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Help w/ acceleration formula

  1. Jan 28, 2004 #1
    This is not strictly a physics problem, but I thought this would be the best place to ask...

    I am having trouble calculating the time between events using an acceleration formula. I have the rate of deceleration, and would like to know the time until the next full unit occurs.
    Code (Text):

    d =  1.0000000 cigarettes
    v =  0.0069444 cig/min
    a = -0.0000017 cig/min/min

    d = v*t + (a*t*t)/2
    Am I using the right formula? Can anyone see a way to isolate time? Maybe suggest a variation on the formula that would have the effect I am looking for?
  2. jcsd
  3. Jan 28, 2004 #2
    Time until the next 'full unit' occurs? What, you mean when Cigs=0?
  4. Jan 28, 2004 #3
    Re: Acceleration Question

    You can solve for time by solving the equation quadratically.

    Your equation is correct; d is actually (x-x0), the 'distance' you travel from start point to end point. Usually it's the number of meters travelled; for your problem it's the number of cigarettes.

    You've got positive velocity and negative acceleration, so the shape of this curve (cigarettes on the vertical axis, minutes on the horizontal axis) is an inverted parabola, like the path of a football.

    If you wish to know the value of t when you've reached '2 cigarettes', then you solve

    (x-x0) = v0t + 0.5at^2


    x = 2 cig
    x0 = 1 cig (x0 is 'x-naught', or 'original x')
    v0 = 0.006944 cig/min ('original v')
    a = -0.0000017 cig/min/min
    t = ???

    1 = (0.006944)t - 0.000000085t^2


    0.000000085t^2 - 0.006944t + 1 = 0

    which is an equation of the same form as

    2t^2 + 6t - 8 = 0


    At^2 + Bt + C = 0

    and the values of t for which it is true are found by the quadratic formula,

    t = -B +/- sqrt(B^2 - 4AC)

    where +/- means (plus or minus),
    sqrt means 'square root of', and
    --------- means 'divided by'.

    For my 2t^2 example, the answers are
    t = 1 and t = 4.

    For the numbers you gave, making sure to remember the minus sign in front of B, I get:

    t = 144 minutes
    t = 8022 minutes which is almost six days

    Remember the curve is like a football flying through the air: it starts at one cigarette (or meter) high, arcs up through 2 cigs, and then higher still, peaks out at about 15 cigs high, and starts getting lower and lower, until it passes through the 2 cig mark (again) late in the fifth day

    Which is the problem. When I read your question, I assumed that the quantity of cigarettes referred to the number smoked *so far*, a cumulative number. The negative acceleration would refer to a person who was *cutting back*, reducing the rate at which they smoke. This is fine; negative accelerations make sense for smoking. BUT negative velocities do not. I cannot be *undoing* the number of cigs I have smoked at so many cigs per minute; it's nonsense.

    Unless you consider a celery stick, or a minute on a treadmill, or whatever, to be equal to one 'negative cigarette'...

    anyway, I put the roots at 144 and 8022; that's when the level hits two cigs. If you want to know where it hits zero cigs, redo the problem with d = (x - x0) = (0 - 1) instead of (2 - 1) like I did above; you should get about 8310 min and -141 min. (Yes, a negative number; extend the curve I've described *backwards* past the t = 0 line and see that it hits the horizontal axis back there, too: it's again a nonsense solution, given by the math but ruled out by the physics (or common sense) of the problem.)

    I've attached a screen capture of the graph I produced in Excel to help me solve this problem; had I had my TI-83 with me, I could've done it in a tenth of the time it took me this way.

    Have I answered your question?


    Attached Files:

    • cigs.jpg
      File size:
      10.1 KB
  5. Jan 30, 2004 #4

    The graph showed me exactly what the formula was doing and I realized that I am using the wrong formula. I should be doing a very basic acceleration formula and wasn't. I should have spent some more time doing some algebra and a little less time looking stuff up in books:

    (I'm not a mathemetician or physicist so forgive poor variable names)
    Code (Text):

      Meaning of variables:
      A  = acceleration
      d  = Change in position (distance)
      Vc = Velocity Change
      Vo = Velocity Original
      Ve = Velocity End
      t  = time

      Which Variables have:
      A  = have
      d  = have
      Vc = don't
      Vo = have
      Ve = don't
      t  = ???

      We know these formulas:
      Vc = Ve - Vo
      Ve^2 = Vo^2 + 2ad
      A  = Vc/t

       a = Vc/t
       a = (Ve-Vo)/t
       a = ( (Vo^2 + 2ad)^(1/2) - Vo )/t
      at = ( (Vo^2 + 2ad)^(1/2) - Vo )
       t = ( (Vo^2 + 2ad)^(1/2) - Vo )/a

      Using some actual numbers I have:

      Vo = 0.0069444 cig/min (~25 per day)
       a = -0.0000017 cig/min/min (~60 days to 0 cig/min)
       d = 1.0000000 cig
       t = ( (Vo^2 + 2ad)^(1/2) - Vo )/a
       t = ( (0.0069444^2 + 2*(-0.0000017)*1.0000000)^(1/2) - 0.0069444 ) / (-0.0000017)
       t = ( (0.0000482 - 0.0000034)^(1/2) - 0.0069444 ) / (-0.0000017)
       t = ( (0.0000448)^(1/2) - 0.0069444 ) / (-0.0000017)
       t = ( 0.00669512 - 0.0069444 ) / (-0.0000017)
       t = (-0.00024927) / (-0.0000017)
       t = 146.63267827

    Based on these numbers, the person attempting to reduce their smoking is allowed a cigarette in 147 minutes.
    Thanks very much for the help.

    Does the new math look correct?
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