Find the position equation from this velocity equation
$$\displaystyle \frac{dr}{dt} = v_{t}\sqrt{1-e^{-v_{t}t}},$$
where $t$= time and $v_t$= constant
I'm wondering if there's a way to solve this without having a hyperbolic function in the solution. Thank you
For the problem you have $$\int dr = \int v_t\sqrt{1-e^{-v_tt}} dt$$ If you let $u = e^{-v_tt}$ then you get a new integral $$ \int dr = -\int \sqrt{1-e^u}du.$$ Again make a substitution and let $s = e^u$ and you again get a new integral $$\int dr = -\int\dfrac{\sqrt{1-s}}{s}ds$$. Make one more substitution and say $p = \sqrt{1-s}$. You will obtain a new integral $$ \int dr = 2\int \dfrac{p^2}{1-p^2}dp = 2\int (\dfrac{1}{1-p^2}-1) dp.$$ You can now use the fact that $$\int \dfrac{dp}{1-p^2} = \dfrac{1}{2}\dfrac{\log(1+p)}{\log(1-p)}$$ so your solution will be $$r = \dfrac{\log(1+p)}{\log(1-p)}-p+constant$$ and then make all the substitutions back in.