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Determining Displacement, Adding Vectors

  1. Aug 10, 2014 #1
    1. The problem statement, all variables and given/known data

    I've been stuck on this problem for hours. Any help you can provide is greatly appreciated.

    From a lookout point, a hiker sees a small lake ahead of her. In order to get around it, she walks 4.5km in a straight line toward the end of the lake. She turns right making a 60° angle with her original path, and walks to a campsite 6.4km in the new direction. Determine her displacement from the lookout point when she has reached the campsite.

    2. Relevant equations

    c2 = a2 + b2 -2ab cosθ
    sin a/a = sin b/b

    3. The attempt at a solution

    c2 = 4.52 + 6.42 – 2(4.5)(6.4)cos60
    c2 = 20.25 + 40.96 – 28.8
    c2 = 32.41
    c = 5.69km

    sin θ = (sin 60/5.69) X 6.4
    sin θ = 0.974
    θ = 76.9°

    θ = 90 – 77 = 13°

    The hiker is 5.7km 13° above horizontal, right of her original heading from the lookout when she reached the campsite.

    The book's answer is 5.8km, 18° away from the horizontal from the lookout.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 10, 2014 #2

    jedishrfu

    Staff: Mentor

    Have you tried graphing the solution and then comparing it to your calculations to see where you may have gone wrong?

    Sometimes it helps with understanding things in a geometrical sense.
     
  4. Aug 10, 2014 #3
    I have. I've literally spent hours on this question. I'm trying to keep this correspondence course rolling with no instructor while working a full-time job and I'm at the point where I think the book is wrong.

    I'm guessing that's not the case but I just can't afford to spend any more time spinning my wheels. Can you see any flaws with my equation?

    It seems to go wrong right off the bat. This equation doesn't appear to be difficult but my answer for R=5.69km isn't the same as the book's answer of 5.8km. The book must be wrong. It's a very expensive book so if it's wrong, I'd like some cash back!

    Thank you for your time.
     
  5. Aug 10, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If you goes, on the second leg, 60 degrees from the from the first leg, the angle inside your triangle is 180- 60= 120 degrees, not 60.
     
  6. Aug 10, 2014 #5
    Thank you but the diagram in the book shows the inside angle is 60 degrees. Unless I've misunderstood what you're suggesting.

    I'd post a pic but I don’t know how.
     
  7. Aug 16, 2014 #6
    figured out how to post a pic and also realized that I messed up the formatting of the "2"s in my attempt at a solution so here it goes again:

    c2 = 4.52 + 6.42 – 2(4.5)(6.4)cos60
    c2 = 20.25 + 40.96 – 28.8
    c2 = 32.41
    c = 5.69km

    sin θ = (sin 60/5.69) X 6.4
    sin θ = 0.974
    θ = 76.9°

    θ = 90 – 77 = 13°

    The hiker is 5.7km 13° above horizontal, right of her original heading from the lookout when she reached the campsite.

    The book's answer is 5.8km, 18° away from the horizontal from the lookout.

    DSC_0317.jpg

    Can anyone help?

    thank you!
     
  8. Aug 16, 2014 #7

    NascentOxygen

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    Staff: Mentor

    Your triangle work is right. We don't have any information that relates the triangle to North or East, so you can't give any bearing relative to the cardinal points.

    There is perhaps a reference if we can take the grid pattern on the mud map to be aligned N upwards, but no such correspondence is indicated, so it would be unwise to blindly assume it. Usually.

    Perhaps as an exercise you could assume the vertical lines are, in fact, aligned N-S and see what the displacement angle is relative to East? You'll observe that the 4.5 km leg is not precisely northerly, but the grid can help you.
     
  9. Aug 16, 2014 #8

    NascentOxygen

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    Staff: Mentor

    The point others have made that the second leg is actually at 120 degrees to the direction of the first should be heeded, and brought to the attention of the person who set this homework task.
     
  10. Aug 17, 2014 #9
    This is a correspondence course of which there is no instructor. I don't believe they are looking for an answer relative to the cardinal points as the book's answer is "5.8km, 18° away from the horizontal from the lookout."

    My answer is "The hiker is 5.7km 13° above horizontal, right of her original heading from the lookout when she reached the campsite."

    Why are the answers different if my triangle work is right?
     
  11. Aug 17, 2014 #10

    NascentOxygen

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    Staff: Mentor

    My answer was perhaps unclear. Let's approach this another way....

    At what angle to the vertical is that 4.5 km leg?
     
  12. Aug 18, 2014 #11
    According to my protractor; 7 degrees from vertical.

    I think if I could figure out why my answer is 0.1 km different than the book's answer, the rest would work itself out.
     
  13. Aug 19, 2014 #12

    NascentOxygen

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    Staff: Mentor

    Okay, we'll go with that for the moment.

    Now knowing that the 4.5 km leg is not exactly aligned with the vertical grid lines, does that change your answer for the angle to the horizontal of the displacement line?

    Your answer for the magnitude is correct.
     
  14. Aug 21, 2014 #13
    ahh I see your point.

    I'll change my response to "When at the campsite, the hiker is 5.7km, 77° to the right of the lookout with respect to the 4.5 km path".

    Thank you very much for the perspective.
     
  15. Aug 22, 2014 #14

    NascentOxygen

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    Staff: Mentor

    BUT the textbook wants your answer to be expressed as degrees above the horizontal. So what would your answer be now?
     
  16. Aug 23, 2014 #15
    Well that's how they answered the question so I suppose it wouldn't hurt to follow suit.

    I know the hyp is 5.69 but I'm having a difficult time trying to figure out one of the other angles.

    DSC_0332.jpg

    Can you give me a hint?
     
  17. Aug 23, 2014 #16

    NascentOxygen

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    Staff: Mentor

    I thought I had? :wink:

    How many degrees is that acute angle between the "not quite vertical" path and the horizontal grid line at its base?
     
  18. Aug 23, 2014 #17
    lol

    I thought there must be a more scientific and accurate way than using my protractor.
     
  19. Aug 23, 2014 #18

    NascentOxygen

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    There usually is. You generally can't rely on diagrams being sketched accurately to scale, instead you should use the lengths marked on. For example, if you can see what the x-step is for a particular y-step, you can use trig and say the angle is that whose tan = Δy / Δx. In your case, the map is marked on a grid to allow you to estimate distances off that grid. A protractor should give a similar answer because the figure here is sketched to scale.
     
  20. Aug 23, 2014 #19
    Please excuse me but I'm just not certain I understand correctly.

    Are you saying that I should use the protractor answer or is there another way? We seem to have already established that my answer isn't going to match the answer provided in the back of the book (5.8km vs 5.7km).

    If I use my protractor to determine the angle from horizontal, it says 27 degrees above horizontal.

    I can't believe how long it's taking me to answer this one.... I'm going to need to bring a sleeping bag for my exam.
     
  21. Aug 23, 2014 #20

    NascentOxygen

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    Staff: Mentor

    Instead of using a protractor to measure the angles, you could use another means, viz., trigonometry, to calculate the angles. This involves using sine, cosine, or tangent to find angles when you know lengths of the sides. To determine those lengths of sides, you can "count squares" (and partial squares) on the grid.

    However, this seems new to you, so we are going to have to accept that using your protractor will suffice here. It's an improvement on your original answer, anyway. :wink:

    The reason you don't get the book's answer is because they deliberately drew it not to scale, so that you won't get close to the right answer if you just take measurements off their drawing.

    Good luck with your exam. Coffee and a cut lunch might be more appropriate?
     
    Last edited: Aug 23, 2014
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