Having trouble with finding this displacement vector

  • Thread starter ericcy
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  • #1
ericcy
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Homework Statement:
An airplane flies with a heading of [N58W] from Sydney, NS to Newcastle, NB, a distance of 618km. The airplane then flies 361km on a heading of [E35S] to New Glasgow, NS.

a) Determine the displacement of the airplane for the trip

b) In what direction will the plane have to fly in order to return directly to Sydney?

books answer was 4.0x10^2km [E28N] and W28S, couldn't figure out how to get these answers, not sure if they're wrong or if I'm missing something.
Relevant Equations:
a^2+b^2=c^2, sinTHETA=opp/hyp, cosTHETA=adj/hyp
Broke it into its components finding d1x, d1y, d2x, etc... Using those components I found drx to be 228.38km and dry to be 120.429km. Did Pythagoras to get 258km as the resultant displacement, heading N62W. I'm honestly lost. I'm doing the question the correct way, I just don't know what I'm missing. I'm supposed to solve by breaking it into components.
 

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  • #2
PeroK
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I must admit I don't know what ##N58W## means. Could you explain?

Can you post a diagram of your work?
 
  • #3
jbriggs444
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I must admit I don't know what ##N58W## means. Could you explain?

Can you post a diagram of your work?
A trip to Google suggests that a notation of "N58°W" denotes an angle 58 degrees west of due north.
 
  • #4
ericcy
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I must admit I don't know what ##N58W## means. Could you explain?

Can you post a diagram of your work?
I apologize if my work is messy or hard to read. Reply above explains what it means, thanks.
 

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  • #5
PeroK
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I apologize if my work is messy or hard to read. Reply above explains what it means, thanks.
What about ##E35S##?
 
  • #6
jbriggs444
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What about ##E35S##?
In general, you would start with one compass direction. Due North, East, South or West. In this case, east. Then you would shift by the indicated number of degrees in one of the two compass directions at right angles to the first. In this case, southward. So E35S is 35 degrees south of due east.
 
  • #7
PeroK
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In general, you would start with one compass direction. Due North, East, South or West. In this case, east. Then you would shift by the indicated number of degrees in one of the two compass directions at right angles to the first. In this case, southward. So E35S is 35 degrees south of due east.
With that assumption the answer looks wrong. You have a plane going roughly NW then back roughly SE and ending up NE?

I get roughly what the OP gets.
 
  • #8
PeroK
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I apologize if my work is messy or hard to read. Reply above explains what it means, thanks.
What you have is similar to my rough sketch.
 
  • #9
jbriggs444
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I apologize if my work is messy or hard to read. Reply above explains what it means, thanks.
As I read it, you start by obtaining the components for the first flight leg, ##d_1##.

You evaluate 618 km times the sin of 58 degrees for the x component to obtain 524 km. This seems correct. However, I see no attempt to try to apply a sign convention.

You evaluate 618 times the cosine of 58 degrees for the y component to obtain... something.

You proceed to do the same for the x and y components of the second leg, ##d_2## and get (295, 203)

You add component-wise to obtain ##d_R## as (328, 120)

You apply the pythagorean theorem to obtain a magnitude of 288 km.

It would be so much easier to check the math if I could actually read the numbers.
 
  • #10
ericcy
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You evaluate 618 km times the sin of 58 degrees for the x component to obtain 524 km. This seems correct. However, I see no attempt to try to apply a sign convention.
What do you mean a sign convention?

Also, is everything else alright? I can clarify some numbers for you if you need it.
 
  • #11
jbriggs444
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What do you mean a sign convention?
Like positive y = north, negative y = south, positive x = east, negative x = west.

You can choose to encode the direction in the sign that way. Or you can explicitly state the direction. Leaving the direction indication out entirely leaves us guessing at your intent. You should be writing for the reader.
Also, is everything else alright? I can clarify some numbers for you if you need it.
How about if you transcribe the whole thing (minus diagrams) for us rather than make us do it for you? I do not know that there are many numbers there that I can read.
 
  • #12
PeroK
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I think it's clear that if you go ##618km## roughly NW and back ##360km## roughly SE, then you end up roughly ##250km## NW and there's no way you can end up ##400km## roughly NE.
 
  • #13
ericcy
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d1x= sin58(618)=-524.093km, d1y=cos58(618)=327.49km

d2x= cos35(361)= 295.713km, d2y= sin35(361)=-207.061km

drx= -524.093+295.713= -228.38km
dry= 327.49-207.061= 120.429km

Using pythagoras with these values I get dr= 258km and an angle (using tan) N62W

@PeroK
 
  • #15
jbriggs444
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Have you looked on a map. Sydney to Newcastle is 160 km, not 618. And it's northeast (E58N maybe), not northwest.
 
  • #16
ericcy
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That's the data given in the question, so I dunno...
 
  • #17
PeroK
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Have you looked on a map. Sydney to Newcastle is 160 km, not 618. And it's northeast (E58N maybe), not northwest.
This is Sydney, Nova Scotia; not Sydney, Australia.
 
  • #18
jbriggs444
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This is Sydney, Nova Scotia; not Sydney, Australia.
Dang. Figured NS for New South Wales.
 
  • #19
PeroK
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That's the data given in the question, so I dunno...
I'd say you've done the problem correctly with the data given. Just move on.
 
  • #20
Steve4Physics
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258km as the resultant displacement, heading N62W.

Hi. You have found the displacement correctly. The book-answer is wrong (it happens sometimes). The displacement has magnitude 258km and direction N62W (or if preferred W28N).

But note that for part b) you are not asked for the direction of displacement. You are asked for the direction of the return journey.
 

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