Determining Displacement Using Vector Components

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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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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?
 
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