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Deceptively simple kinematics

  1. Sep 13, 2009 #1
    So this problem is giving me some issues, my algebra may be rusty so it has me stumped.

    A Rocket experiences constant acceleration upward for 16s and then stops accelerating. It continues upward and at time 20s the distance from the ground is 5100 meters. Find the acceleration experienced during the 16s. What is the rockets speed as it passes a cloud at 5100m.


    The best I could do was get the equation 5100-x = .5a*162

    Where a is the acceleration during the 16s and 5100-x is equal to the distance traveled during that time.

    all this coming from the basic equation Sf=S0+V0t+.5at2

    This leads me to think i should use a system of equations to solve it but im really stumped as how to begin


    Any suggestions would be helpful, I'd really like to solve it on my own though so no solutions plz
     
  2. jcsd
  3. Sep 14, 2009 #2

    rl.bhat

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    Time 20 second is from the beginning or after 16 seconds?
     
  4. Sep 14, 2009 #3
    from the beginning so 4 seconds after the rocket stops
     
  5. Sep 14, 2009 #4
    Nevermind, figured it out
     
  6. Sep 14, 2009 #5

    rl.bhat

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    If x is the distance traveled in 4 seconds, then
    x = vf*4 - 0.5*g*16.........(1)
    vf = vi + 16*a...(2) Here vi = 0.
    So x = 64*a - 8*g....(3)
    You have already written
    5100 - x = 0.5*a*16^2...(4)
    Substitute the value of x and solve for a.
     
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