Solving Displacement: A Route Analysis

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Homework Help Overview

The problem involves calculating the displacement of a carrier driving from her office to two towns, with specific directions and distances provided. The subject area pertains to vector analysis and trigonometry, particularly in the context of displacement and triangle properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the shape formed by the carrier's route, identifying it as a triangle. There are inquiries about the appropriate use of the cosine rule and Pythagorean theorem for non-right triangles. Questions arise regarding the angles involved and the correct interpretation of the given directions.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the angles and sides of the triangle formed by the carrier's route. Some guidance has been provided regarding the use of the cosine rule and sine law to find the lengths and angles, but there is no explicit consensus on the final answers yet.

Contextual Notes

Participants are navigating through potential misunderstandings about angles and triangle properties, with references to diagrams that illustrate their reasoning. There is an emphasis on ensuring the correctness of calculations and interpretations.

Eathrock
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Homework Statement



A carrier drives from her office 22.0 km northerly direction to town. Drives in direction of 60.0 degrees south of east for 47.0 km to another town. Whats her displacement from office?

Homework Equations





The Attempt at a Solution



I drew a diagram of her route but I do not know what to do next.
 
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What type of shape do you get?
 
A Traingle
 
Eathrock said:
A Traingle

cool. so do you know which side it is whose length you need? do you know the cosine rule?
 
Hypotunese is what I need, right?
Cosine rule is to determine the sides/anlges of a non right angle.
Could I use Pythagorean theorem?
 
Last edited:
Eathrock said:
Hypotunese is what I need, right?
Cosine rule is to determine the sides/anlges of a non right angle.
Could I use Pythagorean theorem?

In this case you don't have a right triangle, so no hypoteneuse... so you can't use pythagorean theorem. But you can use the cosine rule to get the length of the third side of that triangle...

Then you need the direction also... so you need the angle of that third side (measured from north, south, east or west... whichever is most convenient).
 
Okay. Let me try to work it out.
 
a^2=b^2+c^2-2bccosA
 
40.731
 
  • #10
Eathrock said:
40.731

careful... what angle do you have in the triangle... opposite the side you're calculating?
 
  • #11
I did a^2 = 22^2 + 47^2 - 2(22)(47)cos60 to get 40.731 and then I did
40.731^2 = 47^2 + 22^2 - 2(47)(22)cosB to get me 60
 
  • #12
Eathrock said:
I did a^2 = 22^2 + 47^2 - 2(22)(47)cos60 to get 40.731 and then I did
40.731^2 = 47^2 + 22^2 - 2(47)(22)cosB to get me 60

Yes, but in my diagram, the 60degrees, isn't the angle in the triangle... I get 90-60=30 degrees...
 
  • #13
How do you get 90 degrees?
 
  • #14
Eathrock said:
How do you get 90 degrees?

90 degrees is the angle between the path (22.0 km north) and the east axis... she drives 60 degrees south of east... that means her path is 60 degrees below the east axis... or in other words 30 degrees east of south...

For example... 10 degrees south of east, is 80 degrees east of south...
 
  • #15
Oh okay. So I should replace 60 with 30 in the formula I showed you earlier.
 
  • #17
Ok. I see. Make so much more sense now.
Is the side 30.530 and is that the answer?
 
  • #18
Eathrock said:
Ok. I see. Make so much more sense now.
Is the side 30.530 and is that the answer?

I'm getting 30.034km... you should also get the direction.
 
  • #19
Yes. You are right. 30.034 km.

How would you find the direction?
 
  • #20
Eathrock said:
Yes. You are right. 30.034 km.

How would you find the direction?

You can find the angle of the displacement from any of the 4 axes (north south east west). I've drawn in one here as x:

http://www.imagevimage.com/gallery.php?entry=images/triangle2.gif

So when you find x... the direction would be x degrees west of south... hope that makes sense...

To find x... you'd use the sine law...

I've drawn x in two places... because those two angles are equal (since the lines are parallel)... hope this makes sense...
 
  • #21
30.034^2 = 47^2 + 22^2 - 2(47)(22)cosB to get the angle?
 
  • #22
Eathrock said:
30.034^2 = 47^2 + 22^2 - 2(47)(22)cosB to get the angle?

By B, are you referring the angle I called X on the diagram? What is the side opposite X...
 
  • #23
Yeah...
Is this one better?

47^2 = 30.0394^2 + 22^2 - 2(30.0394)(22)cosB to get the angle
 
  • #24
Eathrock said:
Yeah...
Is this one better?

47^2 = 30.0394^2 + 22^2 - 2(30.0394)(22)cosB to get the angle

yup. that works. you can also use sine rule to get the angle... ie: sinB/47 = sin30/30.04
 
  • #25
I got 128.491, is that it for the problem?
 
  • #26
Eathrock said:
I got 128.491, is that it for the problem?

Yes, but I made a couple of mistakes in my drawing... first I pointed the vector in the wrong direction(for some reason I was pointing it backwards)... plus I didn't consider that the angle was obtuse... but I fixed it and here's an updated picture:

http://www.imagevimage.com/gallery.php?entry=images/2_triangle4.GIF

It was good that you used the cosine rule, because the sin rule doesn't give the right answer here...

So the magnitude is 30.0394km and the direction is 128.491 degrees east of north... I've labelled the resultant vector as R.

hope I didn't make more errors...

So when you have the option to use the cosine rule over the sine rule... use the cosine rule. :wink:
 
Last edited:
  • #27
Thank you! :)
 
  • #28
Eathrock said:
Thank you! :)

No prob. Don't trust my work... check for yourself... make sure it all checks out to you...
 

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