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Homework Help: Finding magnitude and direction of displacement.

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data
    In wandering, a grizzly bear makes a displacement of 1563m due west, followed by a displacement of 3348m in a direction 32 degrees NW. What are the (a) the magnitude and (b) the direction of the displacement needed for the bear to return to its starting point? Specify the direction relative to due east.

    2. Relevant equations
    R2 = A2 + B2 – 2ABcosθ

    3. The attempt at a solution
    a. magnitude= (1563^2)+ (3348^2)-2(1563)(3348)cos32=2185.53 m
    b. SE
    I don't know if i am right though cuz i don't understand what they mean by "due east"
  2. jcsd
  3. Jul 26, 2011 #2


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    welcome to pf!

    hi bling-bling! welcome to pf! :smile:
    i don't understand what you mean by "32 degrees NW" :confused:

    "due east" means east …

    for example, south-east would be 45° south of due east, and north would be 90° north of due east :wink:
  4. Jul 26, 2011 #3
    i have no idea what they mean either, but thats what it says in my assignment. do u want me to post a copy of the assignment so u can look at it?
  5. Jul 26, 2011 #4


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    "32 degrees NW" is meaningless

    they must mean either "32 degrees North of West" or "32 degrees West of North" :confused:
  6. Jul 26, 2011 #5
    i thought it was like the bear went 1563m west then turned 32 degrees NW and headed 3348 m in that direction?
  7. Jul 26, 2011 #6


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    maybe :smile:

    can't tell :redface:

    the bear probably didn't know anyway :rolleyes:

    i expect he just saw a sign saying "bear right" :biggrin:
  8. Jul 26, 2011 #7
    haha. but are my answers right?
  9. Jul 26, 2011 #8


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    for the magnitude, shouldn't θ be greater than 90° ?

    and to get the direction, you'll need to use the sine formula

    (and i'm off to bed :zzz:)
  10. Jul 26, 2011 #9
    for a why would theta be greater than 90? and how do i use the sine formula for the direction???
  11. Jul 27, 2011 #10


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    hi bling-bling;! :smile:

    (just got up :zzz: …)
    θ is the internal angle of the triangle, and that is greater than 90° …

    the bear goes West, then he "bears" right to go North-West-ish, not North-East-ish

    another way of looking at it is that if θ was exactly 90°, the it would be Pythagoras, a2 = b2 + c2

    if you make the angle more than 90°, then that increases a2, so you need a plus, so -cosθ has to be positive, ie cosθ has to be negative, ie θ > 90°​

    (and you use the sine formula sinA/a = sinB/b because you now know all three sides and one angle, and the sine formula is a lot quicker and easier to use than the cosine formula :biggrin:)
    Last edited: Jul 27, 2011
  12. Jul 27, 2011 #11
    so i use the sine formula to find the direction but what i got for magnitude, is that right???
  13. Jul 27, 2011 #12


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    Your magnitude won't be correct since you used 32 degrees rather than [probably] 148 degrees.

    Draw a diagram - [at least in your mind].

    The first line is a long way due west - left.
    The next line is up and to the left at 32 degrees.

    Join the end of the second line to the start of the first [the starting point] and you get a triangle with a 148 degree angle and two very small angles.

    You want that third side - and while it will stretch a long way in the direction "a few degrees North of due West", the answer you want, the return journey, is just as far, a few degrees South of East.

  14. Aug 12, 2011 #13
    I also got this question for my AP Summer Class Work. If you didn't find out the answer yet I think I've managed to get it myself.

    I used the same Trig Ratio you used to get the answer just a different angle.

    So my Equation looked something like this: X^2=(3348^2)+(1563^2)-(2*3348*1563)-Cos(148)

    This gave me the answer of 1785m

    Since the question didn't ask for an angle of the displacement I just said that the displacement was south of east.

    Hopes this helps anyone else that has problems with this question.
  15. Aug 17, 2011 #14
    caleb did u get 1785m as the magnitude? and why did u put a negative sign infront of the cos(148) because that doesnt belong there.. right??
  16. Aug 17, 2011 #15
    because what i have as an example in my hw is that they use the angle that is given. for example, if they say that the bear turned 65 degrees, thats the angle they use in the formula, in my example.
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