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Homework Solutions 7

# Homework Solutions 7 - homework 03 ASH BEN Due Feb 6 2008...

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homework 03 – ASH, BEN – Due: Feb 6 2008, 11:00 pm 1 Question 1, chap 3, sect 99. part 1 of 2 10 points A ship cruises forward at v s = 6 m / s rel- ative to the water. On deck, a man walks diagonally toward the bow such that his path forms an angle θ = 19 with a line perpen- dicular to the boat’s direction of motion. He walks at v m = 2 m / s relative to the boat. Draw the vectors to scale on a graph to determine the answer. θ v s v m At what speed does he walk relative to the water? Correct answer: 6 . 91474 m / s (tolerance ± 5 %). Explanation: v s v v m Scale: 1 m φ θ α β Let : v s = 6 m / s v m = 2 m / s v = 6 . 91474 m / s θ = 71 α = 55 . 1286 β = 161 , and φ = 15 . 8714 . 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 = v 2 m + v 2 s 2 v m v s cos β v = bracketleftbig v 2 m + v 2 s 2 v m v s cos β bracketrightbig 1 / 2 = bracketleftbig (2 m / s) 2 + (6 m / s) 2 2(2 m / s)(6 m / s) cos(109 )] 1 / 2 = bracketleftbig (2 m / s) 2 + (6 m / s) 2 +(7 . 81363 m 2 / s 2 ) bracketrightbig 1 / 2 = 6 . 91474 m / s . Alternatively: We can analyze the vector addition using the components of the vectors. Note: vectorv = vectorv s + vectorv m , or v x = v s + v m sin θ = (6 m / s) + (2 m / s) sin(19 ) = 6 . 65114 m / s and v y = v m cos θ = (2 m / s) cos(19 ) = 1 . 89104 m / s Hence the speed of the man with respect to the water is v = radicalBig v 2 x + v 2 y = radicalBig (6 . 65114 m / s) 2 + (1 . 89104 m / s) 2 = 6 . 91474 m / s . Question 2, chap 3, sect 99. part 2 of 2 10 points At what angle to his intended path does the man walk with respect to the water? Correct answer: 55 . 1286 (tolerance ± 5 %). Explanation: The law of sines can be used to compute the requested angle α , which is the angle opposite the ship’s path and velocity. sin α v s = sin β v

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homework 03 – ASH, BEN – Due: Feb 6 2008, 11:00 pm 2 sin α = v s v sin β α = arcsin bracketleftBig v s v sin β bracketrightBig = arcsin bracketleftbigg 6 m / s 6 . 91474 m / s sin(109 ) bracketrightbigg = 55 . 1286 . Alternate Solution: Using vector compo- nents from Part 1, we have tan φ = v y v x φ = arctan parenleftbigg v y v x parenrightbigg = arctan parenleftbigg 1 . 89104 m / s 6 . 65114 m / s parenrightbigg = 15 . 8714 . Therefore the angle between vectorv m and vectorv is α = 90 θ φ = 90 (19 ) (15 . 8714 ) = 55 . 1286 . Question 3, chap 3, sect 99. part 1 of 2 10 points A superhero flies 281 m from the top of a tall building at an angle of 21 below the horizontal. Draw the vectors to scale on a graph to determine the answer. a) What is the horizontal component of the superhero’s displacement? Correct answer: 262 . 336 m (tolerance ± 5 %). Explanation: 262 . 336 m 100 . 701 m 281 m Scale: 100 m 21 Basic Concept: x = d (cos θ ) Given: d = 281 m θ = 21 Solution: x = (281 m)[cos( 21 )] = 262 . 336 m .
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Homework Solutions 7 - homework 03 ASH BEN Due Feb 6 2008...

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