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Formula for average acceleration

  1. Feb 7, 2005 #1
    A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the jet is moving with a speed of 400 m/s.What is the jet's acceleration, assuming it to be a constant acceleration?

    first i converted 4km to m. then i said that since at first he was traveling 300m/s, it must of taken him about 13.33 seconds to reach 400m/s. then i followed the formula for average acceleration, which is vf-vi/tf-ti, and i get a wrong answer. what am i doing wrong?
     
  2. jcsd
  3. Feb 7, 2005 #2

    dextercioby

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    HINT:Use Galieo Galilei's formula
    [tex]v^{2}_{fin}=v^{2}_{init}+2ad [/tex]

    Daniel.
     
  4. Feb 7, 2005 #3

    xanthym

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    You didn't properly account for acceleration and computed the elapsed time incorrectly. You can use the Galieo formula indicated in the previous msg, but you may not remember that formula on a test. Sometimes it's best to concentrate on a few "basic" formulas which are fairly inuitive and easy to remember. Thus the following solution, although longer, may also have a place in your suite of mathemagical tools. We start with the following basic formulas valid for constant acceleration A:

    [tex] V_f = V_i + A*t [/tex]
    [tex] D = V_i*t + (1/2)*A*t^2 [/tex]

    where "D" is distance, "Vi" is INITIAL velocity, "Vf" is FINAL velocity, and "t" is Elapsed Time. For your problem:
    D = (4 km) = (4000 m)
    Vi = (300 m/s)
    Vf = (400 m/s)

    Thus, we can substitute values and set up a system of 2 equations in 2 unknowns ("A" and "t"):
    (400) = (300) + A*t ::: Eq #1
    (4000) = (300)*(t) + (1/2)*A*(t^2) ::: Eq #2

    Solving for "t" in Eq #1 and placing the resulting expression in Eq #2:
    t = 100/A
    (4000) = (300)*(100/A) + (1/2)*A*((100/A)^2)
    40 = 300/A + (1/2)*A*(100/A)*(1/A)
    40 = 300/A + 50/A

    A = (8.75 m/sec^2)
     
  5. Feb 7, 2005 #4
    thank you very much for the replies guys, it was alot of help.
     
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