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Vectors and Components-Part Two

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data
    A ship leaves Miami. It travels 1000 miles east, then 400 miles north, then 200 miles west. It ends up at Bermuda. How far and in what direction(heading) did it end up from it's starting point?


    2. Relevant equations
    Non applicable at this point.


    3. The attempt at a solution

    I drew a diagram of the three vectors. From my diagram, it is easy to tell that it ended up northeast of where it started. Other than that, I'm having a hard time getting beyond the "draw a diagram" part. What's the next step? The examples the teacher gave aren't really similar.

    My knowledge is limited for this topic, as this is high school physics, so thanks in advance for a descriptive answer! :smile:
     
  2. jcsd
  3. Dec 29, 2013 #2

    PhanthomJay

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    now that you have a sketch roughly to scale, draw in the resultant vector as an arrow between Miami and Bermuda. What is tte x comp of that vector as determined from your graph? Y comp? Use Pythagoras and your basic sohcahtoa trig equations to get the distance and angle from start to end point.
     
  4. Dec 29, 2013 #3
    Ah, I think I see the problem here. My sketch is awful. I thought I may be able to use the Pythagorean theorem, only my sketch looks like an incomplete rectangle. Shall I attach an image?

    Thanks! :smile:
     
  5. Dec 29, 2013 #4

    PhanthomJay

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    OK attach a sketch.... let's repeat...draw it to scale roughly...and be sure to draw a vector fom the start point of the first vector to the endpoint of the 3rd vector.....draw it in a different color than the others...that is the resultant vector you are looking for...from its x and y components, use Pythagoras to get the hypothenuse or displacement from start to finsh...use the tan function to get the angle..
     
  6. Dec 29, 2013 #5
  7. Dec 29, 2013 #6

    PhanthomJay

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    Sketch is fine. See that blue vector you drew from Miami to Bermuda...consider it the hypotenuse of a right triangle with height of 400 and base dimension length of ?? ?
     
  8. Dec 29, 2013 #7
    Ah, okay. So just divide it so that it IS a triangle. Is the base simply 800 miles?
     
  9. Dec 29, 2013 #8

    PhanthomJay

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    Yes!
     
  10. Dec 29, 2013 #9
    Okay, that seemed alarmingly easy. If I used the Pythagorean theorem, the resultant would be approximately 894.4 mi. Is that all for the first half of the problem?

    Thanks!
     
  11. Dec 30, 2013 #10

    PhanthomJay

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    yes that value is correct, but it should be rounded off a bit. Now you need to calculate the heading....
     
  12. Dec 30, 2013 #11
    So just 894 mi, then. Am I calculating the top angle of the triangle?
     
  13. Jan 5, 2014 #12
    Could you please confirm the next step? Thanks!
     
  14. Jan 5, 2014 #13
    894.4 miles should be fine .

    Now you need to calculate the angle which the line joining the initial and final positions(resultant vector) makes with the horizontal .
     
    Last edited: Jan 5, 2014
  15. Apr 1, 2014 #14
    Wow, I forgot about this thread for quite some time. I am so glad I attached the sketch. Would it be a 30 degree angle? Thank you!
     
  16. Apr 2, 2014 #15

    PhanthomJay

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    I think your sketch has long since disappeared, but if the y comp is 400 and the x comp is 800 and the hypotenuse is 894.4, you are not calculating the angle correctly. Use sohcahtoa! And it is best to calculate theta using the x and y comps for the trig calc, to avoid round off errors.
     
  17. Apr 2, 2014 #16
    I found it in post #5. :) I forgot what angle we're calculating. I think I have it straight now. Could I use sin(inv)=800/894.4?
     
  18. Apr 2, 2014 #17

    PhanthomJay

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    By doing it that way, what angle are you calculating? You have got to be sure of that. Then the actual direction of the resultant vector should be specified by a heading, for example, so many degrees East of North.
     
  19. Apr 2, 2014 #18
    I would be calculating the acute angle in between the resultant vector and the base. The direction would be NE. Affirmative?
     
  20. Apr 2, 2014 #19

    PhanthomJay

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    Negative. Use soh cah toa....focus on toa...tangent of angle the resultant makes with base is opposite side over adjacent side....direction is not exactly northeast, northeast would be a 45 degree angle with horiz or vert.
     
  21. Apr 2, 2014 #20
    Okay, so right angle, but wrong equation and direction?
    tan=o/a
    tan=400/800
    =26.6.

    NNE?
     
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