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Velocity component questions

  1. Jun 9, 2007 #1
    1. The problem statement, all variables and given/known data
    At an air show, a jet plane has velocity components v_x = 625 km/h and v_y = 415 km/h at time 3.85 s and v_x2 = 838 km/h and v_y2 = 365 km/h at time 6.52 }.

    2. Questions
    Question A.
    For this time interval, find the x component of the plane's average acceleration.
    {I have already used all of my tries on this one and have moved onto question B}

    Question B.
    For this time interval, find the y component of the plane's average acceleration.

    I tried 81.9 but it says it's wrong. ((415/3.85)+(362/6.52))/2 = 81.9??

    Question C.
    For this time interval, find the magnitude of its average acceleration.

    Question D.
    For this time interval, find the direction of its average acceleration.

    Please help, I have no idea how to go about solving the above questions. I'm not looking for the answer, just a little guidance.


    Last edited: Jun 9, 2007
  2. jcsd
  3. Jun 9, 2007 #2
    first define average acceleration.
    then, state the formula for average acceleration explicitly, then plug in the values

    can't understand what u are trying here ((415/3.85)+(362/6.52))/2 = 81.9 ???

    why divide the speed by time? acceleration is something different.

    actually first define acceleration here...
  4. Jun 9, 2007 #3
  5. Jun 9, 2007 #4
  6. Jun 9, 2007 #5
    When I enter "-18.73" it tells me it's the wrong answer. That was the last attempt at the question so I need to move on to part C.

    In part C I have no idea where to begin as the answer to part B was wrong....

    I'm lost.....

  7. Jun 9, 2007 #6
    but u forgot the units, ur equation is correct but the value at the end would not be 18,73 because the units do not correspond. one is km/h and the other is seconds
  8. Jun 9, 2007 #7
    Argh, I caught that on a few other problems. Dang it...

    Thanks for the help, it's really appreciated.

  9. Jun 9, 2007 #8
    can u now do the other two parts? C and D?
  10. Jun 9, 2007 #9
    Well, I am doing some research on them right now, trying to figure out how I can find the magnitude and direction from the two time's. I did find this list (http://www.grc.nasa.gov/WWW/K-12/airplane/translations.html) but it seems to just tell how to calculate average acceleration.

    I know the velocity is a vector, indicating it has both a magnitude and a direction. Part C is asking to find the magnitude in km/h indicating a straight velocity. I think I know everything to finish part C and D but something is not clicking.

    I'm a little lost. Could you possibly point me in the right direction?

    After converting my units (_s>_hr) I came up with the following answers:
    Part A: 287191.01 km/hr^2
    part B: -67415.73 km/hr^2
    Last edited: Jun 9, 2007
  11. Jun 9, 2007 #10
    after u calculate both of the accelerations (ax and ay) use pythagorean theorem to find the resultant acceleration and its direction
  12. Jun 9, 2007 #11
    Hmmm.... let me see if I got this right.

    To find the magnitude:
    x^2 = (287191.01 _km/_hr^2)^2 + (-67415.73 _km/_hr^2)^2
    x = .036172991 _m/_s^2 or -.036172991 _m/_s^2
    x = 486.802 _km/_hr^2 or -486.802 _km/_hr^2
    Which one should I use? I'm thinking 486.802 _km/_hr^2 because it's (+).

    Also, to find the direction I used:
    sin(AavgX/AavgY) = sin(287191.01/-67415.73) = -.07 deg.

    Are the above correct or am I way off?
  13. Jun 9, 2007 #12
    x comp of average acc:
    ((838-625)/3,6)/(6,52-3,85)=22,16 m/s^2 in (+x)

    y comp. of average acc.:
    ((365-415)/3,6)/(6,52-3,85)=5,202 m/s^2 in (-y)

    magnitude of average acc.:
    sqrt[(22,16^2)+(5,202^2)]=22,76 m/s^2

    direction of average acc.:
    arccos (22,16/22,76)=13,2 degrees
    13,2 degrees S of E
  14. Jun 9, 2007 #13
    Wow, I do suck at this stuff....

    Thanks for all the help and time. Just one little question, when calculating the x and y average acceleration, you used 3.6. Where did this number come from?

    Again, thank you for all of your help!
  15. Jun 10, 2007 #14
    when converting km/h to m/s u divide by 3,6
    1 km =1000 m
    and 1 h= 3600 sec.
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