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Kinematic problem

  1. Mar 18, 2012 #1
    Hello,

    I'am doing some selfstudy to aqcuire more experience with mechanics. I got a geat book I'am currently reading. But got stuck in a question I can't answer...


    I hope you guys can get me on the good track, just need an approach to tackle te problem. Did already solve problems,but this one seems odd to me...So I wait for good tips. Thank you.

    11.65.png
     
  2. jcsd
  3. Mar 18, 2012 #2
    you have an initial velocity/position 200*10^3, 600
    you know that deceleration is constant
    you have a final velocity/position 50e3, 586

    find the time it takes to go 14 feet with constant deceleration from 200e3 to 50e3

    then, at 30 meters, he decelerates again, to zero m/s
    so find that time, and sum them up
     
  4. Mar 19, 2012 #3
    [tex]a=constant, v(t)=-a.t+v_{0}, x(t)=-a\frac{t^{2}}{2}+v_{0}.t+x_{0}[/tex]
    Initial conditions:
    [tex]v_{0}=55.56 m/s[/tex]
    [tex]x_{0}=0 m[/tex]
    Situation after [tex]t_{1} sec.[/tex]:
    [tex]v(t_{1})=13.89 m/s[/tex]
    [tex]x_(t_{1})=14 m[/tex]

    Rewrite equations:
    [tex]-a.t_{1}=13.89-55.56=-41.67 m/s[/tex]
    [tex]14=(-41.67).\frac{t_{1}}{2}+55.56.t_{1}=(-20.84+55.56).t_{1}[/tex]

    Solution:
    [tex]t_{1}=0.4032 sec.[/tex]
    This answer is the input to the velocity equation to get the constant decceleration, wich is [tex]103.3m/s^{2}[/tex]

    Travel time to reach the 30m position: [tex]\frac{556m}{13.89 m/s} = 40.03 s[/tex]

    Total travel time so far: [tex]40.43 s[/tex]
    The decceleration time tot reach ground level can be get with the same procedure.
    Total travel time: [tex]44.73 s[/tex]

    Thank for the good help, relative easy problem actually.
     
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