Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord!

Like

Report

Carnegie Mellon University

Like

Report

Problem 20 Medium Difficulty

A cruise ship sails due north at 4.50 $\mathrm{m} / \mathrm{s}$ while a Coast Guard patrol boat heads $45.0^{\circ}$ north of west at 5.20 $\mathrm{m} / \mathrm{s}$ . What are (a) the $x$ -component and $(\mathrm{b})$ y-component of the velocity of the cruise ship relative to the patrol boat?

Answer

(a) $V _ { \mathrm { CG } _ { \mathrm { x } } } = 3.68 \mathrm { ms } ^ { - 1 }$
(b) $V _ { \mathrm { CG } _ { \mathrm { y } } } = 0.82 \mathrm { m } \mathrm { s } ^ { - 1 }$

Discussion

You must be signed in to discuss.

Video Transcript

so we can say that four part I had the velocity vector for the cruise ship relative to the Coast Guard. The cruise ship relative to the guard is gonna be equaling the velocity of the cruise ship plus the velocity of the guard. And so the velocity of the cruise ship guard in the ex direction would be equaling the sub sea X plus a piece of G x. And this is gonna be equaling. Well, this would be zero. And so this would be equaling negative 5.20 meters per second, multiplied by co sign of 45 degrees. This is equaling negative 3.677 meters per second or weaken just round the velocity. The X component of the velocity of the cruise ship relative to the guard is equaling negative 3.6 eight meters per second for Part B. Then the velocity of the cruise ship relative to the Coast Guard in the UAE direction would be equaling the velocity of the cruise ship. And then why direction? Plus the velocity of the Coast Guard in the wind direction. This would be equaling 4.50 meters per second and then again, plus 5.20 meters per second sign 45 degrees. And so we find that the velocity of the cruise ship relative to the guard in the wind direction is equaling 8.18 meters per second. This would be our final answer for Part B and our final answer for part A. That is the end of the end of the solution. Thank you for watching.

Carnegie Mellon University
Top Physics 101 Mechanics Educators
Elyse G.

Cornell University

Farnaz M.

Other Schools

Zachary M.

Hope College

Jared E.

University of Winnipeg