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Sailboat Direction - Kinematics in Two Directions

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A sailboat is traveling east at 5.1m/s . A sudden gust of wind gives the boat an acceleration = (0.80 m/s2}, 40° north of east).

    What is the boat's direction 6s later when the gust subsides?


    2. Relevant equations
    Vf = V0 + at


    3. The attempt at a solution

    To solve I simply created X and Y components of the velocity after those six seconds, in order to use the arctan function to then find my angle:

    Vfx = Vox+at
    Vfx = 5.1 m/s + (6s)* (0.80cos40) = 8.777013327 m/s (x)

    Vfy = 0 m/s +(6s) * (0.80sin40) = 3.085380526 m/s (y)

    To find the angle, I then simply input these values into the arctan function:

    tan-1(vy/vx) = θ
    tan-1(3.085/8.777) = 19°, north of east

    This answer, however, is incorrect. Do you have any advice on my approach, or anything that I may be negating my calculations?
     
  2. jcsd
  3. Sep 16, 2012 #2
    Ok here is just a suggestion. Try using 50 as the angle and recalculate the velocities. This I think they were trying to say that the angle of the wind gust it makes with the north is 40 degrees.
     
  4. Sep 16, 2012 #3
    I appreicate that, but there is a first part of the question that asks for the velocity after 7.7 seconds, in which I used 40° as my angle of reference. I arrived at the correct answer for that. It also says "north of east," which I am sure means north of the negative x-axis (east). I believe that were it phrased otherwise (east of north), that would mean 40° east (to the left) of the positive y-axis, so I don't believe the angles are the issue, unless someone else feels otherwise.
     
  5. Sep 16, 2012 #4

    HallsofIvy

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    No, "40 degrees N of east" means starting from the east (the x-axis) measure toward the north (counter clockwise) 40 degrees. That is what bimbambaby is doing.
     
  6. Sep 16, 2012 #5
    Halls of Ivy -

    Any idea why the method I employed wouldn't work? I've calculated my velocity components correctly, and that should provide the correct answer, in theory.
     
  7. Sep 16, 2012 #6
    The situation is not like of projectile or free fall.
    Once the engine and the wind stops, we assume it to be stopped(maybe it takes a few seconds)

    I think the question ask is "where", in direction of x and y.
     
  8. Sep 16, 2012 #7
    Azizlwl -

    I'm not quite sure that I follow your meaning? We're determining the direction of the sailboat in terms of angle relative to the negative (east) x-axis. That is the directional component of the vector.
     
  9. Sep 16, 2012 #8
    We assume the engine is giving the boat a constant velocity due east.
    The wind too giving a force that equivalent to acceleration.

    We can find the position of the boat after 6s.

    After the gust subsides surely the boat is heading east.
     
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