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Help with the acceleration

  1. Oct 5, 2004 #1
    please help me with this one

    A speed skater moving across frictionless ice at 8.70 hits a 4.90-m-wide patch of rough ice. She slows steadily, then continues on at 5.80 .

    What is her acceleration on the rough ice?
     
  2. jcsd
  3. Oct 5, 2004 #2

    Diane_

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    Homework Helper

    I assume she's moving at 8.70 m/s initially, and that we're dealing with algebra-based physics.

    First step is to identify the known information: you have an initial velocity of +8.70 m/s, a final velocity of +5.80 m/s, and a displacement of +4.90 m. You're looking for an acceleration.

    There is a direct approach to the answer, but let's take this from the basics. You know that acceleration is the change in velocity over time:

    a = (vf - v0)/t

    You could use this one if you had the time, but the only other piece of information you have is the displacement.

    You may recall that displacement under constant acceleration is given by:

    s = v0 * t + (1/2)a*t^2

    Not useful, as it requires the acceleration and time both. However, there's another fundamental formula that involves the displacement: by definition, the average velocity is

    Va = s/t

    Again, we need the time. Average velocity under constant acceleration can also be obtained by averaging the initial and final velocities:

    Va = (v0 + vf)/2

    So:

    s/t = (v0 + vf)/2

    t = 2s/(v0 + vf)

    At this point, you could plug in the numbers to find the time, then plug that into the definition of acceleration to find the answer you need. Let me suggest, though, that you do the algebra first - i.e. substitute this relation into the definition of acceleration to get a relation giving acceleration in terms of v0, vf, and s. The result will probably be familiar to you. Also remember that the result will assume constant acceleration, since one of our equations also assumes that.

    As a check - we're representing the vector nature of acceleration as a plus or minus here, and defining the direction she's currently travelling as '+'. Your answer for acceleration, then, should be negative. If you don't get that, then there's a problem somewhere.

    Hope this helps.
     
  4. Oct 5, 2004 #3
    wow, thank you very much, i used this equation, it's really helpful though
    (vf)^2=(vi)^2+2a(xf-xi)
    :)
     
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