# hw 7 - homework 07 – HETTIGER CHRISTOF – Due Feb 3 2008...

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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 θ = v re sin θ = v re v br θ = 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 θ = (14 km / h) cos(21 . 8037 ◦ ) = 12 . 9985 km / h . and ∆ t min = ∆ x v be = ∆ x v br cos θ = 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 θ = 25 ◦ with a line perpen- dicular to the boat’s 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. θ 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 φ θ α β Let : v s = 3 m / s v m = 5 m / s v = 6 . 83217 m / s θ = 65 ◦ α = 23 . 4506 ◦ β = 155 ◦ , and φ = 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 β = 90 ◦ + θ . 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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hw 7 - homework 07 – HETTIGER CHRISTOF – Due Feb 3 2008...

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