rest undergoes an acceleration of 3 . 2 m / s 2 to the right and 4 . 4 m / s 2 up. a) What is its speed after 6 . 4 s? Correct answer: 34 . 8198 m / s. Explanation: Basic Concepts The direction of the motion depends only on the horizontal and vertical components 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 002 (part 2 of 2) 10.0 points b) What is its direction with respect to the horizontal at this time? Answer between − 180 ◦ and +180 ◦ . Correct answer: 53 . 9726 ◦ . Explanation: The vertical component v y is the side oppo site the angle θ and the horizontal component v x is the side adjacent to the angle, so tan θ = v y v x θ = arctan v y v x The angle θ must be expressed in degrees. 003 (part 1 of 2) 10.0 points A particle travels to the right at a constant rate of 6 m / s. It suddenly is given a vertical acceleration of 3 . 6 m / s 2 for 3 . 1 s. What is its direction of travel after the acceleration with respect to the horizontal? Answer between − 180 ◦ and +180 ◦ . Correct answer: 61 . 736 ◦ . 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 θ and the horizontal com ponent v x is the side adjacent to the angle, so tan θ = v y v x θ = arctan parenleftbigg v y v x parenrightbigg = arctan parenleftbigg 11 . 16 m / s 6 m / s parenrightbigg = 61 . 736 ◦ 004 (part 2 of 2) 10.0 points What is the speed at this time? Correct answer: 12 . 6707 m / s. Explanation: so v = radicalBig v 2 x + v 2 y = radicalBig (6 m / s) 2 + (11 . 16 m / s) 2 = 12 . 6707 m / s yindeemark (rry82) – Homework 3 – chelikowsky – (59005) 2 005 (part 1 of 2) 10.0 points A cannon fires a 0 . 367 kg shell with initial velocity v i = 9 . 9 m / s in the direction θ = 45 ◦ above the horizontal. Δ x Δ h 9 . 9 m / s 4 5 ◦ Δ y y The shell’s trajectory curves downward be cause of gravity, so at the time t = 0 . 617 s the shell is below the straight line by some verti cal distance Δ h . Your task is to calculate the distance Δ h in the absence of air resistance....
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This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
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