solution physic 2425

solution physic 2425 - ebadi (pe2985) Chapter 2 Sherman...

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Unformatted text preview: ebadi (pe2985) Chapter 2 Sherman (74201) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A particle moves according to the equation x = (10 m / s 2 ) t 2 where x is in meters and t is in seconds. Find the average velocity for the time in- terval from 2 . 01 s to 3 . 37 s. Correct answer: 53 . 8 m / s. Explanation: Let : t 1 = 2 . 01 s and t 2 = 3 . 37 s . v av = x t = (10 m / s 2 ) ( t 2 2- t 2 1 ) t 2- t 1 = (10 m / s 2 ) [(3 . 37 s) 2- (2 . 01 s) 2 ] 3 . 37 s- 2 . 01 s = 53 . 8 m / s . 002 (part 2 of 2) 10.0 points Find the average velocity for the time interval from 2 . 01 s to 2 . 11 s. Correct answer: 41 . 2 m / s. Explanation: Let : t 1 = 2 . 01 s , and t 3 = 2 . 11 s . v ave = x t = (10 m / s 2 ) ( t 2 3- t 2 1 ) t 3- t 1 = (10 m / s 2 ) [(2 . 11 s) 2- (2 . 01 s) 2 ] 2 . 11 s- 2 . 01 s = 41 . 2 m / s . 003 (part 1 of 2) 10.0 points An acceleration (in m/s 2 ) has the time de- pendence shown on the graph below. The particle starts from rest ( v = 0 m/s) at the origin ( x = 0 m). 1 2 3 4 5- 3- 2- 1 a (m/s 2 ) t (s) Acceleration vs Time Find the velocity at t = 5 s. 1. v t =5 =- 4 m / s 2. v t =5 =- 6 m / s 3. v t =5 =- 1 m / s 4. v t =5 =- 8 m / s 5. v t =5 =- 3 m / s correct 6. v t =5 =- 2 m / s Explanation: Let :...
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This note was uploaded on 09/06/2010 for the course PHYS 2425 taught by Professor Sherman during the Spring '08 term at Collin College.

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solution physic 2425 - ebadi (pe2985) Chapter 2 Sherman...

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