1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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=(vo)^2+2a(x-xo)
    ...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

    diazona

    User Avatar
    Homework Helper

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

    LowlyPion

    User Avatar
    Homework Helper

    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

    diazona

    User Avatar
    Homework Helper

    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,

    3.8=(.030)t,

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

    LowlyPion

    User Avatar
    Homework Helper

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

    It is decelerating after all.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook