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2-Dimensional Kinetic Motion

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A sailboat is traveling east at 5.0 m/s. A sudden gust of wind gives the boat an acceleration of 0.80 m/s^2 at 40 degrees north of east. What are the boat's speed and direction 6.0 s later when the gust subsides?


    2. Relevant equations
    xf=xi+vix(delta t)+1/2ax(delta t)^2
    vfx=vix+ax(delta t)


    3. The attempt at a solution
    This is really giving me a tough time.

    I started by trying to determine the velocity of both the x and y values.

    vfx=vix+ax(delta t)
    vfx=5m/s+(0)(6)
    vfx=5m/s

    vfy=5m/s+(.8)(6)
    vfy=9.8 m/s

    v=sqrt(vx^2+vy^2)
    v=sqrt(5^2+9.8^2)
    v=11.002 m/s

    I really need help determining the position after 6s.
     
  2. jcsd
  3. Oct 14, 2007 #2

    learningphysics

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    Homework Helper

    You don't have ax and ay right...

    The acceleration is 0.80m/s^2 at 40 degrees north of east... use trig to get ax and ay...

    Then as you did

    vfx = vix + (delta t)ax

    vfy = viy + (delta t)ay

    And just use your formula

    xf=xi+vix(delta t)+1/2ax(delta t)^2

    for the position... xi = 0. and use the corresponding equation in the y-direction.
     
  4. Oct 15, 2007 #3
    Thank you very much for your help...
    I think this is correct... could you please review it for me?

    cos40(.8)= ax= 0.613 m/s^2
    sin40(.8)= ay= 0.514 m/s^2

    Vfx= 5+(0.613)(6)= 8.68 m/s
    Vfy= 5+(0.514)(6)= 8.08 m/s
    V= sqrt(8.68^2+8.08^2)= 11.86 m/s

    xf=0+5(6)+.5(.613)(6)^2= 41.034 m
    yf=0+5(6)+.5(.514)(6)^2= 31.542 m

    tan-1(31.542/41.034)= 37.55 degrees NE
     
  5. Oct 15, 2007 #4

    learningphysics

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    Homework Helper

    Viy = 0. so

    Vfy= 0+(0.514)(6)= 3.084 m/s

    so V = sqrt(8.68^2+3.084^2)= 9.21m/s.

    and the direction is just arctan(3.084/8.68) = 19.56 degrees north of east.

    Reading over the question, i don't think you need position... just speed and direction...
     
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