# Velocity over time applying constant power and considering air drag

I wanted to understand how velocity changes over time when applying a constant power to a body of mass $m$. I figured out that in this case, since $$P=d/dt(KE)=d/dt(\frac{1}{2}mv^2),$$ integrating I get finally that $v(t)=\sqrt(\frac{2Pt}{m})$. Very easy here.

Now I want to find the same when at the same time a constant power is applied there is an opposite force, like drag $F_d$, that depends on the velocity of the body. Considering air densit $\rho$, drag coefficient $C_d$ and area $A$, in time $t$ that power would be $$P_d(t)=v(t)F_d(t)= (C_d A \frac{\rho}{2}v^2(t))v(t).$$ Since that power depends on the velocity, I cannot just say $P_{net}=P-P_d$ and proceed as before to obtain a speed over time formula. Could anyone provide any insight on this?

You will obtain a first-order nonlinear differential equation:

$$P=P-P_d = P-D v^3 = \frac{m}{2} \frac{d}{dt}(v^2)$$

$$\to \frac{d}{dt} (v^2) + \frac{2D}{m} v^3 - \frac{2P}{m} = 0$$

If we define $u(t)=v(t)^2$ , $2D/m = A$ and $-2P/m = B$, we obtain

$$\frac{d}{dt} u(t) + A \ u(t)^{3/2} + B = 0$$

When $A$ is small this reduces to the result you obtained without considering drag.

Sadly though, I don't know any easy way to solve this equation!

• Can easy be simulated :) – DilithiumMatrix May 16 '16 at 15:19
• I know, actually I tried first to discretize time and I get reasonable results from that, but I wanted to find a more elegant solution. I'm starting to believe is not possible. Surely is not straightforward, phew!! – user2427894 May 16 '16 at 17:00

'valerio92' is correct and for:

$$\frac{d}{dt} u(t) + A \ u(t)^{3/2} + B = 0$$

... there probably are no analytical solutions (wolfram alpha's DSolve for instance yielded nothing).

But $F_d=C_d A \frac{\rho}{2}v^2$ is not the only model for drag forces. At low speeds:

$$F_d=kv$$

... can be used, so linear dependence of the drag force on speed.

The differential equation then becomes:

$$u'+Au+B=0$$

With $A=\frac{2k}{m}$ and $B=-\frac{2P}{m}$.

This solves well: $$-\int \frac{du}{Au+B}=\int dt$$

$$-\frac{1}{A}\ln|Au+B|=t+c$$ Assuming boundary conditions: $$t=0,u=0$$ $$-\frac{1}{A}\ln|B|=c$$ $$\ln|Au+B|=-At+\ln B$$ $$Au+B=e^{-At+\ln B}=Be^{-At}$$ $$u=\frac{B(e^{-At}-1)}{A}$$ Finally, back substituting $u$, $A$ and $B$:

$$\large{v=\sqrt{\frac{P}{m}\Big(1-e^{-\frac{2k}{m}}\Big)}}$$

Note that for $t \to +\infty$, then $v=\sqrt{\frac{P}{m}}$.