ODEs/Physics help

[Kirobaito]

I'm the only person in my ODEs class to have not taken a physics class, and my teacher always forgets that. Trying to do it myself, I can't separate this equation and my head isn't in a variable-separating mode. Help plz.

A rocket of mass 100 kg is fired straight up with a net force of 1,000,000 newtons. It experiences air friction f = -100(ns^2/m^2)v^2. At t=0, v=0, y=0. Assume the mass of the rocket does not change.

a) Write Newton's second law for the rocket.
b) Solve analytically for V(t).
c) Use the answer in part b) and let v = dy/dt. Solve for y(t) using the ODE mode on the calculator. Neatly sketch y(t) for [0,20].

I can do C, except I'm not sure if I'm doing a) right and I can't seem to do b).

a) f = ma

Would you just subtract the air friction from the net force, so it's 1000000 - 100v^2 = 100dv/dt? That's what I did.

b) 1000000 = 100(dv/dt + v^2)
10000=dv/dt + v^2

I can't seem to separate it from here. I'm probably just MISSING something completely obvious, but I can't seem to think straight.

Any help, from those who have taken advanced Calc?

[nik0tine]

Wait.. You're doing this for fun?

[Kirobaito]

Wait.. You're doing this for fun?
...no. This is a homework problem. He gave it to us because he THINKS we've all taken physics and naturally know the stuff.

[eestlinc]

how are you taking differential equations without ever taking physics?

[Kirobaito]

how are you taking differential equations without ever taking physics?
Because they didn't make physics a pre-requisite, and the only reason I signed up for the course in the first place is because there was nothing else to take.

[eestlinc]

well, Newton's second law is sigma(f) = ma so you calculate the sum of all forces acting on the rocket. Mass is constant as stated in the problem (even though it really wouldn't be if it was burning fuel).

forces on the rocket:
1000000 N straight up
gravity straight down
air resistance straight down

so add up all the forces, plug in the mass, and solve for the acceleration.

just remember in physics to add up all the forces and that velocity (thus acceleration) is directional.

[Kirobaito]

So that'd be 1000000-10-100v^2 = 100dv/dt. That still doesn't help me in separating it.

I still end up with 999990=100(dv/dt + v^2). I still can't separate that.

(He lets us use -10 for acceleration due to gravity).

[eestlinc]

never mind that

[Kirobaito]

Since when? Last time I checked, v=at + v[o].

EDIT: I could easily use an integrating factor on this if it wasn't for the fact that the v in the air friction is squared.

[eestlinc]

yea i haven't done this crap in so long, I'm trying to remember it.

i think the units are important in this.

Newtons are kg*m/s²

velocity is m/s

so the force of air resistance = (-100)(kg*m/s²)(s²/m²)(m²/s²) = -100 kg*m/s² = -100 N

so the air friction force is constant, which means your equation is simply 999890 N = 100 kg dv/dt

which is the same as 9998.9 m/s²=dv/dt

so then dv = 9998.9 m/s² dt

so v = 9998.9 m/s² * t

[Kirobaito]

But what about the v^2?

The air resistance isn't constant. It's based on the velocity, according to the problem.

[eestlinc]

the v² is in there, but the units all cancel out. I don't know, don't listen to me. I can't remember any of this crap anymore. the seconds are t, so maybe plug t in for the seconds and work from there.

[Kirobaito]

That makes sense, but isn't that essentially saying that the thing never stops going up and just keeps on getting faster and faster?

[eestlinc]

yea, the air resistance is definitely dependent upon the velocity of the rocket, and of course the units cancel out to be N.

Can't you reduce your equation to 9999.9 = dv/dt + v²

[Kirobaito]

Yes. But from what I've learned, that appears inseparable. I came up with simplifying it to that point (at least on the right side) hours ago.

[eestlinc]

well, I never took diff equations so I don't know either way.

[Endless]

Newton's second law:

d(mv)/dt = F
mdv/dt+vdm/dt = F, since m is constant, m*dv/dt = F

m*dv/dt = -m*g - k*v² [k=100, m=100]
100*dv/dt = -100 (g+v²)
dv/dt = -v²-g
dv/dt = -(dy/dt)²-g

And then, well, we're still stuck. Maybe with G² = g and a = -1
dv/dt = a*(v-Gi)*(v+Gi), with i as in i² = -1
or
dv/dy = -dy/dt - g* dt/dy?

That page (http://www.profjrwhite.com/diff_eqns/pdf_files/appls_1.pdf or http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/Methodes-Maths/white/deqn/a1/nwt2law/nwt2law.html) solves it for downward motion.

Edit: forgot to count the 1000000. duh.
So yeah, what kb says below. dv/dt = 10000-v²

[Kirobaito]

I finally figured it out. The net force of 1000000 already includes gravity, so it's just

1000000 - 100v^2 = 100dv/dt

Which becomes

dt = dv/(10000-v^2)

You can separate that into

dv/(100+v)(100-v)

And use partial fractions, where both A and B are equal to 1/200...

dt = dv/200(100+v) + dv/200(100-v)

Take the integral and you get

t = (1/200)(ln|100 + v| - ln|100 - v|)

200t = ln|100+v/100-v|

Put e to both sides:

e^200t = (100+v)/(100-v)

100e^200t - ve^200t = 100+v

100e^200t - 100 = v + ve^200t
100e^200t - 100 = v(1 + e^200t)

v(t) = (100e^200t - 100)/(1+e^200t)

v(t) = 100(e^200t-1)/(1+e^200t)

I'm not sure if you can reduce it more.