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Vectors and displacement.

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    The minute hand on a watch is 2.0 cm in length. What is the displacement vector of the tip of the minute hand:
    a) From 8:00 to 8:20 A.M.?
    b) From 8:00 to 9:00 A.M.?


    2. Relevant equations
    N/A.


    3. The attempt at a solution
    So, first I drew a diagram of the clock and its initial position. Assuming 3:00 and 9:00 to be the x-axis, and 12:00 and 6:00 to be the Y-axis, and that 6:00 makes a 90° with 3:00, then when the minute hand points to 4:00 (denoting 8:20), it's angle, theta, would then make an angle of 30° with the negative x-axis. So, the displacement of the minute hand would then be:
    [itex]\vec{d} = 4 * 2.0 cm = 8 cm[/itex] [30° below the horizontal]
    Is this correct?

    As for b, the displacement is equal to 0, since the minute hand ends up exactly where it started after a change in distance from 8:00 to 9:00 A.M., correct?
     
  2. jcsd
  3. Oct 1, 2011 #2

    NascentOxygen

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    Staff: Mentor

    So you have a triangle with two sides of 2 cm, and the third side of 8 cm?

    Can you draw that?
     
  4. Oct 1, 2011 #3
    So, now I would need break the 2 cm side that makes an angle of 30 degrees with the negative x-axis into components, correct?
     
  5. Oct 1, 2011 #4

    NascentOxygen

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    Staff: Mentor

    At 20 mins past, the hand makes an angle with the positive x-axis.

    Any method that gets the right answer is okay. I broke the triangle up into two equal triangles.
     
    Last edited: Oct 1, 2011
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