An animal climbing something (forward/backward slide): is there a generalized solution?

Question

A snail is at the bottom of a 30 foot well. Every hour the snail is able to climb up 3 feet, then immediately slide back down 2 feet. How many hours does it take for the snail to get out of the well?

I know the answer. But I also like generalizing things. Is there a general formula for resolving this kind of puzzle? The formula must work in every single particular case of general distance, speed up, and speed down.

Answer 1

Assuming the animal doesn't leave the well early enough for the problem to be trivial, a working formula would look like this:

$t(h_0+kd+r) = t_0 + kp + τ(r)$

where $h_0$ is the height of the first "peak" (3 feet), $t_0$ is the time at which it occurs (1 hour), $p$ is the time between peaks (also 1 hour), $d$ is the distance between peaks (1 foot), and $τ$ is an auxiliary function $[0,d) \mapsto [0,p)$ that describes the times at which intermediate distances are reached, for cases like "this riddle with 30 feet replaced by 30.5 feet" ($τ(0)=0,τ(0<x<1) = \frac{2+x}{3}$).

Applying this formula yields

$t(3+27×1+0) = 1+27×1+0 = 28$ hours, as desired

Answer 2

To calculate the number of hours it takes for the snail to get out of the well, we can use the following formula:

This is a function of javascript for general cases:
function snail(column, hourClimb, hourFall) {
const days = (column - hourFall) / (hourClimb - hourFall);
return days < 1 ? 1 : Math.ceil(days);
}

Let's apply this formula to the current scenario:

Number of hours = (30 - 3) / (3 - 2) + 1 = 27 / 1 + 1 = 27 + 1 = 28

Answer 3

t = d / (pv - nv)

Where,
t = time
d = distance
pv = positive velocity
nv = negative velocity