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Physics 2d Kinematics Question

  1. Jun 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A sailboat is sailing north with a speed of 12m/s. A wind begins to blow and 72s later the boat's velocity is 18 m/s, 25 degrees north of west. What is the magnitude and direction of the boat's average acceleration during this interval? What is the magnitude and direction of the boat's average velocity and displacement?


    2. Relevant equations

    v = d/t
    sin = opp/hyp
    cos = adj/hyp
    tan = opp/adj
    pythagorean theorem

    3. The attempt at a solution
    Ok I sketched a drawing and made triangles and tried to make them symmetric, but I only confused myself.
    x = (18m/s)(72s) = 1296 m
    sin 25 = x/1296m = 547.71 m
    cos 25 = x/1296m = 1174.57


    This is as far as I can get. Any help would be greatly appreciated. Thanks
     
  2. jcsd
  3. Jun 25, 2007 #2

    Mentz114

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    Gold Member

    To find the average acceleration you need

    (final velocity-initial velocity) / time elapsed.
    Resolve the velocities into N-S and E-W components and do it separately for the components. Then resolve the averages into a single vector.
     
  4. Jun 26, 2007 #3

    andrevdh

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    Homework Helper

    You can calculate the angle, [tex]\theta[/tex], between the 12 and 18 velocity vectors.

    The change in velocity joins the heads of these two vectors thereby completing the triangle.

    use

    [tex]c^2 = a^2 + b^2 - 2ab\cos(\theta)[/tex]

    to calculate the magnitude of the change in velocity.

    Then use

    [tex]\frac{\sin(\theta)}{\Delta v} = \frac{\sin(\alpha)}{18}[/tex]

    to determine its direction
     
  5. Jun 26, 2007 #4
    I'm stumped. i tried drawing another diagram to separate into an x and y component to find the total displacement of the boat. I ended up with 1174.57 for the x displacement and and 1095.44 for the y displacement. The x value may be right but I'm not sure about the y because when I solved for one part of the y value I got 547.71 and then I solved for the other part and got the same thing somehow. So I added the two together to get the total distance for y.
     
  6. Jun 27, 2007 #5

    andrevdh

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    Homework Helper

    The angle between the 12 and 18 velocity vectors is

    [tex]\theta = 90 - 25 = 65[/tex]

    degrees this enables you to claculate the magnitude of the change in velocity vector (it is the other side of the triangle closing the two vectors up - head to head) by using the cosine rule

    [tex]\Delta v^2 = 18^2 + 12^2 - 18 \times 12\ \cos(65^o)[/tex]

    to get the acceleration you just divide the magnitude of [tex]\Delta v[/tex] by 72 s.

    The direction of the average acceleration is also the same as the direction of [tex]\Delta v[/tex]. To obtain the direction of this vector [tex]\Delta v[/tex] just use the sine rule as I suggested in my previous post. The angle that you get from this wil then be the angle that the [tex]\Delta v[/tex] vector makes with the 12 vector. So its direction will be

    [tex]90^o - \alpha[/tex]

    south of east.
     
    Last edited: Jun 27, 2007
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