# Homework Help: Terminal velocity of a body of mass

1. Sep 24, 2008

### musicfairy

A small body of mass m located near the Earth’s surface falls from rest in the Earth's gravitational field. Acting on the body is a resistive force of magnitude kmv, where k is a constant and v is the speed of the body.

a) Write the differential equation that represents Newton's second law for this situation.

a = dv/dt
F = mg - kmv
ma = mg - kmv
a = g - kv
dv/dt = g - kv

I'm pretty sure I got this one right.

b) Determine the terminal speed vT of the body.

F = 0
g - kv = 0
g = kv
vT = g / k

I think I got this one right too.

c) Integrate the differential equation once to obtain an expression for the speed v as a function of time t. Use the condition that v = 0 when t= 0.

What I did so far:

dv/dt = g - kv
dt = dv / (g - kv)

So I would integrate dt = dv / (g - kv). The problem is I don't know how. Can someone please explain how to do this kind of integration, why I would do it that way?

2. Sep 24, 2008

### cepheid

Staff Emeritus
Often times if you have something like:

dv/dt = v,

you'll think to yourself, "oh, the equation is separable", and you'll go:

dv/v = dt

and then "integrate"

to get ln(v) = t + C

notice, however, that you integrated the left side wrt v and the right hand side wrt t, which makes NO sense! So what is actually going on here? Why does this nonsensical operation of splitting dv/dt actually work? It works because you are implicitly using the chain rule as follows

$$\frac{dv}{dt} = v$$

$$\frac{1}{v}\frac{dv}{dt} = 1$$

notice that the left hand side looks like something that *has been* differentiated using the chain rule, and we can write it as that something:

$$\frac{d}{dt}(\ln{v}) = 1$$

Now we can integrate BOTH sides wrt time (which makes waaay more sense)

$$\int { \frac{d}{dt}(\ln(v)) \, dt } = \int{\, dt}$$

$$\ln(v) = t + C$$

In your more complicated example, it is important to realize that this is what is going on, otherwise you'll get stuck.

We have

$$\frac{1}{g-kv}\frac{dv}{dt} = 1$$

We want the left hand side to look like something that has been differentiated by the chain rule. It SORT OF does. The only problem is that the derivative of ln(g-kv) wrt time actually differs from what's on the left hand side by a factor of -k. Once you get that sorted, it's the same procedure.

3. Sep 24, 2008

### musicfairy

Thanks for the explanation. I'm starting to understand this more.

What do you mean that the left differs by a factor of -k?

I have the answer. It's -(1/k)ln(g - kv) = t + C

Where did that come from? Where did the other k come from?

I thought doing this problem would help me understand how these weirder intergrations work, but it's confusing me even more.

4. Sep 24, 2008

### cepheid

Staff Emeritus
alright, here goes:

The goal is to recognize that the left hand side can be expressed as an exact derivative, so that integrating it becomes trivial.

the equation has a term 1/(g-kv) multiplied by v'(t), which suggests that it is the result of differentiating ln(g-kv) with respect to time using the chain rule. If that were true, then the left hand side would reduce to the time derivative of the natural log of g-kv, and integrating it would be easy (integral of a derivative is just the function itself...so you're just left with the log term). But that's NOT quite right. Check it out:

$$\frac{d}{dt} (\ln(g-kv)) = \frac{1}{g-kv} \frac{d}{dt}(g-kv) \frac{dv}{dt}$$

$$= \frac{1}{g-kv} (-k) \frac{dv}{dt}$$

Notice that this doesn't QUITE match the left hand side of our differential equation. There's that extra factor of (-k). Which means that the left hand side of our equation is this thing divided by -k. THAT's where the 1/(-k) comes from.

5. Sep 24, 2008

### musicfairy

I stared at the problem and answer while waiting for an answer and had the same train of thought. That was very helpful. Thanks a lot.