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HW 2 Solutions

HW 2 Solutions - kim(jkk547 – HW02 – markert –(58840...

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kim (jkk547) – HW02 – markert – (58840) 1 This print-out should have 19 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 negative. 2. It is not changing at that time. correct 3. It is positive. 4. Unable to determine. 5. It is zero. Explanation: Solution: By definition, the acceleration a = Δ v Δ t . Since the acceleration is zero, 0 = Δ v Δ t 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. Unable to determine. 2. 4 m/s 2 3. 5 m/s 2 4. 9 m/s 2 5. 7 m/s 2 6. 8 m/s 2 7. 3 m/s 2 8. 0 m/s 2 correct 9. 6 m/s 2 Explanation: Solution: Because the distance traveled each second is constant, its velocity is constant. Therefore, 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 . 33 m / s 2 for 15 . 2 s; (b) Constant velocity for the next 0 . 782 min; (c) Constant negative acceleration of 10 . 1 m / s 2 for 5 . 85 s. What was the total displacement x for the complete trip? Correct answer: 2882 . 86 m. 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 2 004 10.0 points A 500-kilogram sports car accelerates uni- formly from rest, reaching a speed of 30 me- ters per second in 6 seconds. During the 6 seconds, the car has traveled a distance of 1. 30 m 2. 60 m 3. 180 m 4. 90 m correct 5. 15 m Explanation:

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kim (jkk547) – HW02 – markert – (58840) 2 v f = v 0 + at = at since v 0 = 0, so a = v f t . s = s 0 t + v 0 t + 1 2 at 2 = 1 2 at 2 = 1 2 parenleftBig v f t parenrightBig t 2 = 1 2 v f t since v 0 = 0 and s 0 = 0, so s = 1 2 (30 m / s) (6 s) = 90 m . 005 10.0 points An electron, starting from rest and moving with a constant acceleration, travels 5 . 4 cm in 14 ms. What is the magnitude of this acceleration in km/s 2 ? Correct answer: 0 . 55102 km / s 2 . Explanation: An electron starting from rest with con- stant acceleration will travel a distance of d = 1 2 at 2 in t seconds. Given the time and distance traveled we solve for a : a = 2 d t 2 making any necessary unit conversions along the way. 006 (part 1 of 2) 10.0 points The initial speed of a body is 2 . 22 m / s. What is its speed after 3 . 35 s if it accelerates uniformly at 4 . 04 m / s 2 ?
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