The tunnelling speed of a machine is shown is modelled as $$v=\frac{k}{s}$$ where k is the constant with the units $m^2/hr$ and $s$ is the length of the tunnel already dugout.
I am to find a reasonable lower and upper bound for the time it takes to dig 5 meters when 1 meter is already dugout and $k=2$.
My attempt:
So I need to find time and I can do that by: $$v=\frac{k}{s}$$ $$\frac{ds}{dt}=\frac{k}{s}$$
Rearranging and integrating:
$$\int(s)ds = k\int dt$$ $$\frac{s^2}{2}=kt$$ $$t=\frac{s^2}{2k}$$
$k = 2$, therefore: $$t=\frac{s^2}{4}$$
$1 m$ was already dugout, and the machine moves $5m$ more. Therefore, time from $s=1$ to $s=6$ is: $$t=\frac{6^2}{4}-\frac{1^2}{4}=8.75 hours$$
However, I doubt the correctness of this.