# Finding the graph formula with know points and other equation.

 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 = $\frac{P}{mv}$-$\frac{CDpAv2}{2m}$ Since F = $\frac{P}{v}$ I am trying to graph this as a velocity-time graph, however, I don't know how to do it. There is no time variable that I can replace with x, and the y-value (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(1-e-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 $\sqrt[3]{\frac{2P}{CDpA}}$, so the equation would be something like: y = $\sqrt[3]{\frac{2P}{CDpA}}$(1-e-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*106 and CD = 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)
 P: 425 Given ##a=\frac{dv}{dt}##, what you have is a separable ordinary differential equation $$\frac{dv}{dt}=\frac{\alpha}{v}-\beta v^2$$ 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.
 P: 98 I have never done differential equations before, and I tried to just find the anti-derivative, 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|)-(CDpAv3t)/(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.
$a = \frac{P}{v} - \frac{ (C_d p A v^2) }{2}$
 P: 98 Oh ok, I'll give it a go: a = ($\frac{Pt}{m}$)ln(|v|)-$\frac{C_{D}pAv^{3}t}{6m}$ Seems to work, thanks.