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Unformatted text preview: muir (cem2577) – HW2 – Niu – (58205) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points On a drive from your ranch to Austin, you wish to average 44 mph. The distance from your ranch to Austin is 110 miles. However, at 55 miles (half way), you find you have averaged only 33 mph. What average speed must you maintain in the remaining distance in order to have an overall average speed of 44 mph? Correct answer: 66 mph. Explanation: Let t denote the total time, d the total distance, and v the average velocity over the total distance. Let t 1 denote the time over the first half, d 1 the first half distance, and v 1 the average velocity of the first half. Let t 2 denote the time over the second half, d 2 the second half distance, and v 2 the average velocity of the second half. The change in velocity occurs at d 1 , therefore d 2 = d − d 1 . (1) Average speed is v ≡ Δ d Δ t = d t . Thus t 1 = d 1 v 1 (2) t = d v . (3) The total time t = t 1 + t 2 , so t 2 = t − t 1 = d v − d 1 v 1 (4) since d 2 = v 2 t 2 v 2 = d 2 t 2 ≈ 66 mph . Notice that the formula for average velocity when acceleration is constant: v = v 1 + v 2 2 yields: v 2 = 2 v − v 1 = 2(44 mph) − 33 mph = 55 mph an incorrect answer!!!! Do you know why this formula is incorrect? What is the acceleration at the halfway point? What is the acceler- ation over the rest of the distance? Is the acceleration constant? For constant acceler- ation v = v + at . The graph of velocity vs time is a straight line. For this problem the graph of velocity vs time is a step function. 002 10.0 points In order to qualify for the finals in a racing event, a race car must achieve an average speed of 243 km / h on a track with a total length of 1490 m. If a particular car covers the first half of the track at an average speed of 205 km / h, what minimum average speed must it have in the second half of the event in order to qualify? Correct answer: 298 . 293 km / h. Explanation: Let t t be the maximum time to complete the trip. t t = total distance needed average speed = d v av The time spent to the first half, t 1 is t 1 = half distance v av 1 = d/ 2 v av 1 Thus, the maximum time that can be spent second half of the trip is t 2 = t t − t 1 and the required average speed on the second half is v av 2 = d/ 2 t 2 003 10.0 points An electron travels 2 . 15 m in 5 . 57 × 10 − 8 s. How fast is it traveling? Correct answer: 3 . 85996 × 10 7 m / s. Explanation: muir (cem2577) – HW2 – Niu – (58205) 2 Speed and velocity have units of distance per unit time, so v = s t = 2 . 15 m 5 . 57 × 10 − 8 s = 3 . 85996 × 10 7 m / s 004 (part 1 of 2) 10.0 points A particle moves according to the equation x = (10 m / s 2 ) t 2 where x is in meters and t is in seconds....
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This note was uploaded on 02/03/2009 for the course PHY 58205 taught by Professor Niu during the Spring '09 term at University of Texas at Austin.
- Spring '09