# 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: 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 Ve^2 = Vo^2 + 2ad 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?