HW4_Solutions - hiarker (srh959) homework 04 Turner (58220)...

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Unformatted text preview: hiarker (srh959) homework 04 Turner (58220) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A car is moving at constant speed on the freeway. The driver sees a patrol car at time t 1 and rapidly slows down by around 10 miles per hour. After continuing at this speed for a few minutes, the driver at time t 2 returns to the earlier constant speed. Which graph correctly describes the cars acceleration a ( t ) as a function of time? Take the forward direction of motion as positive. 1. t 1 time a t 2 correct 2. t 1 time a t 2 3. t 1 time a t 2 4. t 1 time a t 2 5. t 1 time a t 2 6. t 1 time a t 2 7. t 1 time a t 2 8. t 1 time a t 2 Explanation: Analyze the acceleration over each part of the trip: 1) Moves at constant speed: a = 0 2) Rapidly slows down: a < 0 briefly 3) Continues at this speed: a = 0 4) Returns to earlier speed: a > 0 briefly 5) Original constant speed: a = 0 002 10.0 points A record of travel along a straight path is as follows: (a) Start from rest with constant accelera- tion of 2 . 96 m / s 2 for 18 . 4 s; (b) Constant velocity of 54 . 464 m / s for the next 0 . 751 min; (c) Constant negative acceleration of- 8 . 69 m / s 2 for 5 . 52 s. What was the total displacement x for the complete trip? Correct answer: 3123 . 46 m. Explanation: This trip is divided into three sections: acceleration from rest: x a = 1 2 a t 2 = 1 2 (2 . 96 m / s 2 ) (18 . 4 s) 2 = 501 . 069 m ; hiarker (srh959) homework 04 Turner (58220) 2 constant velocity motion: x b = v t = (54 . 464 m / s)(0 . 751 min) 60 s 1 min = 2454 . 15 m ; and deceleration: x c = v t + 1 2 a t 2 = (54 . 464 m / s)(5 . 52 s) + 1 2 (- 8 . 69 m / s 2 )(5 . 52 s) 2 = 168 . 247 m . Therefore, x tot = x a + x b + x c = 501 . 069 m + 2454 . 15 m + 168 . 247 m = 3123 . 46 m . 003 (part 1 of 2) 10.0 points An electron has an initial speed of 3 . 72 10 5 m / s. If it undergoes an acceleration of 1 . 8 10 14 m / s 2 , how long will it take to reach a speed of 6 . 82 10 5 m / s? Correct answer: 1 . 72222 10- 9 s. Explanation: Let : v = 3 . 72 10 5 m / s , a = 1 . 8 10 14 m / s 2 , and v = 6 . 82 10 5 m / s . v = v + a t t = v- v a = 6 . 82 10 5 m / s- 3 . 72 10 5 m / s 1 . 8 10 14 m / s 2 = 1 . 72222 10- 9 s . 004 (part 2 of 2) 10.0 points How far has it traveled in this time? Correct answer: 0 . 000907611 m. Explanation: x = x + v t + 1 2 a t 2 = (3 . 72 10 5 m / s) (1 . 72222 10- 9 s) + 1 2 (1 . 8 10 14 m / s 2 ) (1 . 72222 10- 9 s) 2 = . 000907611 m ....
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This note was uploaded on 04/23/2010 for the course PHY 303K taught by Professor Nui during the Spring '09 term at University of Texas-Tyler.

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HW4_Solutions - hiarker (srh959) homework 04 Turner (58220)...

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