Solving the Last Practice Problem: A Coast Guard Challenge

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Homework Help Overview

The discussion revolves around a physics problem involving vector addition and subtraction related to a boat's distress call and its subsequent location. The problem requires determining the components of a vector representing the displacement from the distress call location to the boat's found position.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss resolving vectors into components, with one participant sharing their calculations for the components of the vectors involved. Questions arise about the correctness of the approach and the interpretation of vector directions.

Discussion Status

Some participants are providing feedback on the original poster's attempts, suggesting that there may be errors in the vector addition method used. There is an emphasis on the importance of drawing diagrams and reconsidering the angles used in the calculations.

Contextual Notes

One participant mentions being unable to reference class notes due to a recent hospital stay, which may impact their understanding of the problem setup and vector operations.

anuman
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I'm studying for an exam and can't seem to get this last practice problem! I've tried many ways but still keep getting the wrong answer. The professor has already given us the correct answer to the problems, but I want to know HOW to do it for the exam! Thanks in advance :)

A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat was later found is B = 30.0 km, 15.0° north of east.

12. What are the components of the vector from the point where the distress call was made to the point where the boat was found? In other words, what are the components of vector C = B - A?

x component y component
A)17.3 km, east 35.7 km, south
B)35.7 km, west 17.4 km, north
C)40.6 km, east 51.2 km, south
D)17.3 km, west 51.2 km, south
E)40.6 km, east 35.7 km, north

13. How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector C?
A)65.3 km B)39.7 km C)26.5 km D)54.0 km E)42.5 km
 
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How did you attempt the problem?

Did you first resolve each of the given vectors into their two components?

You should show us your attempt at the problem so we can see where you may have gone wrong.
 
This is how I attempted it...I have been in the hospital for a while so I don't have class notes.

45sin15=11.64 30sin15=7.76
45cos15=43.46 30cos1515=28.97

11.64+7.76=19.4 y-component
43.46+28.97=72.43 x-component
 
anuman said:
This is how I attempted it...I have been in the hospital for a while so I don't have class notes.

45sin15=11.64 30sin15=7.76
45cos15=43.46 30cos1515=28.97

11.64+7.76=19.4 y-component
43.46+28.97=72.43 x-component

Careful. Draw a diagram.

I don't think the terms combine in the way that you attempted.

The first vector is 15° to the right of North. This could also be written as 75° north of East if it would help you not to mix the Sines and Cosines.

Also you did a vector addition not a subtraction.

The easiest way to subtract is to add the inverse of the second vector.
 

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