# Help w/ acceleration formula

by kavius
Tags: acceleration, formula, w or
 P: 3 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. 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?
 P: 949 Time until the next 'full unit' occurs? What, you mean when Cigs=0?
P: 3

## Help w/ acceleration formula

Thank-you.

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)
  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
A  = Vc/t

Therefore:
^^^^^^^^^
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?

 Related Discussions Introductory Physics Homework 3 General Physics 2 Introductory Physics Homework 14 Introductory Physics Homework 2 General Physics 5