farleyknight said:
Ack! Okay, my Calc II teacher really didn't cover separable differential equations as well as he could have. He especially didn't cover their applications. I'll probably have to pay for his incompetence and study it better when I have the time, but for right now, here's what I have:
F_{net} = m*\frac{dv}{dt}, by definition. Since the motor is turned off, F_{app} = 0, and the opposing frictional force, f_k(v) = 73v. We then have
m*\frac{dv}{dt} = 0 - 73v
or
\frac{m*dv}{-73v} = dt
Integrating both sides
\frac{m}{-73} \int \frac{1}{v} dv = \int dt = \frac{m}{-73} \ln |v| + C = t
When t = 0, v = 27.7, so solving for C
\frac{-1150}{73}* \ln(27.7) = -C = -52.32
Now we must find t = t_f where v = 13.78.
\frac{-1150}{73}* \ln(13.78) + 52.32 = t_f = 10.99
But this answer doesn't work.
Either a) I screwed up in my calculations, or b) my lack of experience with differentials, I don't have something set up properly.
Just as a rule of thumb, work parametrically. It's easy to lose your constants when you work numerically, and as a result, it's very hard for someone looking at your work to follow what you were doing. I'm afraid I got lost along the way.
You made a mistake when you took the indefinite integral instead of a definite one (Not that it's a
mistake, it just means you need to do a lot more work):
-k\frac{dv}{V}=dt
For the RHS, I'll integrate from V=V_0 to V=V(t), for the LHS, I'll integrate from t=0 to t=t
-k \int_{v_0}^{v(t)} \frac{dv}{V}=\int_{0}^{t} dt
-k \left[\ln{V}\right]_{v_0}^{v(t)}=\left[t\right]_{0}^{t}
-k(\ln{v(t)}-\ln{v_0})=t
In more compact form:
k\cdot\ln{\frac{v_0}{v(t)}}=t
And from there you can get your final answer.
Developing that expression further we can find:
v(t)=v_0\cdot e^{-\tfrac{t}{k}}
I'll now show the second method I mentioned in my first post.
Let's look at the first differential equation we got:
\frac{dv}{dt}=-\tfrac{1}{k}v
Rewriting using Newton's notation we get:
\ddot x = -\tfrac{1}{k}\dot x
This is a differential equation that asks the question, what function of time, when differentiated once with respect to time, gives itself times a negative constant?
An easy solution to guess would be the exponential we got at the end of the integration above. Checking for the initial values would confirm that it is the true solution and that we haven't missed anything with our educated guess.
You may be familiar with the procedure from how you derive SHM in class (You get the following differential equation \ddot x = -\omega^2 x and guess the harmonic function as a solution to the diff. equation)
Or if you're not, now you'll have a head-start once you get to that subject. :)
Oh, and you said a little something that bugged me. You said that \Sigma \vec F = m\vec a
by definition
The above equation is Newton's Second Law. It should not be taken for granted. :)
