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Question concerning displacement and vectors

  1. Aug 26, 2014 #1
    [1] A car is driven 215 km west and then 85 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.



    [2] I have no idea what equations are necessary for this problem. I reread the chapter, but this problem in particular is stumping me.

    [3] Originally, I thought I could make a triangle by using the vector of the 215 km west, the 85 km southwest, and the distance between the origin and the head of the 85 km southwest vector as sides. I thought if I could split the big triangle into two right triangles, I could use trigonometric functions to find the displacement, to no avail...
     
    Last edited by a moderator: Aug 26, 2014
  2. jcsd
  3. Aug 26, 2014 #2

    Nathanael

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    Welcome to physics forums. I would suggest focusing on the "85 km southwest" part first.

    Can you break it up in to it's southern and western components?
     
  4. Aug 26, 2014 #3

    SteamKing

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    It's a problem in trigonometry, if you drew the correct figure. Remember to use the directions of the individual vectors to find at least one angle in the triangle. You already have the lengths of two of the sides:

    http://en.wikipedia.org/wiki/Solution_of_triangles
     
  5. Aug 26, 2014 #4
    This is great advice so far!

    While you could do vector addition by creating a triangle using the 215km W, 85km SW, and a hypotenuse connecting the origin to the SW tip, this could be tricky without CAD or unneeded trig functions.

    I also recommend breaking up the SW vector into its S and W components. This way, when doing vector addition, you'll have a nice right triangle.

    Do you see how to break the SW vector up? Imagine that's all you have for a moment, having it start at the origin. You know it's purely SW, so you know what angle it makes from the x axis. Knowing this angle and the length if the vector, you can draw (and calculate the lengths of) sides to create a triangle with the vector as the hypotenuse. Keep in mind you want one side purely in an x direction, and one in a y!

    This divides the vector into its components.
     
  6. Aug 26, 2014 #5
    I just figured out how to reply, sorry about that! And @Nathanael I can make it 85 km south and 85 km west correct?
    @SteamKing thanks for the link, I'm checking it out now!
    and @amadinger I thiiiiink I understood that a bit, thanks! :D
     
  7. Aug 26, 2014 #6

    jbriggs444

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    If you have a huge square that has sides that are 85 km long, how far is it from one corner to the corner that is diagonally opposite?
     
  8. Aug 26, 2014 #7
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