Recent content by Ahmedbasil

Owh, okay, I assumed it is the current OUT of the capacitor. I see :). My bad, and thank you very much.

I suppose that dVc/dt doesn't direction into account. Why doesn't it though? At first I considered this, but then I dismissed it, knowing that it's perfectly fine to have a rate of change w.r.t another variable negative. The equation should be: I_{c} = C |\frac{dV_{c}}{dt}| Anyway, thanks very...

Homework Statement You have a capacitor, of capacitance C farads, with charge Q coulombs. It is connected in series with a resistor of resistance R ohms. Derive an expression for the potential difference over the capacitor at any time t.2. Homework Equations and theorems I_{c} =...

A force is a vector, it cannot have a negative magnitude. Although what you see as -3000 Newtons is really just a contradiction of the direction. The 3000 Newtons is acting in the direction opposite to the one you assumed it's acting in. F=ma, should actually be |F|=m|a|, or \vec{F} = m \vec{a}...

Well, the answer is clearly 220 meters :D. This one made me laugh

No problem at all. Regarding "my idea" of substituting z=1/v, that's not my idea/method :P, it's a textbook method. Well, regarding your question "how the derivative of z wrt t becomes 1*v^-2 dv/dt". It's simple. You have z = v^{-1}, where v is a function of t. As we are differentiating w.r.t...

I haven't gone through the solution either, but I suppose as t -> infinity, v(t) -> 0. That's how it should be afaik. EDIT: had a quick look at the second term: c(e^kt/m) in z(t). I didn't solve the integral for z(t), but as t->infinity, then c(e^kt/m) also goes to infinity, unless the constant...

EDIT: If "Write a differential equation for the system, as a function of the cars decceleration." that's the question, then you don't even have to solve the differential equation... anyhow: From what I can see you're having problems solving the actual differential equation for a function of...