hw4 - This print-out should have 11 questions....

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Unformatted text preview: 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. Let us plot the acceleration of the car as a function of time; take the forward direction of motion as positive. Which graph correctly describes the cars acceleration 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 8. t 1 time a t 2 correct 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. Dur- ing that time period its acceleration is nega- tive. 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 (so its ac- celeration is positive for a short while) to the original constant speed at which it remains (so its acceleration goes back to zero). 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 . 59 m / s 2 for 13 . 7 s; (b) Constant velocity for the next . 782 min; (c) Constant negative acceleration of- 10 . 3 m / s 2 for 3 . 96 s. What was the total displacement x for the complete trip? Correct answer: 1967 . 67 m. Explanation: This trip is divided into three sections: homework 04 Turner (58185) 1 hinojosa (jlh3938) homework 04 Turner (58185) 2 (a) Acceleration from rest: x a = 1 2 at 2 (b) Constant velocity motion: x b = vt (c) Deceleration: x = vt + 1 2 at 2 003 (part 1 of 2) 10.0 points An electron has an initial speed of 4 . 47 10 5 m / s. If it undergoes an acceleration of 1 . 9 10 14 m / s 2 , how long will it take to reach a speed of 5 . 79 10 5 m / s? Correct answer: 6 . 94737 10- 10 s. 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 = 5 . 79 10 5 m / s- 4 . 47 10 5 m / s 1 . 9 10 14 m / s 2 = 6 . 94737 10- 10 s . 004 (part 2 of 2) 10.0 points How far has it traveled in this time? Correct answer: 0 . 0003564 m. Explanation: Plugging t into our formula for x gives us x = x + v t + 1 2 a t 2 = 0 . 0003564 m . 005 (part 1 of 2) 10.0 points A motorist is traveling at 15 m / s when he sees a deer in the road 50 m ahead....
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hw4 - This print-out should have 11 questions....

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