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Magnitude and direction of acceleration

  1. Aug 14, 2008 #1
    1. The problem statement, all variables and given/known data

    An aircraft in level flight at a speed 210ms^-1 and traveling due north turns 14 degrees east. If the manoeuvre takes 1.2s to complete what are the magnitude and direction of the acceleration?



    2. Relevant equations

    a = v/t
    magnitude = (sqrt)x^2 +y^2



    3. The attempt at a solution

    210/1.2=175
     
  2. jcsd
  3. Aug 14, 2008 #2
    Your answer was 175 m/s^2? Everyone in the plane would be dead. Apply the equation in two dimensions, the X and Y directions and represent the velocity's in vector notation, not just as a magnitude.
     
    Last edited: Aug 14, 2008
  4. Aug 14, 2008 #3
    ok tried this but got an even bigger answer...

    Xold= 210ms^-1 Xnew= 0
    Yold= 0 Ynew=210sin(14)=50.80

    then I devided by 1.2 for each so: 210/1.2 and 50.80/1.2 then stuck those figures in the magnitude equation to get 180ms^-2
     
  5. Aug 14, 2008 #4
    Remember that acceleration is the change in velocity. That is dV/dt. First find the magnitudes of the change in the X and then the change in the Y. Xnew does not equal 0.

    I found an answer of 42.6 m/s^2, or about 4.3g's. Must be a fighter plane.
     
    Last edited: Aug 14, 2008
  6. Aug 14, 2008 #5
    how do I know/workout what Xnew is?
     
  7. Aug 14, 2008 #6
    Well, you know the original direction, and if the aircraft maintains the same speed(NOT VELOCITY), but the direction changes by 14 degrees, then some simple trig will show you its Vx = Vo*sin(angle). This is assuming that the plane is originally traveling in the Y direction. Have you drawn a picture? Draw the path of the plane with arrows, then join the arrow together tip to tail and it will make a triangle. It should make visualizing the math a little easier.

    Remember change in velocity is dv = (Vi - Vf)
     
  8. Aug 14, 2008 #7
    Remember vectors are a length and a direction. They can also be written in component form. Have you been taught to write vectors as a component times a unit vector along x-axis plus a component times a unit vector along the y-axis? If so, then try writing the before velocity vector and the after velocity vector in this form. You should have been taught how to get the components using trig. Once you have the before and after velocity vector in component form, then use what Topher925 has said about the definition of acceleration as the derivative of velocity to find average acceleration in component form. Once you have that, you can find the magnitude and direction from the components.
     
  9. Aug 14, 2008 #8
    Hmm, I must be having a ditsy moment. I did this question, by splitting the initial and final velocities into components of X and Y, and approximating the acceleration by (Vf-Vi)/t and came up with an acceleration of 8.3m/s^2.

    Only way my approach differs from yours (Topher) is that I use final value minus initial value when approximating small changes.

    Only sanity check I have is if it turns 14 degrees in 1.2 seconds, then it turns full circle in just over 30 seconds, doesn't sound much like a fighter jet pulling 4G's?

    You reckon I've done something wrong?
     
  10. Aug 14, 2008 #9
    I also note by your second post that you appear to be thinking along the right path, but you are messing up the calculation of your components. Are you making a drawing of the vectors before and after? Both X and Y for the after vector should change from the first vector.
     
  11. Aug 14, 2008 #10
    Thats can't be it, f-i or i-f it doesnt matter whats you take the magnitude.

    Maybe I'm the ditz, Ill check my numbers.
     
  12. Aug 14, 2008 #11
    I got the same answer as Topher. My question on your sanity check would be, How big is the circle that he went in those 30 seconds?
     
  13. Aug 14, 2008 #12
    Problem was my fault! Sorry if I added any confusion, one day I'm gonna learn how to use a calculator properly!:blushing:

    Regards,

    Barny
     
  14. Aug 15, 2008 #13
    ok i started again with this:

    I drew a diagram as suggested. We are starting off in the Y direction at 210ms^-1

    so Yold= 210ms^-1 Ynew=210cos(14)=204ms^-1
    Xold=0 Xnew=210sin(14)=50.80ms^-1

    then magnitude = (sqrt) 204^2 + 50.80^2 =210ms^-1 ??? same as what we started with......

    please help I cant get my head round this and I have an exam on monday!
     
  15. Aug 15, 2008 #14

    Redbelly98

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    Of course the magnitude is the same, since the speed doesn't change.

    Calculate the x component of acceleration:

    ax = (Xnew - Xold) / t

    and similarly for ay, the y component of acceleration.
     
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