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Average acceleration problem

  1. Dec 1, 2007 #1
    Hi! I am new here and can't for the life of me figure out what I am doing wrong solving this problem, if someone could help that would be great!

    QUESTION:
    A bird takes 8.5 s to fly from position A (va= 4.4 m/s (31 ° S of E) to position B(vb= 7.8 m/s ( 25° N of E) along the path. Find the bird's average acceleration.


    MY ANSWER
    vax/va = cos31
    vax = 3.8 m/s

    vay/va = sin31
    vay = -2.3 m/s

    vbx/vb = cos25
    vbx = 7.1 m/s

    vby/vb = sin32
    vby = 3.3 m/s

    ax = vbx - vax / t
    ax = 7.1 m/s - 3.8 m/s / 8.5s
    ax = 0.4 m/s(2)

    ay = vby - vbx / t
    ay = 3.3 m/s + 2.3 m/s / 8.5s
    ay = 0.7 m/s(2)

    a(2) = ax(2) + ay(2)
    a(2) = 0.16 + 0.49
    a = 0.8 m/s(2)

    tan(theta) = ay / ax
    tan(theta) = 0.7 m/s(2) / 0.4 m/s(2)
    tan(theta) = 1.75
    theta = 60 degrees

    Therefore the bird's acerage acceleration is 0.8 m/s 60 degrees N of E.
     
  2. jcsd
  3. Dec 2, 2007 #2
    What's the correct answer?
     
  4. Dec 2, 2007 #3
    The correct answer isn't actually listed so I'm not sure what it is, my answer just does not seem right to me as the angle seems wrong.
     
  5. Dec 2, 2007 #4
    Hmm, I came up with .8 m/s/s also, and I did that even before I looked at your answer, so I don't know.
     
  6. Dec 2, 2007 #5
    Well I thought I was correct with the .8 m/s/s but what I really wasn't sure about was the 60 degrees N of E. Thanks for trying it out btw!
     
  7. Dec 2, 2007 #6
    I also used v=at and x=.5at^2. I now came up with 1.5 m/s/s at 60 degrees. So now I'm stuck between which answer it is.. But I do believe it's 60 degrees.
     
  8. Dec 2, 2007 #7
    Hmm how did you come up with 1.5 m/s/s if I may ask?
     
  9. Dec 2, 2007 #8
    I took the x and y components of each vector and used x=vt. I then found the x and y components of the distance. I found the change in distances of each vector to get the resultant vector (distance). I then used x = .5at^2 and used the time above to get 1.5 m/s/s at 60 degrees. But I think I did something wrong with the math because you'd think it'd come out to be .8 m/s/s again.
     
  10. Dec 2, 2007 #9
    Great thanks so much for the help, I'll do it over again just to see if I get 0.8 or 1.5, but I'm glad we come out with the same angle.
     
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