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Deflection of sailboard

  1. Jan 16, 2014 #1

    You're sailing at 6.5ms^-1 when a wind gust hits, lasting 6.3s accelerating your board at 0.48ms^-2 at 35° to your original direction.
    Find the magnitude and direction of your displacement during the gust.

    3. The attempt at a solution


    Attached is the magnitude of the displacement. How should I go about determining the direction of the sail board?

    I ought to have been clearer in my question: I have determined the angle of displacement (6.394 degrees from the x-axis but I am unable to visualize a schematic as to what is going on)
    could someone give me a leg-up?
    Last edited: Jan 16, 2014
  2. jcsd
  3. Jan 16, 2014 #2
    You numeric results are correct. What sort of visualization do you have in mind?

    You could take a piece of paper, and draw a rectangular system of axes on it. Let at the beginning of the gust the boat be at the origin. Plot the trajectory during the gust.
  4. Jan 16, 2014 #3

    Where should the x and y component of the velocity go? what about the angle of displacement?
  5. Jan 16, 2014 #4
    Do not plot the velocity. Plot the position. That will give a visualization of the trajectory.
  6. Jan 16, 2014 #5
    Does this make sense?
  7. Jan 16, 2014 #6
    I would normally expect axes too be oriented differently, but that is not essential. The initial and final points are indicate correctly. But the trajectory should not be a straight line.
  8. Jan 16, 2014 #7

    Should it be curve due to the acceleration? If so, how should it be imposed- below the hypothenus and connecting the hypothenus?
  9. Jan 16, 2014 #8
    You do not have to guess. Take a number of times between zero and 6.3 seconds and compute the position at those times. Then connect them on your diagram.
  10. Jan 16, 2014 #9

    I get it. Thanks!
  11. Jan 16, 2014 #10
    Here is another consideration for you. Let the y axis be parallel to the direction of acceleration due to the gust, and x be perpendicular to it. What will the trajectory look like in these coordinates?
  12. Jan 16, 2014 #11
    Given s = vt;

    s = (48.749ms^-1i , 5.4633ms^-1j) 1.26s = 61.42m i, 6.88mj
    s = (48.749ms^-1i , 5.4633ms^-1j) 2(1.26s) = 122.8474 i, 13.76mj
    s = (48.749ms^-1i , 5.4633ms^-1j) 3(1.26s) = 232.16m i, 26mj
    s = (48.749ms^-1i , 5.4633ms^-1j) 4(1.26s) = 309m i, 34.67mj
    s = (48.749ms^-1i , 5.4633ms^-1j) 5(1.26s) = 386.946m i, 43.34mj
  13. Jan 16, 2014 #12
    isn't it similar to the one I just did?
  14. Jan 16, 2014 #13
    s = vt when there is no acceleration. Which is not the case in this problem. Besides, (48.749ms^-1i , 5.4633ms^-1j) is the final displacement, not the velocity.
  15. Jan 16, 2014 #14
    What a blunder to have taken displacement x time.

    If the acceleration is not constant, none of the kinematics equation can be used, isn't it?
  16. Jan 16, 2014 #15
    Why is the acceleration not constant? As I said, you got the correct numeric result originally. What can't you repeat the same thing, but for different times?
  17. Jan 16, 2014 #16

    Ok. I read your post wrongly. I read it as "no constant acceleration" instead of "no acceleration".

    The solution is then;

    ax = 0.393ms^-2
    vx = 6.5ms^-1

    ay = 0.2753ms^-2
    vy = 0ms^-1

    s = vit + 0.5at^2

    s = (6.5ms^-1, 0ms^-1)([t=0,1.26,6.3]s + 0.5(0.393ms^-2,0.2753ms^-2)([t=0,1.26,6.3]^2

    The above would give me the position of (i,j) from [t = 0,1.26,6.3] →from t=0 to 6.3 in segments of 1.26s
  18. Jan 16, 2014 #17
    Yes, that seems OK.

    But pay attention to #10 as well. You do not need to compute anything there. Just think.

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