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Help quick physics problem!

  1. Jul 3, 2008 #1
    1. The problem statement, all variables and given/known data

    A rock is pushed from rest across a sheet of ice, accelerating at 2.0 m/s2 over a distance of 20 cm. The rock then slides at a constant speed for 15.0 s until it reaches a rough section that causes the rock to stop in 5.5 s. (Assume two significant digits.)

    1. What is the speed of the rock when it reaches the rough section?
    2. At what rate does it slow down once it reaches the rough section?
    3. What is the total distance that the rock slides?

    2. Relevant equations
    V2=v1+a (change in t)

    3. The attempt at a solution
    v2=3000cm =30m

    Thats all i understood could some1 show me step by step how to do a b c pleaseee!
    thank you!
  2. jcsd
  3. Jul 3, 2008 #2


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    Hi Rade1990! :smile:
    But you don't know the change in t … you only know the change in x.
    The 15 seconds is when there's no change in v.
  4. Jul 3, 2008 #3
    i already did that:S?
  5. Jul 3, 2008 #4
    I suggest that you try to break it up into three sections. The accelerating section, the non-accelerating section and finally the rough (de)accelerating section. First step is to find the speed that the rock gains as it is pushed in the first section.

    As tiny-tim pointed out, you do not know the time that this happened for so you cannot use the equations listed. I suggest that you write down all the forms of the kinematic equations, what you know and what you are looking for, and based on that find the final speed.

    After you find this speed, I'll give you a hint on what to do next.
  6. Sep 28, 2008 #5
    I did this problem and got the speed for section one to 0.4m/s and i did not know how the find the next speed of section two we only given the time. thks
  7. Sep 29, 2008 #6


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    Welcome to PF!

    Hi hamd22! Welcome to PF! :smile:

    0.4? :confused:

    Show us what formula you used. :smile:
  8. Sep 29, 2008 #7
    sorry! i used v2^2=v1^2+2ad then i got the speed to be 0.8m/s for the first section now how do i find the speed for the next section we only have t=15s. thks
    Last edited: Sep 29, 2008
  9. Sep 29, 2008 #8


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    No … its √(0.8), isn't it?

    For the next section, you need an equation with v and a and t. :smile:
  10. Sep 29, 2008 #9
    the √(0.8) is 0.8, so do we use this as initial v and we have time too.Also we need a, is a zero? how do i find them? thks
    Last edited: Sep 29, 2008
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