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PHY 303K - Homework 7 - homework 07 FIERRO JEFFREY Due Feb...

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homework 07 – FIERRO, JEFFREY – Due: Feb 9 2008, 11:00 pm 1 Question 1, chap 4, sect 6. part 1 of 2 10 points The pilot of a small plane maintains an air speed of 74 knots (or nautical miles per hour) and wants to fly due North (0 ) with respect to the Earth. If a wind of 33 knots is blowing from the east (90 ), calculate the heading (azimuth) the pilot must take. Correct answer: 26 . 4839 (tolerance ± 1 %). Explanation: Let : vectorv pa = 74 knots , and vectorv w = 33 knots . Let vectorv pg be the velocity of the plane relative to the ground, vectorv pa be the velocity of the plane relative to the air and vectorv ag be the velocity of the air relative to the ground. East North vectorv pg vectorv pa v w θ Headings are measured clockwise from due North. vectorv pg = vectorv pa + vectorv ag Because of the right triangle, sin θ = v ag v pa θ = sin 1 bracketleftbigg v ag v pa bracketrightbigg θ = sin 1 bracketleftbigg 33 knots 74 knots bracketrightbigg = 26 . 4839 . Question 2, chap 4, sect 6. part 2 of 2 10 points What is the speed of the plane relative to the ground? Correct answer: 66 . 2344 knots (tolerance ± 1 %). Explanation: cos θ = v pg v pa v pg = v pa cos θ = (74 knots) cos(26 . 4839 ) = 66 . 2344 knots . Question 3, chap 4, sect 6. part 1 of 1 10 points Two boat landings are 3 . 1 km apart on the same bank of a stream that flows at 5 . 1 km / hr. A motorboat makes the round trip between the two landings in 2 . 5 hr. What is the speed of the boat relative to the water? Correct answer: 6 . 48858 km / hr (tolerance ± 1 %). Explanation: vectorv bs = vectorv bw + vectorv ws , where vectorv bs = velocity of boat relative to shore, vectorv bw = velocity of boat relative to water, and vectorv ws = velocity of water relative to shore. Given : L = 3 . 1 km , vectorv ws = 5 . 1 km / hr , and t = 2 . 5 hr . Going upstream, v bw v ws v bs v bs = v bw - v ws . Going downstream,
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homework 07 – FIERRO, JEFFREY – Due: Feb 9 2008, 11:00 pm 2 v bw v ws v bs v bs = v bw + v ws . The total time for a round trip is thus t tot = t up + t down = L v bw - v ws + L v bw + v ws t tot ( v 2 bw - v 2 ws ) = L ( v bw + v ws ) + L ( v bw - v ws ) t tot v 2 bw - 2 L v bw - v 2 ws t tot = 0 v bw = 2 L ± radicalbig 4 L + 4 v ws t 2 tot 2 t tot Since 4 L 2 - 4 v 2 ws t 2 tot = 4 (3 . 1 km) 2 + 4 (5 . 1 km / hr) 2 (2 . 5 hr) 2 = 688 . 69 km 2 , we have v 2 bw = 2 (3 . 1 km) ± 688 . 69 km 2 2 (2 . 5 hr) = 6 . 48858 km / hr . Question 4, chap 5, sect 2. part 1 of 3 10 points Suzie (60 kg) is roller-blading down the sidewalk going 28 miles per hour. She notices a group of workers down the walkway who have unexpectedly blocked her path, and she makes a quick stop in 0.5 seconds. What is Suzie’s average acceleration? Correct answer: - 7 . 82153 m / s 2 (tolerance ± 1 %). Explanation: Let : v i = 28 mi/h , t = 1 . 6 s , and v f = 0 m / s . First we need to convert Suzie’s initial speed into meters per second: 20 miles hour × 1 hour 3600 s × 1609 m 1 mile = 12 . 5144 m / s a = Δ v t = v f - v i t = 0 m / s - 12 . 5144 m / s 1 . 6 s = - 7 . 82153 m / s 2 .
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