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I'm a programmer and my maths skills were probably never sufficient for this problem. I have a nice curve that describes the velocity ($v$) of a "falling" object over time ($t$) that accelerates to a terminal velocity (note, this is not a physical simulation, but essentially a sort of "tweening" curve).

$$v = \frac{1}{-e^{2t}} + 1$$

From this, I'm hoping to find two expressions that:

  1. calculates the distance at time $t$
  2. calculates the time at which the object reaches distance $d$

My problems are basically that even if I knew how to express the problem correctly mathematically (and frankly I don't), I never completely understood how to go from an integral to a nice, simple expression. And compilers only care for expressions. Oh, and I've chosen an equation that features $e$.

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  • $\begingroup$ wolframalpha.com/input/?i=integrate+1%2F(-e%5E(2t))+%2B+1 $\endgroup$ – Matti P. Jul 17 '19 at 10:09
  • $\begingroup$ In addition to the integral, you'd also need the initial conditions (location and velocity in the beginning) $\endgroup$ – Matti P. Jul 17 '19 at 10:09
  • $\begingroup$ @MattiP I'm assuming starting with $v = 0$ and $d = 0$ $\endgroup$ – Andrew Jul 17 '19 at 10:18
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We have $$v(t) = \frac{dx}{dt} = 1-\frac{1}{e^{2t}}$$ So $$ x(t) = \int 1-{e^{-2t}} dt = t + \frac{1}{2}{e^{-2t}} + K $$ where $K$ is a constant that you can find solving $x(0)=x_0$.

The time at which the object is at the position $x$ is given by $$ x = t + \frac{1}{2}{e^{-2t}} + K $$ This is a bit tricky for me but Wolfram gives $$ t = \frac{1}{2}(W_n(-e^{2K-2x}) + x - K) $$ where $W_n$ is the analytic continuation of the product log function.

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