Newtonian Mechanics – Drag Force Proportional to $v^2$, Physical Interpretation

dragnewtonian-mechanics

Suppose we have an object with mass $m$, initial velocity $u_0$ and the drag force is $-λu^2$.

Solving for $u$ in Newton's 2nd law we get that:
$u(t)= \frac{u_0}{1+ku_0t}$
where $k=\frac{λ}{m}$.

We can clearly see that there was a time $t_0 =-\frac{1}{ku_0}$ (before what we considered as $t=0$) when the velocity was infinite.

I'm not sure what this means. The only way that the object has a finite velocity at a time t is that it had infinite velocity at $t_0$?.

If that's the case, why isn't that a problem for this particular model of drag force?

Also, does this anomaly have to do with the non linearity of the ode (Newton's 2nd law)?

Best Answer

It's basic math: when you're solving a differential equation, you're doing it in a given domain, and the solution is limited to this domain.

In this example, you used $t=0$ as the initial time, so you solved the equation on $\mathbb{R}^+$. You simply cannot extend the solution to negative times, which are outside the resolution domain.

Physically speaking, just because you see velocity decrease doesn't mean it decreased from infinity before you started looking.

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