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Unformatted text preview: Niu (qn269) HW01 Tsoi (58020) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a safety belt can withstand an acceleration of 300 m / s 2 . How thick should barriers be to safely stop a car that hits the barriers at 86 km / h? Correct answer: 0 . 951133 m. Explanation: The final velocity v f = 0 and the deceler ation is negative. A maximum deceleration will be experienced in this collision, so v 2 f = v 2 2 a d v 2 = 2 a d d = v 2 2 a = (86 km / h) 2 2 (300 m / s 2 ) (1000 m / km) 2 (0 . 000277778 h / s) 2 = (23 . 8889 m / s) 2 2 (300 m / s 2 ) = . 951133 m . which is the minimum thickness of the safety barrier (a maximum value in a denominator will insure a minimum quotient). Dimensional analysis for d : parenleftBig km h parenrightBig 2 m s 2 = km 2 h 2 s 2 m parenleftBig 1 h 3600 s parenrightBig 2 parenleftBig 1000 m 1 km parenrightBig 2 = 0 . 07716 m 002 (part 1 of 3) 10.0 points Consider the velocity curve of a one dimen sional motion along the xaxis. The initial position is x = 10 m. The scale on the horizontal axis is 2 s per division and on the vertical axis 2 m/s per division. b b b b b a b c time (2 s) per division velocity (2m/s)perdivision The position at t = 4 s is given by 1. x = 30 m correct 2. x = 14 m 3. x = 18 m 4. x = 10 m 5. x = 24 m 6. x = 22 m 7. x = 12 m 8. x = 16 m 9. x = 28 m 10. x = 20 m Explanation: Change in position is the area under the velocity vs time graph, so x = x + x o = 1 2 (10 m / s) (4 s) + (1 s) (10 m / s) = (20 m) + (10 m) = 30 m . Niu (qn269) HW01 Tsoi (58020) 2 Recall that the area of a triangle is 1 2 base height. 003 (part 2 of 3) 10.0 points The average velocity v between 0 s and 6 s is given by 1. 5 m / s < v 6 m / s 2. 1 m / s < v 2 m / s 3. 8 m / s < v 9 m / s 4. 7 m / s < v 8 m / s 5. 3 m / s < v 4 m / s 6. 4 m / s < v 5 m / s 7. 6 m / s < v 7 m / s correct 8. 9 m / s < v 10 m / s 9. 2 m / s < v 3 m / s 10. 0 m / s < v 1 m / s Explanation: The definition of average velocity is v = x t . During the time interval 0 to 6 s, x = 1 2 (10 m / s) (4 s) + (2 s) (10 m / s) = (20 m) + (20 m) = 40 m . Hence the average velocity between 0 to 6 s is v = (40 m) (6 s) 6 . 667 m / s . 004 (part 3 of 3) 10.0 points The average acceleration a between 14 s and 16 s is given by 1. 0 m / s 2 < a 1 m / s 2 2. 3 m / s 2 < a 2 m / s 2 3. 3 m / s 2 < a 4 m / s 2 4. 5 m / s 2 < a 4 m / s 2 5. 2 m / s 2 < a 1 m / s 2 correct 6. 1 m / s 2 < a 2 m / s 2 7. 1 m / s 2 < a 0 m / s 2 8. 2 m / s 2 < a 3 m / s 2 9. 4 m / s 2 < a 3 m / s 2 10. 4 m / s 2 < a 5 m / s 2 Explanation: The definition of average acceleration is...
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This note was uploaded on 09/01/2009 for the course PHY M taught by Professor Staff during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Staff
 Physics

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