
#1
Jan312, 06:44 PM

P: 4

1. The problem statement, all variables and given/known data
The problem is to find the velocity as a function of time, given the following; the object is in free fall and its initial velocity is zero. The formula given for the air resistance is: a = g  kv, where g is acceleration d/t gravity, k is a constant, and v is velocity. 2. Relevant equations The problem gives the hint to solve using integration by parts, with u = g  kv 3. The attempt at a solution The solution given in the back of the book is v = [itex]\frac{g}{k}[/itex](1e^{kt}) I have no idea how to arrive at this solution. 



#2
Jan312, 07:54 PM

HW Helper
P: 2,874

Set up the differential equation relating velocity and time, to begin with. 



#3
Jan312, 10:11 PM

P: 1,135





#4
Jan412, 12:40 AM

HW Helper
P: 2,874

Free fall with air resistance, must find velocity
You don't have to integrate by parts, BTW. This is an easily separable first order differential equation. Maybe a simple substitution to clarify the form, but that's it.




#5
Jan412, 02:18 PM

P: 4

Thanks for the replies everyone, but I'm still not seeing it. I am not trying to copy and paste an answer here, I'm doing this on my own time, trying to study before I go back to school in the Fall.
Where does the e come from? Is there some basic calculus that I'm overlooking? It looks to me that solving the problem should go something like this: [itex]\frac{dv}{dt}[/itex]= g  kv ∫[itex]^{t}_{0}[/itex]dvdt = ∫[itex]^{t}_{0}[/itex](g  kvt)dt v = gt  kvt → v = [itex]\frac{gt}{1 + kt}[/itex] Which makes sense, but doesn't match the answer in the book. Is the book incorrect, or am I? What am I missing here? Thanks again for the help everyone. Merb 



#6
Jan412, 02:34 PM

P: 262

When you neglect the gravity term, you should immediately see that the solution of
[itex]\frac{dv}{dt}=v[/itex] is [itex]v=e^{t}[/itex] because of the property of exponential functions: [itex]\frac{d(e^{t})}{dt}=e^{t}[/itex] Look up how to solve first order ode's using integrating factors and solving homogeneous and nonhomogeneous equations. Hope this helps. 



#7
Jan412, 03:57 PM

P: 4

Why would you neglect the gravity term?




#8
Jan412, 04:14 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

[tex]\frac{dv}{dt}= g kv[/tex] you get [tex]\int\frac{dv}{g kv}= \int dt[/tex] to integrate on the left, let u= g kv so that du= k dv, dv= (1/k)du [tex]\frac{1}{k}\int\frac{du}{u}= \int dt[/tex] so that [tex]\frac{1}{k}ln u= ln(u^{1/k})= t+ c[/tex] where c is the constant of integration. Taking the exponential of both sides [tex]u^{1/k}= e^{t+ c}= Ce^t[/tex] Take the k th power of both sides to get [tex]u= g kv= Ae^{kt}[/tex] where [itex]A= C^{k}[/itex] Finally, solve for v: [tex]v(t)= \frac{Ae^{kt} g}{k}[/tex] 



#9
Jan412, 06:22 PM

P: 4

Thank you very much Halls, that was a huge help. (It is a definite integral, though, evaluated at an arbitrary time t and assuming v=0 at t=0. I did away with the constant of integration.) I still can't get the answer that is given in the back of the book, however. When I manipulate it to a form similar to the answer above, I get v=[itex]\frac{g}{k}[/itex](1[itex]\frac{1}{g}[/itex]e^{kt}) (the exp should not be divided by g). Another strange thing is that both the answer I got using your method and the answer in the book check out when you differentiate with respect to t and set the result equal to a=gkv, so both appear to be valid, but that can't be right. Any ideas?




#10
Feb2612, 08:56 PM

P: 14

How would you solve differently if you had m(dv/dt)= mg kv^{2} 



#11
Feb2712, 08:52 AM

P: 262

[itex]\int \frac{1}{1ax^2}dx=\frac{1}{\sqrt{a}}arctanh(\sqrt(a)x)[/itex]
There is probably a nice variable transformation to get an easier intermediate integral in case you don't know the above integral. 



#12
Feb2712, 10:26 PM

P: 14

Thanks!



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