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Required acceleration for takeoff

  1. Sep 9, 2011 #1
    Ive come accross this question, and just pondering on is i am right or completely wrong?! The question is: A jumbo jet must reach a velocity of 360km/h for takeoff. It uses a 1.8km runway. What is the constant acceleration required for it to reach the velocity before takeoff?

    I tried a few different times but sometimes I get different answers. So, I have the required velocity, the distance and so far the acceleration is x.
    So, first I calculated the time from 0 to 360km/h down a 1.8km runway. 360=1.8/t
    =360t=0.005 hours
    = 18 seconds to move down the runway.
    And now i do Velocity- acceleration x time
    360=A18
    360/18=A
    A=20km/hour or 5.56m/s

    Is this completely wrong:(?
     
  2. jcsd
  3. Sep 9, 2011 #2
    Hi...
    I think the proper course would be to recall the proper kinematic properties of constant acceleration and displacement...
    You are, no dobut acquainted with: [tex] v^2 = {v_0}^2+2as [/tex], where a is the constant acceleration, and s is the distance travelled.
    Try working up from here,
    Daniel
    P.S
    Check the units for acceleration, that should also point you in the proper direction
     
    Last edited: Sep 9, 2011
  4. Sep 9, 2011 #3
    i tried that before but my answer came to be 36,000!?
    As initial velocity was 0, i did: 360^2=0+2a(1.8)
    = 129,600=3.6a
    a=36,000
    What am I missing?
     
  5. Sep 9, 2011 #4
    Actually, yes...
    If you take heed to my post-script, you'd find that units must be conserved.
    You're given entries in Km & Km/h, but you're obviously looking for results in [tex] \frac{m}{s^2} [/tex]-This being the acceleration.
    Are you aware as to how this conversion should take place?(From Km-m, Km/h, m/s)...
    Plugging in the data properly should land you exactly where you need to be..
    Daniel
     
  6. Sep 9, 2011 #5
    still dont get it:( does that mean that 36,000 is right but just in the wrong units??
     
  7. Sep 9, 2011 #6
    Yes, that's actually what had happened.
    Look, in 1 km there are 1000m, right?
    And in one hour, there should be 3600 seconds(60sec*60min).
    If you then use this scale on your results(by multiplying by 1000 and dividing by 3600) you should be in the clear...
    However, be sure that you do this BEFORE you paste in your variables. First, convert your V and then your s(or x). Otherwise, you'd end-up with Km/h^2 which is a different and more complex story...
    Was that helpful?
    Good luck,
    Daniel
     
  8. Sep 10, 2011 #7
    ohh im sorta getting it now. thanks for all your help, much appreciated:)
     
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