hw4_s - homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am 1...

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Unformatted text preview: homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am 1 Question 1, chap 2, sect 4. part 1 of 1 10 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. Let us plot the acceleration of the car as a function of time; take the forward direction of motion as positive. Which of the following graphs correctly describes the cars accelera- tion a ( t )? 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: The car is at first moving at a constant speed and therefore its acceleration is zero. After seeing a patrol car it starts slowing down i.e. decelerating for a short while. During homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am 2 that time period its acceleration is negative. The acceleration soon goes back to zero and stays at zero while the car moves at a constant (but lower) speed once again. At time t 2 the car briefly accelerates (and so its acceleration is positive for a short while) to the original constant speed at which it remains (and so its acceleration goes back to zero). Question 2, chap 2, sect 4. part 1 of 1 10 points A record of travel along a straight path is as follows: (a) Start from rest with constant accelera- tion of 2 . 97 m / s 2 for 15 . 4 s; (b) Constant velocity for the next 0 . 91 min; (c) Constant negative acceleration of- 9 . 91 m / s 2 for 5 . 61 s. What was the total displacement x for the complete trip? Correct answer: 2950 . 12 m (tolerance 1 %). Explanation: This trip is divided into three sections: (a) Acceleration from rest: x a = 1 2 at 2 (b) Constant velocity motion: x b = vt (c) Deceleration: x = vt + 1 2 at 2 Question 3, chap 2, sect 5. part 1 of 2 10 points An electron has an initial speed of 260000 m / s. If it undergoes an acceleration of 2 . 1 10 14 m / s 2 , how long will it take to reach a speed of 522000 m / s? Correct answer: 1 . 24762 10- 9 s (tolerance 1 %). Explanation: Basic Concepts: v = v + a t x = x + v t + 1 2 a t 2 Solution: Assuming the acceleration given is an average acceleration, we can use v = v + at and solve for t : t = v- v a = 522000 m / s- 260000 m / s 2 . 1 10 14 m / s 2 = 1 . 24762 10- 9 s ....
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This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw4_s - homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am 1...

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