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Homework Help: Grade 12 Calculus plane deceleration question

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    When an F-18 jet lands on the runway, it's position is defined by s(t)=189t-t^7/3 meters after t seconds

    calculate the braking time

    calculate the deceleration when the jet stops ( i think she means immediately before?)

    2. Relevant equations

    derivative

    3. The attempt at a solution

    s'(t)=189-7/3t^4/3
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2010 #2

    rock.freak667

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    What about the derivative, you want to find a t for which s'(t) is what value?
     
  4. Apr 6, 2010 #3
    @rock.freak667

    well i dunno wouldn't taking the derivative help me find how long it takes to stop?
     
  5. Apr 6, 2010 #4

    rock.freak667

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    Ok what does the derivative represent ?
     
  6. Apr 6, 2010 #5
    the derivative would represent the deceleration wouldn't it? because the function represents the time.
     
  7. Apr 6, 2010 #6
    ***distance/time
     
  8. Apr 6, 2010 #7

    rock.freak667

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    Which is velocity right? So if it is braking, what should the final velocity be?
     
  9. Apr 6, 2010 #8
  10. Apr 6, 2010 #9

    rock.freak667

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    Right, so what do you get for t when you equate s'(t) to zero?
     
  11. Apr 6, 2010 #10
    81=t^4/3

    so t=0.92448s

    so it takes 0.92448s for the plane to stop.

    now for to find the deceleration at an instantaneous pnt i have to take second derivative?
     
  12. Apr 6, 2010 #11

    rock.freak667

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    You have two differing statements. Check it over.
     
  13. Apr 6, 2010 #12
    well s'(t)=189-7/3t^4/3

    0=189-7/3t^4/3

    -189/(-7/3)=t^4/3

    81=t^4/3

    4/3root of 81= t

    t=0.92448s

    does that make sense?
     
  14. Apr 6, 2010 #13
    no it doesnt. 1.3333333333333root of 81 = 27s
     
  15. Apr 6, 2010 #14

    rock.freak667

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    Yes 27s works better.

    Now how would you get the acceleration from s'(t)?
     
  16. Apr 6, 2010 #15
    second derivative?

    s''(t)=(-28/9)t^1/3
     
  17. Apr 6, 2010 #16

    rock.freak667

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    and if it takes 27 seconds to stop, what is the deceleration at that instant?
     
  18. Apr 6, 2010 #17
    (-84/9) m/s^2

    thank you so much by the way for everything!

    is that right?
     
  19. Apr 6, 2010 #18
    (-84/9) m/s^2

    Thanks for everything by the way!

    is that right?
     
  20. Apr 6, 2010 #19

    rock.freak667

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    Yes that is correct.
     
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