# Kinematics of Falling Object

1. Aug 1, 2011

### sn0wxboarder

First, let me preface this question that I have been out of academia for about 5 years now, and am just starting to get back into it, although it seems my Calc/Kinematics is a little rusty.

b]1. The problem statement, all variables and given/known data[/b]

An object of weight W is falling through a medium such that the object's drag force is proportional to its velocity. Express the velocity in terms of time if the initial velocity of the object is zero.

2. Relevant equations

F=ma
F=KV
a=dV/dt

3. The attempt at a solution

Ok, so just from the problem statement alone, I know we have an object on which two forces are acting; the weight of the object due to gravity and the drag force on the object.

Fnet = mg - KV

Since g is always a constant, it's the acceleration of the drag force we are trying to solve for, correct? Since F=KV, and F=ma, by association we have ma=KV, or a=KV/m, so now we have a function of acceleration in terms of velocity. If we plug this into a=dV/dt, we end up with:

KV/m = dV/dt

or

dt = (m/KV)dV

Integrating both sides, we end up with:

$\int$dt = (m/K)$\int$(1/V)dV or t = (m/K)ln(V)+C

Now if we solve for V in terms of t we get:

ln(V) = (tK/m)-C or V = e^[(tk/m)-C]

So this is as far as I have been able to get with this problem, because when you plug in initial velocity to solve for C, you get 0 = e^(0-C), but e^x is undefined when x = 0.

Am I making a wrong assumption at the beginning of the problem? I've got about 6 pages of scratch work and I feel this is the closest I've gotten to the actual solution. Any advice would be helpful. Thanks in advance.

Last edited: Aug 1, 2011
2. Aug 2, 2011

### ehild

NO. The body has acceleration, not the force. The acceleration caused by the individual forces add up. The acceleration of the body is proportional to the total force Fnet. Fnet=ma. As a = dv/dt,

m (dv/dt ) = mg -Kv.

This is a first-order linear differential equation, solve for v(t). You can solve it by separation of the variables.

ehild

3. Aug 2, 2011

### Andrzej

The relevant equations should be

Ft = ma = W - kV
W = mg
a = dV/dt

m dV/dt = mg - kV (1)

This differential equation has an easy to guess special solution
V = mg/k = const.

The homogeneous equation for the one above is
m dV/dt = -kV (2)

Note that it is similar to yours but the sign. So the solution to (2) is
V = e^(-k/m t + C) = Vo e^(-k/m t)

(BTW. e^0 = 1)

Now you have a special solution to the equation (1), and a general solution to the homogeneous equation. The general solution to (1) is the sum of above
V = mg/k + Vo e^(-k/m t)

Vo may be found remembering that the initial velocity was zero.
0 = mg/k + Vo e^0
Vo = -mg/k

The velocity of the falling object is given by
V = mg/k(1 - e^(-k/m t))

Make a graph of it to see what it means.

4. Aug 3, 2011

### sn0wxboarder

Ok, so here's my best attempt at this:

m(dv/dt)=mg-KV
dv/dt=g-KV/m
1/(g-KV/m)dv=dt (1)

Now from here we can use u-substitution:

u=g-KV/m
du/dv=-K/m or dv=(-m/K)du

Subbing this into equation 1:

(-m/K)(1/u)du=dt or (1/u)du=(-K/m)dt

Now integrate both sides:

ln(u) = (-K/m)(t+C)
e^(ln(u))=e^((-Kt/m)+(KC/m)) or u=e^((-Kt/m)+(KC/m))

Now we sub back in for u, and solve for V:

g-(KV/m)=e^((-Kt/m)+(KC/m))
(KV/m)=g-e^((-Kt/m)+(KC/m))
V=(m/K)(g-e^((-Kt/m)+(KC/m))

Now this is very similar to Andrzej's answer, but somehow he had an additional "g" before his exponential, so he could pull out the term (mg/K). What did I miss that include that? With the included "g", you can use initial conditions to solve for your constant of integration and determine your final formula for V in terms of t.

5. Aug 3, 2011

### ehild

e^((-Kt/m)+(KC/m))= e^(-Kt/m)*e^(KC/m). Find the constant C from the initial condition(v(0)=0. You will find that e^KC/m =g.

ehild

6. Aug 3, 2011

### Andrzej

The reason is, you haven't got the final answer yet, as your formula still has an integrating constant C.
Note that exponential function has a following feature
e^a e^b = e^(a+b).

Find the value of C in your function remembering that for
t = 0
velocity was also zero
V(0) = 0.

7. Aug 3, 2011

### sn0wxboarder

I understand now, thanks so much guys, you rock.