So I'm actually in my honors chemistry class right now, but who cares right? It's not like it's physics... anyways, good to be back!

Yesterday, in my differential equations class, we started section 2.3-- I don't actually remember the title-- at the ungodly early hour of 9 AM. aka, really not that early. Now, if I/ other AP C past/current students remember correctly, early in the year we discussed air resistance on a falling object. According to Newton's 2nd Law, net force or ma equals whatever you determine to be the net force. In this case, using a force diagram, you have the force of mg down minus the effect of air resistance (I'll use kv in this case because we used it in my math class, though last year we used bv and cv^2 I believe). Thus you have ma=mg-kv, and since a=dv/dt, you have m(dv/dt)=mg-kv . This is (wait for it) a differential equation! Yay!

Specifically, it is a linear differential equation, more commonly seen in the form dv/dt+ (something)*v=something. So when Mr. F just skipped over the steps/integration/nothing made sense, that's why. It was a DE.

Note: I don't fully remember how we did the problem, but I think we were just told what the equation came out to, and skipped the actual steps.

Anyways, you put it into the form dv/dt+(k/m)*v=g (the dv/dt can't have a coefficient), and then you do a bunch of really really really cool steps to solve it. You take the stuff in front of the v (in this case, k/m) and set up this: e^(integral of k/m dt). Clearly, this gives you e^(kt/m). You then multiply everything in the equation by this, giving you (e^kt/m)*(dv/dt+ (k/m)*v) = g*(e^(kt/m)). The left side of the equation turns into d/dt of e^(kt/m) times v. We don't actually do anything to get to this, it's just known that that's what it turns into, and you can check it to make sure.

You know have d/dt of e^(kt/m)*v = g*e^(kt/m). you integrate both sides with respect to t, leaving you with:

e^(kt/m)*v=e^(kt/m)*gm/k + Constant ©. Then just isolate velocity.

That gives you v(t) equals gm/k+ C*e^(-kt/m), and you can solve for C pretty easily (either with v(0)=v(subscript)0 or v(0)=0. And that's your air resistance equation! (hopefully)

I'm assuming I messed up a negative sign somewhere, or it should be -gm/k, or something else, but that's the general shape of a) a Linear DE and this force equation. Hopefully it's mostly right, and not overly boring because it's actually kinda cool. So yeah... go physics!

Probably the longest blog post I've done, but I deemed in necessary. Sincerely, your resident Swagmeister

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