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Unformatted text preview: hernandez (ejh742) – homework 04 – Turner – (58120) 1 This printout should have 11 questions. Multiplechoice 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 car’s acceleration a ( t ) as a function of time? Take the forward direction of motion as positive. 1. t 1 time a t 2 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 correct 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 3 . 61 m / s 2 for 17 . 3 s; (b) Constant velocity of 62 . 453 m / s for the next 0 . 989 min; (c) Constant negative acceleration of 9 . 33 m / s 2 for 5 . 74 s. What was the total displacement x for the complete trip? Correct answer: 4450 . 96 m. Explanation: This trip is divided into three sections: acceleration from rest: x a = 1 2 a t 2 = 1 2 (3 . 61 m / s 2 ) (17 . 3 s) 2 = 540 . 218 m ; hernandez (ejh742) – homework 04 – Turner – (58120) 2 constant velocity motion: x b = v t = (62 . 453 m / s)(0 . 989 min) 60 s 1 min = 3705 . 96 m ; and deceleration: x c = v t + 1 2 a t 2 = (62 . 453 m / s)(5 . 74 s) + 1 2 ( 9 . 33 m / s 2 )(5 . 74 s) 2 = 204 . 78 m . Therefore, x tot = x a + x b + x c = 540 . 218 m + 3705 . 96 m + 204 . 78 m = 4450 . 96 m . 003 (part 1 of 2) 10.0 points An electron has an initial speed of 4 . 27 × 10 5 m / s. If it undergoes an acceleration of 3 . 2 × 10 14 m / s 2 , how long will it take to reach a speed of 8 . 43 × 10 5 m / s? Correct answer: 1 . 3 × 10 9 s. Explanation: Let : v = 4 . 27 × 10 5 m / s , a = 3 . 2 × 10 14 m / s 2 , and v = 8 . 43 × 10 5 m / s . v = v + a t t = v v a = 8 . 43 × 10 5 m / s 4 . 27 × 10 5 m / s 3 . 2 × 10 14 m / s 2 = 1 . 3 × 10 9 s . 004 (part 2 of 2) 10.0 points How far has it traveled in this time? Correct answer: 0 . 0008255 m. Explanation: x = x + v t + 1 2 a t 2 = (4 . 27 × 10 5 m / s) (1 . 3 × 10 9 s) + 1 2 (3 . 2 × 10 14 m / s 2 ) × (1 . 3 × 10 9 s) 2 = . 0008255 m ....
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 Spring '08
 Turner
 Physics, Acceleration, Work, m/s

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