# hw4(1) - lenoir(wml297 – homework 04 – Turner –(58220...

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Unformatted text preview: lenoir (wml297) – 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 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 2 . 64 m / s 2 for 19 s; (b) Constant velocity of 50 . 16 m / s for the next 0 . 955 min; (c) Constant negative acceleration of- 12 m / s 2 for 4 . 64 s. What was the total displacement x for the complete trip? Correct answer: 3454 . 25 m. Explanation: This trip is divided into three sections: acceleration from rest: x a = 1 2 a t 2 = 1 2 (2 . 64 m / s 2 ) (19 s) 2 = 476 . 52 m ; lenoir (wml297) – homework 04 – Turner – (58220) 2 constant velocity motion: x b = v t = (50 . 16 m / s)(0 . 955 min) 60 s 1 min = 2874 . 17 m ; and deceleration: x c = v t + 1 2 a t 2 = (50 . 16 m / s)(4 . 64 s) + 1 2 (- 12 m / s 2 )(4 . 64 s) 2 = 103 . 565 m . Therefore, x tot = x a + x b + x c = 476 . 52 m + 2874 . 17 m + 103 . 565 m = 3454 . 25 m . 003 (part 1 of 2) 10.0 points An electron has an initial speed of 3 . 44 × 10 5 m / s. If it undergoes an acceleration of 2 × 10 14 m / s 2 , how long will it take to reach a speed of 7 . 47 × 10 5 m / s? Correct answer: 2 . 015 × 10- 9 s. Explanation: Let : v = 3 . 44 × 10 5 m / s , a = 2 × 10 14 m / s 2 , and v = 7 . 47 × 10 5 m / s . v = v + a t t = v- v a = 7 . 47 × 10 5 m / s- 3 . 44 × 10 5 m / s 2 × 10 14 m / s 2 = 2 . 015 × 10- 9 s . 004 (part 2 of 2) 10.0 points How far has it traveled in this time? Correct answer: 0 . 00109918 m. Explanation: x = x + v t + 1 2 a t 2 = (3 . 44 × 10 5 m / s) (2 . 015 × 10- 9 s) + 1 2 (2 × 10 14 m / s 2 ) × (2 . 015 × 10- 9 s) 2 = . 00109918 m . 005 (part 1 of 2) 10.0 points A motorist is traveling at 17 m / s when he sees a deer in the road 40 m ahead....
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## This note was uploaded on 04/15/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas.

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hw4(1) - lenoir(wml297 – homework 04 – Turner –(58220...

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