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Relative motion problem

  1. Oct 25, 2004 #1
    A sailboat is traveling east at 5. m/s. A sudden gust of wind gives the boat an acceleration =0.80 m/s^2, 40 degree north of east
    1) what is the boat's speed later when the gust subsides?
    2) what is the boat's direction 6.0 s later when the gust subsides?

    I've a question
    I need to find the velocity when the gust wind
    so
    [text]vf = vi +at [\text]
    do the vi =0 or it's 5 m/s though?
    how should i visualize the problem?
    Thanks again
     
  2. jcsd
  3. Oct 26, 2004 #2
    I'm not really able to see how you can solve it if the 6.0s does not also apply to question 1)
    If so then you can calculate the change in velocity.
    You can then vectorize this change and see how it effects the x and y velocity.
    Since you have 5 m/s as x you can see how they affect eachother and you will get a new x vector. Combine the new x vector with the old y vector to get the final vector which has the boat's velocity and direction.
    Maybe there's an easier way to do it? Not sure though, are you working with vectors now?
     
  4. Oct 26, 2004 #3
    Assume the north to be the positive y-direction and east to be the positive x. Resolve the acceleration in both the directions and then apply the equation

    [tex]v = u +at [/tex] for each direction.

    So, for north, you have the inital velocity as zero while it is 5m/s for east. After calculating the final velocities in each direction, use pythagoras theorem and find the final velocity.

    The direction is given by [tex]\theta= tan^{-1}\frac{vy}{vx}[/tex]

    [tex]\Theta[/tex] is the angle made with the east.

    spacetime
    www.geocities.com/physics_all/index.html
     
  5. Oct 26, 2004 #4
    i still don't get how can I resolve to find the acceleration for each of (x, y) direction
    Isn't x acceleration =0 and the y direction is .80 m/s^2 ? I'm confused......... please help
     
  6. Oct 26, 2004 #5
    thank you very much, i got it now .. hihii :)
     
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