physics solution hw 3

physics solution hw 3 - reyes(mr29667 Homework 3...

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reyes (mr29667) – Homework 3 – chelikowsky – (59005) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A particle at rest undergoes an acceleration oF 3 . 3 m / s 2 to the right and 4 . 2 m / s 2 up. a) What is its speed aFter 6 . 4 s? Correct answer: 34 . 1846 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: ±or the horizontal motion, v 0 x = 0, so v x = v 0 x + a x t = a x t ±or the vertical motion, v 0 y = 0, so v y = v 0 y + a y t = a y t The resultant velocity is the hypotenuse oF the triangle Formed by the components, so v = r 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: 51 . 8428 . 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 . 2 m / s. It suddenly is given a vertical acceleration oF 2 m / s 2 For 3 . 9 s. What is its direction oF travel aFter the acceleration with respect to the horizontal? Answer between 180 and +180 . Correct answer: 51 . 5198 . Explanation: Basic Concepts The direction oF the motion depends only on the horizontal and vertical components oF the velocity at any moment. ±or the horizontal motion, a x = 0, so the ve- locity remains the same throughout the mo- tion, and v 0 x = v x = v Solution ±or the vertical motion, v 0 y = 0, so v y = v 0 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 p v y v x P = arctan p 7 . 8 m / s 6 . 2 m / s P = 51 . 5198 004 (part 2 oF 2) 10.0 points What is the speed at this time? Correct answer: 9 . 96393 m / s. Explanation: so v = r v 2 x + v 2 y = r (6 . 2 m / s) 2 + (7 . 8 m / s) 2 = 9 . 96393 m / s
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reyes (mr29667) – Homework 3 – chelikowsky – (59005) 2 005 (part 1 of 2) 10.0 points A cannon Fres a 0 . 24 kg shell with initial velocity v i = 8 . 9 m / s in the direction θ = 55 above the horizontal. Δ x Δ h 8 . 9 m / s 55 Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 494 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. On what does Δ
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physics solution hw 3 - reyes(mr29667 Homework 3...

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