1. ## A physics proof

I have a differential equations problem that I thought someone here may be able to help with. So here goes:
In the motion of an object through a certain medium (air at certain pressures is an example), the medium furnishes a resisting force proportional to the square of the velocity of the moving object. Suppose a body falls due to the action of gravity, through the medium. Let t represent time, and v represent velocity, positive downward. Let g be the usual constant acceleration of gravity, and let w be the weight of the body. Use newton's law, for equals mass times acceleration, to conclude that the differential equation of motion is
(W/g)dv/dt=w-kv^2
where kv^2 is the magnitude of the resisting force furnished by the medium.

2. Originally Posted by Sebastian
I have a differential equations problem that I thought someone here may be able to help with. So here goes:
In the motion of an object through a certain medium (air at certain pressures is an example), the medium furnishes a resisting force proportional to the square of the velocity of the moving object. Suppose a body falls due to the action of gravity, through the medium. Let t represent time, and v represent velocity, positive downward. Let g be the usual constant acceleration of gravity, and let w be the weight of the body. Use newton's law, for equals mass times acceleration, to conclude that the differential equation of motion is
(W/g)dv/dt=w-kv^2
where kv^2 is the magnitude of the resisting force furnished by the medium.
I would not called physics a "proof" but anyways.

By the second law: F=ma

And the equilbrium law: SUM (Forces) = 0 (vec).

Okay, thus we have that when a object is falling down there are two forces: the force of gravity and the resistance force. By the conditions of the problem the resistance force is proportional to the square of speed, in simple terms, kv^2. But! There is also a downard force, w, the weight. Since the sum of the forces is zero, we have,
w-kv^2=0 (since it is going in opposite direction, thus we have a negative).
By Newton's second law the overall force is: a(w-kv^2) where "a" is acceleration. The acceleration in this case is the constant at g. Thus,
F=g(w-kv^2).
But there is another way to find the force.
Again, F=a*m=(dv/dt)*w because m, mass, and weight are presumably the same here. And dv/dt is the acceleration, as you know.
Thus,
w(dv/dt)=g(w-kv^2)
Divide,
(w/g)(dv/dt)=w-kv^2

3. I understand the first part, but this second and third parts are getting me:
b) Solve the differental equation of part A, with the initial conditions that v=v0 when t=0. Introduce the constant a^2=w/k to simpilfy the formulas.

c) There are mediums that resist motion through them with a force porportional to the first power of the velocity. For such a medium, state and solve problems analogous to parts A) and B), except that for convenience a constant b=w/k may be introduced to replace the a^2. Show that b has the dimensions of a velocity.
I realize how good this forum is now, and I'll hang around here to learn some math techniques. Thanks for the help if you can provide it.

4. Originally Posted by Sebastian
I understand the first part, but this second and third parts are getting me:
b) Solve the differental equation of part A, with the initial conditions that v=v0 when t=0. Introduce the constant a^2=w/k to simpilfy the formulas.
The ODE (W/g)dv/dt=w-kv^2 is of variables seperable type so:

int a^2/(w-kv^2) dv = int dt

the left hand side may be rewritten:

k int a^2 / (a^2 - v^2) dv = t + C

and then partial fractions may be used to integrate the left hand side:

k (a/2) int [1/(a+v) - 1/(a-v)] dv = t + C

so:

k a/2 ln[(v-a)/(v+a)] = t + C

(v-a)/(v+a) = exp[2 (t+C)/(ka)]

so:

v = {exp[2 (t+C)/(ka)] +a}/{1-exp[2 (t+C)/(ka)]}

The right hand side probably simolifies some more when you put in the initial
condition etc.

RonL

5. In case you want it, I'll give you a bit more information about these equations and their solutions. But I'll need to stop home and pick up the appropiate text first. I'll try to post later tonight.

-Dan

6. Originally Posted by topsquark
In case you want it, I'll give you a bit more information about these equations and their solutions. But I'll need to stop home and pick up the appropiate text first. I'll try to post later tonight.

-Dan
i was able to go to the teacher today and she showed me a simpler way to do it.
w/g dv/dt=w-kt b=w/k
w/g/k dv/dt=w-kv/k
w/gk dv/dt = w/k - v
bg dv/dt =b-v
b dv/dt =(b-v)g
b/(b-v) dv=g dt
-b/v-bdv =g dt
intr(-b/(v-b)dv)=intrgdt
ln|v=b|=-gt/b+c/b
|v-b|=e^((-gt/b)+c)
v-b=ce^(-gt/b)
v=ce^(-gt/b) +b
v0=ce^0+b
v0=c+b
v0-b=c
v=(v0-b)e^(-gt/b) + b

Thanks for all your help on this problem guys