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Finding Acceleration

  1. Sep 3, 2010 #1
    1. The problem statement, all variables and given/known data
    aircraft has liftoff speed of 130km/hr. What minimum constant acceleration does it require if the aircraft is to be airborne after a run of 271m? answer is in m/s^2


    2. Relevant equations
    V^2 = Vi^2 + 2a(deltax)
    a=(V^2)/Vi^2 + 2(deltax)

    3. The attempt at a solution
    (130^2)/O^2 + 2(271) = a
    a = 28.79 km/hr = 230.24 m/s^2
     
  2. jcsd
  3. Sep 3, 2010 #2

    Filip Larsen

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    The 2nd equation under 2) is not correct. The first equation is ok, so you just need to be careful how you isolate a from it. You should also consider what unit of the values you will use before inserting them (hint: convert km/h to m/s before inserting the value).
     
  4. Sep 3, 2010 #3
    is this right then

    (V^2) - (Vi^2)/(deltax) = 2a
     
  5. Sep 4, 2010 #4

    Filip Larsen

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    Close. You may mean the right thing, but as you have written it, it misses a set of parenthesis around the the two speeds. Alternatively you can insert vi=0 into the first equation before you start isolating for a.
     
  6. Sep 5, 2010 #5
    so if Vi = 0 then the equation is (V^2)/(deltax) = 2a and 130km/hr = 36.11 m/s so
    (36.11 m/s)^2/271m = 2a
    4.812 = 2a
    a = 2.41 m/s

    I feel like i did it right but I need the final units in m/s^2 and isn't this in m^2/s^2
     
  7. Sep 5, 2010 #6

    Filip Larsen

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    The square of velocity in unit m/s has unit m2/s2 which then is divided by distance in unit m, so the resulting unit is m/s2 as it should.
     
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