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Speed at Takeoff and Average Acceleration

  1. May 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Starting from rest an airplane lifts off the ground after 40s and 1.8km. What is its speed at takeoff and its average acceleration during the takeoff run?

    2. Relevant equations
    V= distance/time
    Acceleration= (Vf-Vi)/time

    3. The attempt at a solution
    The airplane travels 1.8km in 40 seconds. This would mean that when substituting in to the formula V=d/t the velocity at takeoff is .045km/s or 45m/s. I then plugged in this value into the acceleration formula a= (45-0)/40s and got an answer of 1.125 m/s^2.
    The answer key says that that answer should be 90m/s and then 2.25 m/s^s and I'm not sure why they are getting that answer?
     
  2. jcsd
  3. May 24, 2017 #2

    phinds

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    Are you sure about that? Think about the actual mechanics of an airplane takeoff from rest at one end of the runway to liftoff at the other end.
     
  4. May 24, 2017 #3
    The airplane would start off slow and then reach a faster speed at takeoff which leads me to believe that the average velocity would be slower than its speed at takeoff, but I'm not sure how to go about calculating that and that doesn't correspond to anything I have calculated.
     
  5. May 24, 2017 #4

    phinds

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    You are on the right track. What happens if you assume a smooth increase in speed from start to takeoff? You know the travel time and the travel distance.
     
  6. May 24, 2017 #5
    distance: x=vi+1/2at^2 is also a relevant equation, and since there is acceleration v=x/t (velocity=distance/time) is Not true
     
  7. May 24, 2017 #6
    Please double-check that equation; it isn't quite right.
    And you are correct that v = x/t does not apply if there is acceleration.

    Edit: Oh, I thought I was responding to the original poster. Oh well, I guess I'll stick with my response. :)
     
  8. May 24, 2017 #7

    haruspex

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    It is true given the right definitions of variables. In that equation, v is the average velocity. The error was in misapplying the equation to a different v.

    It might seem that I am just being pedantic, but I feel students can get confused if they think equations are right in some cases and wrong in others. The thing to stress is that an equation only means anything given the definitions of the variables it mentions. In this case, the student's notes could be emended to "vavg = Δx/Δt".
     
    Last edited: May 24, 2017
  9. May 24, 2017 #8
    you need to use uniformly accelerated motion equations instead. If you don't know what these are I suggest reading through the beginning of your text book.

    the equations you will need are position equation and velocity equation.
     
  10. May 25, 2017 #9
    True, I forgot to multiply velocity with time: x=vi*t+1/2at^2
     
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