# 9 - gonzalez(ng4897 – homework 04 – Turner –(58120 1...

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Unformatted text preview: gonzalez (ng4897) – homework 04 – Turner – (58120) 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 correct 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 3 . 56 m / s 2 for 16 . 4 s; (b) Constant velocity of 58 . 384 m / s for the next 0 . 854 min; (c) Constant negative acceleration of- 9 . 6 m / s 2 for 4 . 45 s. What was the total displacement x for the complete trip? Correct answer: 3635 . 1 m. Explanation: This trip is divided into three sections: acceleration from rest: x a = 1 2 a t 2 = 1 2 (3 . 56 m / s 2 ) (16 . 4 s) 2 = 478 . 749 m ; gonzalez (ng4897) – homework 04 – Turner – (58120) 2 constant velocity motion: x b = v t = (58 . 384 m / s)(0 . 854 min) 60 s 1 min = 2991 . 6 m ; and deceleration: x c = v t + 1 2 a t 2 = (58 . 384 m / s)(4 . 45 s) + 1 2 (- 9 . 6 m / s 2 )(4 . 45 s) 2 = 164 . 757 m . Therefore, x tot = x a + x b + x c = 478 . 749 m + 2991 . 6 m + 164 . 757 m = 3635 . 1 m . 003 (part 1 of 2) 10.0 points An electron has an initial speed of 2 . 97 × 10 5 m / s. If it undergoes an acceleration of 2 . 9 × 10 14 m / s 2 , how long will it take to reach a speed of 5 . 49 × 10 5 m / s? Correct answer: 8 . 68966 × 10- 10 s. Explanation: Let : v = 2 . 97 × 10 5 m / s , a = 2 . 9 × 10 14 m / s 2 , and v = 5 . 49 × 10 5 m / s . v = v + a t t = v- v a = 5 . 49 × 10 5 m / s- 2 . 97 × 10 5 m / s 2 . 9 × 10 14 m / s 2 = 8 . 68966 × 10- 10 s . 004 (part 2 of 2) 10.0 points How far has it traveled in this time? Correct answer: 0 . 000367572 m. Explanation: x = x + v t + 1 2 a t 2 = (2 . 97 × 10 5 m / s) (8 . 68966 × 10- 10 s) + 1 2 (2 . 9 × 10 14 m / s 2 ) × (8 . 68966 × 10- 10 s) 2 = . 000367572 m ....
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## This note was uploaded on 03/10/2010 for the course EE PHY 303K taught by Professor Gordito during the Spring '10 term at École Normale Supérieure.

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9 - gonzalez(ng4897 – homework 04 – Turner –(58120 1...

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