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Homework 2

# Homework 2 - Gozick Brandon – Homework 2 – Due 7:00 pm...

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Unformatted text preview: Gozick, Brandon – Homework 2 – Due: Sep 12 2006, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A body moving with uniform acceleration has a velocity of 11 . 1 cm / s when its x coordinate is 3 . 69 cm. If its x coordinate 1 . 54 s later is- 8 cm, what is the magnitude of its acceleration? Correct answer: 24 . 2739 cm / s 2 . Explanation: Basic Concept: Kinematic Equations x = x + v t + 1 2 at 2 Solution: Solving for a , we get a = 2[ x- x- v t ] t 2 =- 24 . 2739 cm / s 2 2 . keywords: 002 (part 1 of 1) 10 points An automobile and train move together along parallel paths at 23 . 1 m / s. The automobile then undergoes a uniform acceleration of- 4 m / s 2 because of a red light and comes to rest. It remains at rest for 34 . 7 s, then accelerates back to a speed of 23 . 1 m / s at a rate of 1 . 32 m / s 2 . How far behind the train is the automobile when it reaches the speed of 23 . 1 m / s, as- suming that the train speed has remained at 23 . 1 m / s? Correct answer: 1070 . 4 m. Explanation: Let : t rest = 34 . 7 s , v t = 23 . 1 m / s , a 1 =- 4 m / s 2 , and a 2 = 1 . 32 m / s 2 . Basic Concepts: v = v + at x = x + v t + 1 2 at 2 Solution: Let v t stand for the constant speed of the train. For the automobile to decelerate from 23 . 1 m / s to 0 m/s requires a time t 1 = v- v t a 1 =- 23 . 1 m / s- 4 m / s 2...
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Homework 2 - Gozick Brandon – Homework 2 – Due 7:00 pm...

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