# Deriving terminal velocity

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?

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.
• 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.. – mike stone Jul 24 at 22:27
• 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 $v$? – user109923 Jul 24 at 22:57