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PHY 303K - Homework 4

# PHY 303K - Homework 4 - homework 04 FIERRO JEFFREY Due...

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homework 04 – FIERRO, JEFFREY – Due: Jan 28 2008, 11:00 pm 1 Question 1, chap 3, sect 3. part 1 of 2 10 points A boy runs 17 . 1 blocks North, 15 . 3 blocks Northeast, and 1 . 6 blocks West. Determine the length of the displacement vector that goes from the starting point to his final position. Correct answer: 29 . 4014 (tolerance ± 1 %). Explanation: Let ˆ ı = East and ˆ = North. The displace- ment vector is vector R = 17 . 1 ˆ + 15 . 3 cos 45 ˆ ı + 15 . 3 sin 45 ˆ 1 . 6 ˆ ı = R e ˆ ı + R n ˆ  . The length of vector R is | vector R | = radicalBig R 2 e + R 2 n = radicalBig (15 . 3 cos 45 1 . 6) 2 + (17 . 1 + 15 . 3 sin 45 ) 2 = 29 . 4014 . Question 2, chap 3, sect 3. part 2 of 2 10 points Determine the direction of the displacement vector. (Use counterclockwise as the posi- tive angular direction, between the limits of 180 and +180 from East) Correct answer: 71 . 7269 (tolerance ± 1 %). Explanation: The reference angle is θ = arctan vextendsingle vextendsingle vextendsingle vextendsingle R n R e vextendsingle vextendsingle vextendsingle vextendsingle = 71 . 7269 . Since the displacement vector lies in quad- rant 1, the angle is 0 + 71 . 7269 = 71 . 7269 . Question 3, chap 3, sect 1. part 1 of 2 10 points A football player runs directly down the field for 28 m before turning to the right at an angle of 26 from his original direction and running an additional 11 m before being tackled. a) What is the magnitude of the runner’s total displacement? Correct answer: 38 . 1924 m (tolerance ± 1 %). Explanation: 28 m 11 m d 26 Note: Figure is not drawn to scale. Basic Concepts: Δ x = d (cos θ ) Δ y = d (sin θ ) Δ x total = Δ x 1 + Δ x 2 Δ y total = Δ y 1 + Δ y 2 d total = radicalBig x total ) 2 + (Δ y total ) 2 Given: d 1 = 28 m θ 1 = 0 d 2 = 11 m θ 2 = 26 Solution: Δ x 1 = d 1 (cos θ 1 ) = (28 m)(cos 0 ) = 28 m Δ y 1 = d 1 (sin θ 1 ) = (28 m)(sin 0 ) = 0 m Δ x 2 = d 2 (cos θ 2 ) = (11 m)[cos( 26 )] = 9 . 88673 m

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homework 04 – FIERRO, JEFFREY – Due: Jan 28 2008, 11:00 pm 2 Δ y 2 = d 2 (sin θ 2 ) = (11 m)[sin( 26 )] = 4 . 82208 m Δ x total = Δ x 1 + Δ x 2 = 28 m + 9 . 88673 m = 37 . 8867 m Δ y total = Δ y 1 + Δ y 2 = 0 m + ( 4 . 82208 m) = 4 . 82208 m d total = radicalBig x total ) 2 + (Δ y total ) 2 = radicalBig (37 . 8867 m) 2 + ( 4 . 82208 m) 2 = 38 . 1924 m Question 4, chap 3, sect 1. part 2 of 2 10 points b) At what angle to his original displace- ment is his total displacement (with counter- clockwise positive)?
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PHY 303K - Homework 4 - homework 04 FIERRO JEFFREY Due...

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