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Unformatted text preview: homework 07 HETTIGER, CHRISTOF Due: Feb 3 2008, 4:00 am 1 Question 1, chap 3, sect 99. part 1 of 1 10 points A hunter wishes to cross a river that is 1 . 4 km wide and that flows with a speed of 5 . 2 km / h. The hunter uses a small powerboat that moves at a maximum speed of 14 km / h with respect to the water. Draw the vectors to scale on a graph to determine the answer. What is the time necessary for crossing if the boat goes directly across the river to the opposite bank? Correct answer: 6 . 4623 min (tolerance 5 %). Explanation: flow 12 . 9985 km / h 5 . 2 km / h 1 4 k m / h Scale: 10 km / hr 2 1 . 8 3 7 Let : x = 1 . 4 km , v re = 5 . 2 km / h , and v br = 14 km / h . Solution: We need to find the angle to head into the current so that the resultant direction is perpendicular to the current v br sin = v re sin = v re v br = sin 1 parenleftbigg v re v br parenrightbigg = sin 1 parenleftbigg 5 . 2 km / h 14 km / h parenrightbigg = 21 . 8037 . Thus v be = v br cos = (14 km / h) cos(21 . 8037 ) = 12 . 9985 km / h . and t min = x v be = x v br cos = 1 . 4 km (14 km / h) cos(21 . 8037 ) 60 min 1 h = 6 . 4623 min . Alternate Solution: v be = radicalBig v 2 be v 2 re t min = x v be = (1 . 4 km) radicalbig (12 . 9985 km / h) 2 (5 . 2 km / h) 2 = (1 . 4 km) (12 . 9985 km / h) 60 min 1 h = 6 . 4623 min . Question 2, chap 3, sect 99. part 1 of 2 10 points A ship cruises forward at v s = 3 m / s rel ative to the water. On deck, a man walks diagonally toward the bow such that his path forms an angle = 25 with a line perpen dicular to the boats direction of motion. He walks at v m = 5 m / s relative to the boat. Draw the vectors to scale on a graph to determine the answer. v s v m homework 07 HETTIGER, CHRISTOF Due: Feb 3 2008, 4:00 am 2 At what speed does he walk relative to the water? Correct answer: 6 . 83217 m / s (tolerance 5 %). Explanation: v s v v m Scale: 1 m Let : v s = 3 m / s v m = 5 m / s v = 6 . 83217 m / s = 65 = 23 . 4506 = 155 , and = 41 . 5494 . When you complete the parallelogram, the resultant velocity v with respect to the water is the side of the triangle opposite the obtuse angle, which has a measure of = 90 + . Let vectorv s be the velocity of the ship, vectorv m be the velocity of the man, and vectorv be the resultant velocity of the man relative to the water (Earth). By the law of cosines v 2...
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 Spring '10
 Turner
 Physics, Power, Work

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