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

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
This is the end of the preview. Sign up to access the rest of the document.

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 car’s 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online