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Homework Help: Find Acceleration with only Initial Velocity and Final position?

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A hockey puck sliding on a frozen lake comes to rest after 238 m. If its initial velocity is 3.8 m/s what is its acceleration if it is assumed constant? Answer in units of m/s^2

    How long is it in motion? What is its speed after traveling 180 m?

    2. Relevant equations

    x=xo + vot +1/2at^2

    3. The attempt at a solution

    ...v^2=(3.8)^2 + 2a (-238)

    but i don't know the velocity, cause i don't have the time!

    3.8 m/s * 1/238 m = time?

    I find that I can't do anything without the time! I'm so lost and confused. PLEASE HELP.
    Last edited: Jun 10, 2009
  2. jcsd
  3. Jun 10, 2009 #2


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    Homework Helper

    Here's a hint: what does "comes to rest" mean? ;-)
  4. Jun 10, 2009 #3


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    Welcome to PF.

    That equation works. Your final velocity is 0 so ...

    v2 = 3.82 = 2*a*238

    Then you have your acceleration a.

    With that v = a*t so t becomes easy to find.

    To find the speed at 180 m just use the equation you gave above to determine the speed at that distance.
  5. Jun 11, 2009 #4
    But then we have two variables, a, and v, how do I solve then? it doesn't work if you manipulate and substitute v =at because then the other variable, t, comes in.
  6. Jun 11, 2009 #5


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    No you don't, you only have one, a. Look carefully at the equation LowlyPion wrote for you.
  7. Jun 11, 2009 #6
    ok, i tried it, solved for a and I got .030 m/s^2 = a.

    i typed that into my homework answers, and it said it was incorrect (we have multiple times to try it)

    and when I do v=at,


    I get 125.3 seconds....seems a bit too much time. am i doing something wrong?
  8. Jun 11, 2009 #7


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    I trust you entered it as a = -.03 m/s2

    It is decelerating after all.
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