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Unformatted text preview: Version 008/AAACA – makeupquiz919 – Demkov – (59910) 1 This printout should have 5 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A subway train starting from rest leaves a station with a constant acceleration. At the end of 6 . 66 s, it is moving at 10 . 0566 m / s. What is the train’s displacement in the first 4 . 04928 s of motion? 1. 19.7551 2. 39.0296 3. 35.6904 4. 37.9593 5. 23.772 6. 21.8012 7. 12.3795 8. 43.3243 9. 32.976 10. 17.9573 Correct answer: 12 . 3795 m. Explanation: Basic Concepts: Uniform Acceleration of Motion a = v t Displacement of Motion from Rest s = a t 2 2 Solution: The acceleration of the train is a = v t 1 = 1 . 51 m / s 2 So, the displacement in the first 4 . 04928 s is s = a t 2 2 2 = 12 . 3795 m 002 10.0 points Consider the two vectors vector M = ( a, b ) = a ˆ ı + b ˆ , where a = 3, b = 3. 1 2 3 4 5 1 2 3 4 5 1 2 3 4 1 2 3 4 vector N = ( c, d ) = c ˆ ı + d ˆ , where c = 2, and d = 3. 1 2 3 4 5 1 2 3 4 5 1 2 3 4 1 2 3 4 Thus, a and c represent the xdisplacement and b and d represent the ydisplacement in a Cartesian xy coordinate system. Note: ˆ ı and ˆ represent unit vectors; i.e. , vectors of length 1, in the x and y directions, respectively....
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This note was uploaded on 11/10/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Swinney
 mechanics

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