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HW02-solutions

# HW02-solutions - burgin(lcb626 HW02 markert(56480 This...

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burgin (lcb626) – HW02 – markert – (56480) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If the acceleration of an object is zero at some instant in time, what can be said about its velocity at that time? 1. It is not changing at that time. correct 2. It is negative. 3. It is zero. 4. Unable to determine. 5. It is positive. Explanation: The acceleration a = Δ v Δ t = 0 Δ v = 0 . 002 10.0 points An object travels 9 m in the first second of travel, 9 m again during the second second of travel, and 9 m again during the third second. What is the approximate average accelera- tion of the object during this time interval? 1. 0 m/s 2 correct 2. 6 m/s 2 3. 5 m/s 2 4. 7 m/s 2 5. 8 m/s 2 6. 3 m/s 2 7. 9 m/s 2 8. Unable to determine 9. 4 m/s 2 Explanation: Because the distance traveled each second is constant, its velocity is constant, so the acceleration of the object is zero. 003 10.0 points A record of travel along a straight path is as follows: (a) Start from rest with constant accelera- tion of 3 . 05 m / s 2 for 12 . 8 s; (b) Constant velocity of 39 . 04 m / s for the next 0 . 944 min; (c) Constant negative acceleration of 11 m / s 2 for 3 . 79 s. What was the total displacement x for the complete trip? Correct answer: 2530 . 04 m. Explanation: This trip is divided into three sections: acceleration from rest: x a = 1 2 a t 2 = 1 2 (3 . 05 m / s 2 ) (12 . 8 s) 2 = 249 . 856 m ; constant velocity motion: x b = v t = (39 . 04 m / s)(0 . 944 min) 60 s 1 min = 2211 . 23 m ; and deceleration: x c = v t + 1 2 a t 2 = (39 . 04 m / s)(3 . 79 s) + 1 2 ( 11 m / s 2 )(3 . 79 s) 2 = 68 . 9591 m . Therefore, x tot = x a + x b + x c = 249 . 856 m + 2211 . 23 m + 68 . 9591 m = 2530 . 04 m .

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burgin (lcb626) – HW02 – markert – (56480) 2 004 (part 1 of 2) 10.0 points The initial speed of a body is 4 . 91 m / s. What is its speed after 3 . 25 s if it accelerates uniformly at 4 . 62 m / s 2 ? Correct answer: 19 . 925 m / s. Explanation: Let : v 0 = 4 . 91 m / s a 1 = 4 . 62 m / s 2 , and t = 3 . 25 s . v = v 0 + a 1 t , = 4 . 91 m / s + (4 . 62 m / s 2 ) (3 . 25 s) = 19 . 925 m / s . 005 (part 2 of 2) 10.0 points What is its speed after 3 . 25 s if it accelerates uniformly at 4 . 62 m / s 2 ? Correct answer: 10 . 105 m / s. Explanation: Let : a 2 = 4 . 62 m / s 2 . v = v 0 + a 2 t = 4 . 91 m / s + ( 4 . 62 m / s 2 ) (3 . 25 s) = 10 . 105 m / s , which has a magnitude of 10 . 105 m / s .
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