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In my book (Young and Freedman's University Physics) they derive terminal velocity where and use these steps:

...we must find the relationship between velocity and time during the interval before the terminal speed is reached. We go back to Newton's second law, which we rewrite using $a_y = dv_y/dt$: $$m\frac{dv_y}{dt} = mg - kv_y$$ After rearranging terms and replacing $mg/k$ by $v_t$, we integrate both sides, noting that $v_y=0$ when $t=0$: $$\int_0^v \frac{dv_y}{v_y-v_t} = -{k\over m} \int_0^t dt$$ which integrates to $$\ln\frac{v_t-v_y}{v_t} = -{k\over m}t\mathrm{\qquad or \qquad}1-{v_y \over v_t}=e^{-(k/m)t}$$ and finally $$v_y=v_t[1-e^{-(k/m)t}]$$ Note that $v_y$ becomes equal to the terminal speed $v_t$ only in the limit that $t \to\inf$; the ball cannot attain a terminal speed in any finite length of time.

The step I don't understand is how they say that $$\int_0^v \frac{dv_y}{v_y-v_t} = \ln \frac{v_t-v_y}{v_t}$$ By my calculations, we have $$\int_0^v \frac{dv_y}{v_y-v_t} = \ln(v_y-v_t)\rvert_0^v = \ln(v-v_t)-\ln(-v_t) = \ln(\frac{v-v_t}{-v_t}) = \ln(\frac{v_t-v}{v_t}) \neq \ln\frac{v_t-v_y}{v_t}$$ How does the book arrive at its answer and how is my derivation wrong?

For that matter, what do $v$ and $t$ even represent, since their meanings were not defined?

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They have simply renamed $v$ to be $v_y$ after doing the integral. Myself I would have used $v_y$ for the upper limit on the integral and $v$ for the integration variable. Their answer is correct however you read it.

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  • $\begingroup$ How can you rename variables without changing the meaning of the equation? $\endgroup$ – user109923 Jul 24 at 19:07
  • $\begingroup$ The context is everything. You can rename as you wish. Obviously the correct expression has $v_y$ in it as plain $v$ has not been assigned a physical meaning except in an intermediate bit of algebra.. $\endgroup$ – mike stone Jul 24 at 22:27
  • $\begingroup$ Plain $v$, as you say, has not been assigned a physical meaning, but $v_y$ has. To rename $v$ to $v_y$ would be to assert that $v$ can take on the physical meaning of $v_y$. Why is $v$ able to do so? And if it is able to do so, why did we need to introduce a new variable in the first place, when we already have the variable $v_y$ that we could have used instead of $v$? $\endgroup$ – user109923 Jul 24 at 22:57
  • $\begingroup$ For the obvious reason that the author of the book mixed them up when he wrote that bit. $\endgroup$ – mike stone Jul 25 at 12:13

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