Homework 3-solutions

Homework 3-solutions - le (txl65) Homework 3 Spurlock...

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le (txl65) – Homework 3 – Spurlock – (11881) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Initially (at time t = 0) a particle is mov- ing vertically at 7 . 4 m / s and horizontally at 0 m / s. Its horizontal acceleration is 1 . 3 m / s 2 . At what time will the particle be traveling at 48 with respect to the horizontal? The acceleration due to gravity is 9 . 8 m / s 2 . Correct answer: 0 . 658141 s. Explanation: Let : v y 0 = 7 . 4 m / s , g = 9 . 8 m / s 2 , v x 0 = 0 , and a = 1 . 3 m / s 2 , and θ = 48 . v x t v y t v t 48 The vertical velocity is v y t = v y 0 - g t and the horizontal velocity is v x t = v y 0 + a t = a t . The vertical component is the opposite side and the horizontal component the adjacent side to the angle, so tan θ = v y t v x t = v y 0 - g t a t a t tan θ = v y 0 - g t a t tan θ + g t = v y 0 t = v y 0 a tan θ + g = 7 . 4 m / s (1 . 3 m / s 2 ) tan(48 ) + 9 . 8 m / s 2 = 0 . 658141 s . 002 10.0 points Particle 1 is moving on the x -axis with an acceleration oF 3 . 37 m / s 2 in the positive x - direction. Particle 2 is moving on the y - axis with an acceleration oF 5 . 16 m / s 2 in the negative y -direction. Both particles were at rest at the origin at t = 0 s.
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Homework 3-solutions - le (txl65) Homework 3 Spurlock...

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