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Homework Help: Average acceleration

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data
    the problem in a game asked me about average accelration
    [PLAIN]http://keddo.net/up/images/76ld3092n3dq9exr2rst.png [Broken]

    [PLAIN]http://keddo.net/up/images/xhb5m73r6xz5tduy6pe8.png [Broken]

    2. Relevant equations

    i think it is:

    3. The attempt at a solution

    i tried alot but all are wrong last i answered
    a=(2*10)/(50^2) and it was wrong
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 23, 2010 #2
    What are s and d in your equations? The picture shows s = (1/2)at^2, where did your d come from and what happened to time? Try solving for acceleration using their symbols first, then figure out your unknowns. Also, I think the picture should say, "Then I would calculate the time that the piston needs to travel the 5cm to the midpoint."
  4. Jun 23, 2010 #3
    thx for reply
    and those are more information:
    [PLAIN]http://store2.up-00.com/Jun10/B7o30680.png [Broken]
    [PLAIN]http://store2.up-00.com/Jun10/X8E30680.png [Broken]
    Last edited by a moderator: May 4, 2017
  5. Jun 24, 2010 #4
    plz anyone help me
  6. Jun 24, 2010 #5
    First of all, use the given symbols, as PiTHON said...

    [tex]s = \frac12 at^2 \implies a = 2s/t^2[/tex]

    The issue is what is t? This is the time it takes for the piston to move the distance s. (Focus on the third paragraph in the hint.)
  7. Jun 24, 2010 #6

    first find the velocity

    distance = 1000 cm = 10m/60s


    d = vt
    t = d/v
    t = 0.05m/0.166666666m/s
    t = 0.3s

    plug into the eq'n and you get a=2d/t^2 as PiTHON said

    a= 2(5)/0.3^2
  8. Jun 24, 2010 #7
    You know, the more I look at the "help" this computer program is giving you, the more confused I get -- it's quite sloppy. It uses both s and d for distance. It says the movement is "uniform" as though that were more specific than "evenly accelerated" which it is not. Why are we being asked about the "median" (which is different than average) accleration if we are just going to assume it's constant? And to top it all off, why is this problem even using d = 0.5at2? This is really a simple harmonic motion problem.

    Anyway, the formula given is a = 2d/t2. This formula is only valid if the object starts from rest. (:bugeye: -- didn't see that mentioned in the explanation? It's not there.) We are going to assume the hub of the piston starts at the top (where the velocity is momentarily zero). We are going to assume the hub undergoes constant acceleration for the first quarter cycle. (Why? In the first quarter, the hub is speeding up, but in the second quarter it is slowing down. Clearly we can't have constant acceleration through the whole period. I guess we are supposed to assume constant acceleration in the first quarter, then constant deceleration in the second.) The distance covered in the first quarter cycle is 5 centimeters, or 0.05 meters.

    What about the time? This is what I don't get. The second paragraph of the hint says "it is imperative that you figure out how many times the piston rises and falls per second..." Doesn't it just say that above? "The piston ... rises and falls 50 times per second." You can't "figure it out" -- it has to be given.

    Since the hub is moving up and down and up again (a full cycle) 50 times each second, it takes 0.02 seconds per cycle. But remember, we are only interested in one-quarter of that cycle, so t = 0.005 seconds.

    Put it all together and you get 4000 m/s2. Try it and see if it works.

    If I were to do this in a way that seems proper to me, we would assume simple harmonic motion, in which the maximum acceleration would be a = 4π2d/(4t)2 = 4935 m/s2. But this is the maximum. To get the median we should divide by sqrt(2), which is 3489 m/s2. Which is actually pretty close to the other way.
  9. Jun 25, 2010 #8
    thx all very much
    the correct answer:
    4000 m/s2
    thx dulrich

    and i think there are alt of mistakes in describing the problem

    and i ynderstand u thx
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