Determining Displacement Using Vector Components

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Homework Help Overview

The problem involves determining the average velocity of a plane traveling between two locations in Canada, using vector components to analyze its displacement. The journey includes specific compass directions and distances, which are central to the calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations of displacement using trigonometric functions and vector components. There are questions about the accuracy of the compass directions provided and the distances between locations, with some participants suggesting consulting a map for verification.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the provided information. Some guidance has been offered regarding the potential inaccuracies in the compass directions and distances, but no consensus has been reached on the correct answer.

Contextual Notes

Participants note discrepancies between their calculations and the book's answer, as well as differences in distance measurements from external sources. There is a mention of the challenges faced by the original poster in a correspondence course without immediate access to assistance.

killaI9BI
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Homework Statement



This problem has had me puzzled for hours. I really appreciate any help you can provide.

A light plane leaves Shelburne, NS and flies 195km [N15°W] to Saint John, NB. After picking up a passenger, it flies to Moncton, NB which is 149km [N33°E]. The entire trip took 3.75h.

Calculate the average velocity of the plane for the entire trip from Shelburne, NS to Moncton, NB.

Homework Equations



a2 + b2 = c2
tan Θ = y-coordinates/x-coordinates

The Attempt at a Solution



dsjx = (-195) X cos75 = (-50.47)km W
dsjy = 195 X sin75 = 188.36km N
dmx = 149 X cos57 = 81.15km E
dmy = 149 X sin57 = 124.96km N
dx = (-50.47) + 81.15 = 30.68km E
dy = 188.36 + 124.96 = 313.32km N

r2 = 30.682 + 313.322
r2 = 99110.68
r = 314.8
314.8/3.75 = 83.95/h

tan θ = 313.32/30.68 = 10.21
θ = 84.4

The average velocity is 84km/h W 84 N.

The books answer is 48km/h [W29°N]

 
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I agree with your answer. I'd guess one of the given pieces of information has a wrongly stated compass direction, but I haven't been able to figure out exactly what. If these are real places, consult a map.
 
Thank you. I did consult a map which says the distance between Shelburne NS and Moncton, NB is 262km which appears to be different than both mine and the books answers. I've double and triple checked the information provided and it's all accurate.

I'm doing grade 11 physics via correspondence with nobody to ask questions to. Questions like these sort of ruin my flow. Do you think I should submit this answer?
 
How do you interpret the heading W 84 N?
 
killaI9BI said:
Thank you. I did consult a map which says the distance between Shelburne NS and Moncton, NB is 262km which appears to be different than both mine and the books answers. I've double and triple checked the information provided and it's all accurate.
OK. The 33 degrees stated looks a bit low to me. Judging from Google maps it's more like 40 degrees, but that could be deceptive. Anyway, your answer is a lot closer to reality than the book answer. Submit it.
 
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olivermsun said:
How do you interpret the heading W 84 N?


Sorry for the delayed response

84 degrees North of West
 

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