it rotates 3/5 as many times as it did the previous minute for the remaining minutes.
i believe this is the sum of an infinite geometric series.
that formula is S = a / (1 - r)
in that formula, S equals the number of rotations and a = 80 and 1 - r = 1 - 3/5 = 2/5 = .4
the formula becomes S = 80 / .4 = 200.
this says the number of rotations will be 200 before the bicycle stops.
i used excel to determine how minutes it would take to reach 200 rotations.
the results of that analysis is shown below:
the answer has to be greater than 80 since it started at 80 and there would only be additional rotations after that.
the infinite geometric sequence formula is S = a / (1-r)
in your problem, a = 80 and r = 3/5
S = a/(1-r) becomes S = 80 / (2/5) = 80 * 5/2 = 40 * 5 = 200.
the bike wheel will have rotated 200 times before stopping.
the finite geometric series formula is Sn = a * (1 - r^n) / (1-r)
when a = 80, this becomes Sn = 80 * (1 - r^200) / (1-r)
you can see that, after 44 minutes have elapsed the number of rotations remains at 200, meaning that, effectively, the bike wheel has stopped rotating.
you can use the sum of a finite geometric series to see that this is true.
that formula is Sn = a * (1 - r^n) / (1 - r)
when n = 44, formula becomes S44 = 80 * (1 - .6^44) / .4.
solve for S44 to get:
S44 = 200
it doesn't matter how many minutes are elapsed, the number of rotations will remain at 200.
assume 500 minutes have elapsed.
the formula becomes S500 = 80 * (1 - .6^500) / .4
solve for S500 to get:
S500 = 200.
the number of rotations stops at 200 which means the bicycle has stopped rolling when 44 minutes have elapsed.
note that:
r = 3/5 is the same as r = .6
(1 - r) = (1 - 3/5) = 2/5 which is the same as 1-r = .4
(1 - r^n) = (1 - (3/5)^n) = 1 - .6^n