This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 cars 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 ....
View
Full
Document
This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

Click to edit the document details