How Long Does It Take a Raindrop to Reach 63% of Its Terminal Velocity?

[killazys]

Homework Statement

The terminal velocity of a $$2.8*10^{-5}$$kg raindrop is about 14 m/s.
Determine the time required for such a drop, starting from rest, to reach 63 percent of terminal velocity.

Homework Equations

$$F_{g}-F_{D}=-bv$$

The Attempt at a Solution

$$F_{D}=1.96*10\exp{-5}$$
$$mg-b*(0.63V_{T}) = ma\\$$
$$mg-bv = m*\frac{dv}{dt}\\$$
$$\frac{dt}{m} = \frac{dv}{mg-bv}\\$$
$$1/m\int{dt} = \int{\frac{dv}{mg-bv}}\\$$
$$\mbox{plugging into calculator}\\$$
$$1/m*t=50727.1568$$
.994s

Well, that basically. My physics teacher said that there was no need for calculus, so I'm trying to solve without integrating. There are more questions I have problem with, as always on forums I will first do a search, then post.. should I start a new topic if I can't find a solution?
EDIT: decided to use an integral solution. is that the answer?

 

[collinsmark]

Hello killazys,

Welcome to Physics Forums!

Homework Statement

The terminal velocity of a $$2.8*10^{-5}$$kg raindrop is about 14 m/s.
Determine the time required for such a drop, starting from rest, to reach 63 percent of terminal velocity.

Is that the problem, verbatim, as given to you originally? I ask because if you are not supposed to use calculus to solve this problem, something seems missing. For example, are you supposed to assume that the force of friction is a constant force, and not proportional to the raindrop's velocity?

Does this class/coursework involve fluid dynamics? If so, make sure you put some relevant equations in your relevant equations section. For example, in your attempted solution you assumed that the frictional force is directly proportional to the raindrop's velocity. That's not necessarily a bad assumption (and might be a pretty good one), but I want to make sure that that is what you are supposed to assume for this problem, given your coursework.

It also could be that your class/coursework has some equations already given to you that involve exponential decay (equations given to you in your coursework -- that were in fact already derived from differential equations -- but already solved in a form such that you don't have to use calculus to derive them yourself). If so, please put those in your relevant equations section.

Homework Equations

$$F_{g}-F_{D}=-bv$$

Where does the above equation come from? It almost looks similar to Newton's second law, but something doesn't quite look familiar to me. (Maybe I'm misinterpreting what you mean by F_G and F_D).

The Attempt at a Solution

$$F_{D}=1.96*10\exp{-5}$$

What is F_D again? And where does the 1.96 x 10^-5 value come from? I couldn't figure it out from your problem statement.

$$mg-b*(0.63V_{T}) = ma\\$$

Umm..., the above doesn't doesn't look right to me.

$$mg-bv = m*\frac{dv}{dt}\\$$

There you go . That's Newton's second law! And yes, the above is the starting point if you are supposed to assume that the frictional force is proportional to the raindrop's velocity, and if some other equation wasn't already given to you that isn't in differential equation form.

You see, what you have here is a differential equation. And you're not going to be able to solve it without some calculus. So before we go on, I recommend that you verify that your assumptions are correct given your coursework.

$$\frac{dt}{m} = \frac{dv}{mg-bv}\\$$

I'm not sure how you got the above equation, but I'm pretty sure something isn't right. It's not dimensionally correct.

[Edit: oops. It is dimensionally correct. My mistake. I'm still not sure I follow where the equation came from though.]

$$1/m\int{dt} = \int{\frac{dv}{mg-bv}}\\$$
$$\mbox{plugging into calculator}\\$$
$$1/m*t=50727.1568$$
.994s

I solved the differential equation myself, but came up with a different final answer. But I'm getting the impression that all the information about what one should or should not assume hasn't been given in the problem statement.

Well, that basically. My physics teacher said that there was no need for calculus, so I'm trying to solve without integrating. There are more questions I have problem with, as always on forums I will first do a search, then post.. should I start a new topic if I can't find a solution?
EDIT: decided to use an integral solution. is that the answer?

I came up with a different answer.

 

[collinsmark]

$$\frac{dt}{m} = \frac{dv}{mg-bv}\\$$

Okay, I see where you got that now.

$$1/m\int{dt} = \int{\frac{dv}{mg-bv}}\\$$

The above approach is also correct. Sorry about not seeing it earlier. I attribute my original confusion to the fact that I used a completely different method for solving the differential equation. However, your approach (separation of variables) is equally valid as the method that I used.

$$\mbox{plugging into calculator}\\$$
$$1/m*t=50727.1568$$
.994s

Okay, now here is where you start to lose me. 0.994 = -ln(1 - 0.63). But that's not the final answer and the units are not in seconds.

Let's go back to your last equation, and maybe I can help you with the integral.

$$\int \frac{dt}{m} = \int \frac{dv}{mg-bv}$$,

$$\frac{t}{m} = \int \frac{dv}{mg-bv}$$,

Making the substitution that that u = mg - bv, and du = -bdv, and noting that dv = -(1/b)du,

$$\frac{t}{m} = - \frac{1}{b}\int \frac{du}{u},$$

$$\frac{t}{m} = - \frac{1}{b} \ln u + C,$$

where C is an arbitrary constant.

Reversing the substitution, and doing a little algebra gives us

$$t = - \frac{m}{b} \ln (mv - bv) + C,$$

But it's important to note that the arbitrary constant C is not zero. You can use your initial conditions to solve for C noting that v(0) = 0.

Once you have that you can solve for v as a function of t. Then find what v is as t --> ∞. That's the terminal velocity as a function of m, g, and b. Knowing that, set v = 0.63 times that value, and solve for t.

Of course, as you can see the above requires calculus. If you're supposed to do the problem somehow or another without calculus, there seems to be something missing -- something like a already-given equation or different assumption about the frictional force (such as it being constant instead of proportional to the velocity).