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Homework Help: Airplane on the runway

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A plane accelerates from rest at a constant rate of 5.00 m/s^2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time t_{TO} needed to take off?

    2. Relevant equations
    i think the formula i'm supposed to use is d = vt + 1/2at

    3. The attempt at a solution
    not exactly sure if this is the correct formula but thats what i thought i have to use. we know that d = 1800m and we know that initial velocity = 0 and a = 5 m/s^2 so then we isolate for t. thus we get t = 1800 / 2.5 = 720. however i think this might be wrong because i don't think its realistic for a plane to take 12 minutes on the run way to take off. any ideas?

    Also part of the question was What is the distance d_last traveled by the plane in the last second before taking off?

    I tried to use the same formula d = vt + 1/2at and since we figured out time from part a we sub (t - 1) into here and get distance traveled in the last second before taking off. not sure if any of this is right. can someone verify?

    Thank you
  2. jcsd
  3. Sep 15, 2007 #2


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    Homework Helper

    your equation is wrong. it should be:

    d = v0*t + (1/2)at^2

    you didn't have the t^2...

    pluggin in t-1... gives the distance travelled in the first t-1 seconds... not the last 1 second. you're close...
  4. Sep 15, 2007 #3
    oops ya i forgot the square. so then i get 26.8 seconds. does that make sense for a plane to take amount of time for lift of? i think it makes sense to me. not sure. also does this mean that we don't care about the required velocity for takeoff?

    hmmm the second part is tricky. i thoght that t - 1 would mean that lets say it took 10 seconds then t = 10 and the second before it is t - 1 = 9. i'm a bit confused on this part.
  5. Sep 15, 2007 #4


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    Your time looks correct.

    Suppose you know the distance travelled in the first 10 seconds... and the distance travelled in the first 9 seconds... you can get the distance in the last second by subtracting the two...
  6. Sep 15, 2007 #5
    Wow thanks for the fast reply. And thank you very much for all your help. i got both answers correct thanks to you. :)
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