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Finding the graph formula with know points and other equation.by Pharrahnox
Tags: acceleration, air resistance, equation, formula, graph, points, velocity, velocitytime graph 
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#1
Mar1114, 01:33 AM

P: 98

I have an equation for determining the acceleration of an object being propelled by a constant power source, that is affected by air resistance:
a = [itex]\frac{P}{mv}[/itex][itex]\frac{C_{D}pAv^{2}}{2m}[/itex] Since F = [itex]\frac{P}{v}[/itex] I am trying to graph this as a velocitytime graph, however, I don't know how to do it. There is no time variable that I can replace with x, and the yvalue (velocity) is mixed into the equation already. I remember the equation given to me for a similar sort of thing, without air resitance, but instead just a constant friction force, that was something like this: y = k(1e^{ax}) Where k is a constant, which is the maximum speed, and a is another constant which represents the force of air resistance. The maximum speed in this case is [itex]\sqrt[3]{\frac{2P}{C_{D}pA}}[/itex], so the equation would be something like: y = [itex]\sqrt[3]{\frac{2P}{C_{D}pA}}[/itex](1e^{ax}) But that's as far as I've gotten. By the use of iteration, I have determined the velocity at several different times, here's a few, just in case it helps: (0,0) (25,83.4762) (50,118.1195) (75,126.1601) (100,127.7624) (125,128.0709) This is for variables of values: p = 0.001, A = 1900.933, m = 1900.933, P = 1*10^{6} and C_{D} = 0.5 Any help would be greatly appreciated, and if you need any more information, just let me know. EDIT: the equations don't seem to be formatting correctly, so I'll redo them down here: a = P/v  (Cd*p*A*v^2) /2 F = P/v max speed = ( (2*P) / (Cd*p*A) )^1/3 y = ( (2*P) / (Cd*p*A) )^1/3 * (1  e^ax) 


#2
Mar1114, 01:10 PM

P: 425

Given ##a=\frac{dv}{dt}##, what you have is a separable ordinary differential equation [tex]\frac{dv}{dt}=\frac{\alpha}{v}\beta v^2[/tex] where ##\alpha=\frac{P}{m}## and ##\beta=\frac{C_DpA}{2m}## have been substituted to make the equation (slightly) more manageable. Hopefully that's enough to get you going.
I'm not sure where you got your force equation. Are you not using ##F=ma##? Furthermore, I don't know that it's even useful here. I reckon it's doubtful that the equation that works in a constant friction situation applies here. So you're likely barking up the wrong tree there. 


#3
Mar1214, 04:15 AM

P: 98

I have never done differential equations before, and I tried to just find the antiderivative, but my teacher said that I had done it incorrectly, as I differentiated dv instead of dt, or the other way around. Anyway, here is what I got:
v = (Pt/m)*ln(v)(C_{D}pAv^{3}t)/(6m)+c Unfortunately from there, if it is correct, I don't know where to go  how to get v by itself without v on the other side. What do I do from here, or from the start if that isn't correct? Thanks for your response. 


#4
Mar1214, 10:06 AM

Sci Advisor
P: 3,252

Finding the graph formula with know points and other equation.
[itex] a = \frac{P}{v}  \frac{ (C_d p A v^2) }{2} [/itex] 


#5
Mar1214, 03:11 PM

P: 98

Oh ok, I'll give it a go:
a = ([itex]\frac{Pt}{m}[/itex])ln(v)[itex]\frac{C_{D}pAv^{3}t}{6m}[/itex] Seems to work, thanks. 


#6
Mar1614, 06:41 AM

P: 98

Does anyone have any advice on what I should do to get time into the equation, without velocity on both sides? I don't even know what to type into google to find information on it. Any information would be much appreciated.



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