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Homework Help: Direction of a vector (help with part c)

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A model sailboat is slowly sailing west across a pond at .33 m/s a gust of wind blowing at 28 degrees south of west gives the sailboat a constant acceleration of magnitude .30 m/s^2 during a time interval of 2.0 s (a) if the net force on the sailboat during the 2.0 s interval has magnitude .375 N, what is the sailboat's mass? (b) what is the new velocity of the boat after the 2.0-s gust of wind? (c) what is the new direction of the boat after the 2s gust of wind... ___ south of west.

    2. Relevant equations

    3. The attempt at a solution
    i have already determined a using f=ma .375 / .3 which is 1.25 kg

    i have part b... by using vf= vo + at which is .930 m/s

    but im very confused with c! i think that i should be using tan^-1 (Vy/Vx) but i have no idea what to plug in.
  2. jcsd
  3. Sep 20, 2009 #2
    Use superpostion. And ad the vectors up. One vector has a magnutiude of .33m/s @ 0 degrees. The second is the wind blowing on the boat then use your arctan to find the angle
  4. Sep 20, 2009 #3
    what do you mean superposition? well i understand that cos(0)(.33) is .33, sin(0)(.33) is 0. and then i dont understand what to do with the 28 degrees
  5. Sep 20, 2009 #4
    You are on the right path. The cos(0)*.33=.33 that gives you the x component of V1. Sin(0)*.33 gives you the y component of V1. Now take the magnutiude of V2 (.3) and break it down into its X & Y component using -28 degrees (its negative because it south aka below the x axis). The add the X component of V1 and V2 together and do the same for the Y component and uses those vaules to find the new direction.
  6. Sep 20, 2009 #5
  7. Sep 20, 2009 #6
    that makes sense.. and i tried what you said but it is incorrect! any other ideas?
    i added .33 with .264 and 0 with -.1408 i took the arctan of (.594/-.1408) and got -76.66
  8. Sep 20, 2009 #7
    try negative.33 and negative .3 since I would consider west negative x
  9. Sep 20, 2009 #8
    that did not work either :( please help!
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