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A320 Jetliner- Takeoff distance.

  1. Aug 26, 2008 #1
    I was able to answer the first 3 questions. I provided them and their answers just in case those can be used in the fourth question (which I believe uses all three).

    1. The problem statement, all variables and given/known data
    Assume the takeoff speed for an Airbus A320 jetliner is 75 m/s. Velocity data measured during takeoff are as follows:

    t(s).......................V(m/s)
    0.......................... 0
    10.........................23
    20.........................46
    30 .........................69

    (a) What is the takeoff speed in miles per hour?
    167.77 mph


    (b) What is its acceleration?
    2.3 m/s2

    (c) At what time do the wheels leave the ground?
    32.6 s

    (d) For safety reasons, in case of an aborted takeoff, the runway must be three times the takeoff distance. What is the minimum runway length from which an A320 can safely take off?
    m


    2. Relevant equations
    at-V1= -V2
    X=V0*t + (at^2)/2
    (V1)^2= V0^2 + 2ax


    3. The attempt at a solution

    Well, by getting the rest of the answers I kind of assumed I would be able to figure this one out as well, but for some reason I keep hitting a dead-end. I have calculated the total distance moved, except I forgot to account for speed up time. Yet then I came to the dilema of not knowing how to find speed up time. How are we to know the acceleration (though I know it to be 2.3 m/s^2 but I don't know how to use it) if the velocity graph involves just a single slope (therefore making acceleration a straight line). How can we know acceleration. I'm guessing that the velocity vs time graph has no constant for apparently it continues speeding up until 75 where it takes off. Any suggestions?
     
  2. jcsd
  3. Aug 26, 2008 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    You know the intial speed, the takeoff speed and the acceleration - you want distance
    so rearrange (V1)^2= V0^2 + 2ax
     
  4. Aug 27, 2008 #3
    Thanks so much! I solved for x and did the calculations, but the answer continued to come up wrong. FINALLY I realized that I was forgetting to multiply by 3, probably the easiest step since it is given.

    Thanks again!

    ~Phoenix
     
  5. Sep 10, 2008 #4
    i'm having a problem with this also.. i have initial speed as 0, take off as 75, and acceleration as 2.3... when i get x its equal to 1,222.8 then multiplied by 3 equal to 3668.5.... whats the problem???
     
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