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Unformatted text preview: brown (twb493) extra credit 01 turner (57340) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points When Maggie applies the brakes of her car, the car slows uniformly from 14.2 m/s to 0 m/s in 2.36 s. How far ahead of a stop sign must she apply her brakes in order to stop at the sign? Correct answer: 16 . 756 m. Explanation: Let : v i = 14 . 2 m / s , v f = 0 m / s , and t = 2 . 36 s . The equation simplifies to x = v i + v f 2 t = 1 2 v i t = 1 2 (14 . 2 m / s) (2 . 36 s) = 16 . 756 m . 002 10.0 points Consider the plot describing motion along a straight line with an initial position of 10 m. 1 1 2 3 1 2 3 4 5 6 7 8 9 b b b b b time (s) velocity(m/s) What is the position at 9 seconds? Correct answer: 23 . 5 m. Explanation: b b b b b b b b 1 2 3 4 5 6 7 8 9 1 2 3 1 time (s) velocity(m/s) The acceleration during the first 2 seconds is a = v t = 3 m / s 0 m / s 2 s 0 s = 1 . 5 m / s 2 . The position at 2 seconds is 10 m plus the area of the triangle (shaded above) x = 10 m + 1 2 (2 s 0 s)(3 m / s 0 m / s) = 13 m ; however, it can also be calculated: x = x i + v i ( t f t i ) + 1 2 a ( t f t i ) 2 = 10 m + (0 m / s) (2 s 0 s) + 1 2 (1 . 5 m / s 2 )(2 s 0 s) 2 = 13 m . The acceleration during the second time interval is a = v t = 3 m / s 3 m / s 6 s 2 s = 0 m / s 2 ....
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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