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Homework Help: Where is my mistake?

  1. Jan 14, 2005 #1
    The Warrior can take off when its airspeed (speed of the air flowing over the wing) is equal to 100 km/hr. What is the length of runway, in meters, required for the plane to take off if there is a 12 km/hr head wind? Note: the shortest runway at the Rochester International Airport has a length of 1000 m. Take the acceleration from the the answer of the previous question. The acceleration from past problem was .444444444444444m/s^2

    Okay so first I set
    100=-12+.444444444t
    then 112=.444444444t
    convert 112 to m/s
    31.1111111=.4444444444t
    t=70
    then I use that time in
    x=-3.333t+(1/2)(.44444444)t^2
    and solve for x right? and I get
    855.57m as the length of the runway but this is incorrect so where exactly is my mistake?
     
  2. jcsd
  3. Jan 14, 2005 #2

    BobG

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    Science Advisor
    Homework Helper

    A head wind means the wind is hitting the windshield (and the front of the wings). A tail wind means the wind is hitting the tail (and the back of the wings).

    In other words, while sitting still, the plane has an airspeed of 12 km/hr (not -12 km/hr)
     
  4. Jan 14, 2005 #3
    so then...
    100=12+.44444t
    88=.444444t
    88/3.6/.4444444=t
    t=55
    so
    x=3.33333(55)+(1/2)(.444444)(55)^2
    and
    x=855.5
    lol same answer and it is still wrong so what is my mistake now.
     
  5. Jan 14, 2005 #4
    Okay nm I figured it out the wind speed was only providing lift it was not acting on the plane as if the plane was a particle so after solving for T=55 I simply needed to plug it into x=(1/2)(.44444)t^2.

    That was tricky
     
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