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Lost physics student problem with vectors

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    These are two problems my professor has given me to do. I'm not very good at physics and could use a little help.

    1. Sir Killalot's foe Dead Metal is due north of Sir Killalot's position and at a distance of 12m. Dead Metal moves 15m at an angle of 30 degress to a new location. How far and in what direction should Sir Killalot move to intercept Dead Metal?

    He gave us the answers, but I'm not sure how to find the solution
    Answer: 23.4m @ 56.3 degrees

    2. A bug has a displace ment along a line which make an angle of 70 degrees to the x axis. (a) How far from the origin must the bug be located so that its displacement from the origin has an x component of 45cm? (b) What is the y component of the bug's displacement?
    Answer: (a) 132cm; (b) 124cm

    My professor doesn't really care to elaborate. I'm just wondering how I would go about solving these two problems? Any help would be much appreciated.
    Thanks

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 29, 2013 #2

    tiny-tim

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    Welcome to PF!

    Hi Wats! Welcome to PF! :wink:

    Draw the triangle, then use geometry (cosine and sine rules) to find the length and/or angle you're looking for.

    (alternatively, use x and y coordinates, if you know what that means)


    Show us what you get. :smile:
     
  4. Jan 29, 2013 #3
    I worked the both of them. I got number 2. I'm still having a little trouble with number 1. I'm not sure how to get the angle or direction he is going. Also I got 25.2 instead of 23.4. I like using the trig functions, but my professor, for whatever reasons, frowns upon them. He believe in adding and subtracting vectors, but adding and subtracting doesn't make much sense to me. I think the term he used for solving vector addition/subtraction problems was resolution components?
     
  5. Jan 29, 2013 #4
    This is what my professor prefers- Dxy=Dx+Dy or Dx=Dx1+Dx2 and Dy=Dy1+Dy2
    I don't particularly care for it, but I guess I have to play by his rules.
     
  6. Jan 29, 2013 #5

    tms

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    Place SK at the origin or your coordinate system. Draw a vector from there to DM's initial position. Call that [itex]\vec P_1[/itex]. What are the components of that vector? Now draw a vector from DM's initial position to his final position using the information given, and call it [itex]\vec P_2[/itex]. What are the components of that vector relative taking [itex]\vec P_1[/itex] as the origin? Now add the two vectors.
     
  7. Jan 30, 2013 #6

    tiny-tim

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    Hi Wats! :smile:

    (just got up :zzz:)​
    yes, that's what i call adding components (or adding coordinates)

    it's probably slightly more reliable …

    but not so pretty​
    show us what you did :smile:
     
  8. Jan 30, 2013 #7
    IMG_8157.jpg

    IMG_1770.jpg

    Sorry for such the large pictures. Tiny tim, could you explain to me how this coordinate process works? That's the part I don't understand about adding/subtracting vectors.
     
  9. Jan 30, 2013 #8

    tiny-tim

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    Hi Wats!

    In (x,y) coordinates, the first vector is (0,d1), and the second vector is (d2cosθ,d2sinθ).

    Then just add the coordinates. :wink:
     
  10. Jan 30, 2013 #9
    Oh ok. I think I'm beginning to understand. I should of drawn another triangle with the unknown vector being the hypotenuse. How would i find alpha though?
     
  11. Jan 30, 2013 #10

    tms

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    Consider a plane with the usual x-y coordinate system. Any point in the plane thus has two coordinates: x and y. A vector in the plane can be treated the same as a point: put its tail at the origin and its head will be at some point that will have x and y coordinates. These are called rectilinear coordinates.

    You can also identify any point (or vector) using polar coordinates, that is, a distance and an angle (conventionally measured counter-clockwise from the x axis), conventionally called r and θ.

    In this problem you are given the length and angle of a vector (the second vector), and you have to convert from polar to rectilinear coordinates in order to easily add the two vectors. The conversion is
    [tex]\begin{align}
    x &= r \cos\theta\\
    y &= r\sin\theta.
    \end{align}[/tex]
    All you have to do is convert the second vector and then add it to the first, then convert back to polar coordinates. That conversion should be obvious from the above equations.
     
  12. Jan 30, 2013 #11

    tiny-tim

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    You mean the angle of the third vector?

    If it's (A,B), then tanα = B/A. :wink:
     
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