Terminal velocity

1. Nov 28, 2004

Cyrus

looking back at a problem that puzzled the heck out of me.

integrate this: dv/(v-vt) = -k/m (dt).

The book says you get: ln ( (vt-v)/Vt ) = -k/m t

the -k/mt part is fine, because you integrate from 0 to t. The ln part is tricky though. When I integrate, I get ln(v-vt), vt is just a constant. If you do the derivative of my anwser or the books they are the same. How did the book arrive at this anwser and not mine.

2. Nov 28, 2004

cepheid

Staff Emeritus
Can you rewrite the problem more carefully? I don't understand how vt can be a constant if it has 'v' in it, or why a lowercase v suddenly becomes a capital V somewhere, or why v-vt suddenly becomes vt-v. Thanks.

3. Nov 28, 2004

Tide

Just evaluate the definite integral to get the desired result:

$$\int_{0}^{v} \frac {dv'}{v'-v_t} = -\int_{0}^{v} \frac {dv'}{v_t-v'} = \ln{(v_t-v)} - \ln{v_t} = \ln \frac {v_t-v}{v_t}$$

4. Nov 28, 2004

Cyrus

but why even to bothing factoring out a minus sign tide? It just complicateds the integral. Its much simpler to just integrate and get ln(v-vt), no?

when you do the limits you get, ln(v-vt)-ln(v). so thats equal to ln ((v-vt)/v).

Oh my gosh. I completely poo-pooed that integral. I worked it out my way and got the same anwser. Boy do I feel like an idiot. it should work out to be,
ln(v-vt) from 0 to v, which is ln(v-vt)-ln(-vt) = to ln(1-v/vt). Thats just Embarrassing.

Last edited: Nov 28, 2004
5. Nov 28, 2004

Tide

Cyrus,

You factor out the -1 because (a) v < vt and (b) you avoid the nastiness of dealing with logarithms of negative numbers.