{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PHY 303K - Homework 7

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

This preview shows pages 1–3. Sign up to view the full content.

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,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern