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hw1solutions - Niu(qn269 HW01 Tsoi(58020 This print-out...

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Niu (qn269) – HW01 – Tsoi – (58020) 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 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 0 2 a d v 2 0 = 2 a d d = v 2 0 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 ) = 0 . 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 x -axis. 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. a b c time × (2 s) per division velocity × (2 m/s) per division 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 = Δ x + x o = 1 2 (10 m / s) (4 s) + (1 s) (10 m / s) = (20 m) + (10 m) = 30 m .
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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 = Δ x Δ t . During the time interval 0 to 6 s, Δ 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 a = Δ v Δ t . Note: The acceleration is constant from 12 s to 18 s. So it suffices to find the acceleration during this time interval. a = ( 8 m / s) (0 m / s) (18 s) (12 s) = ( 8 m / s) (6 s) 1 . 333 m / s 2 .
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