Recent content by jfrank1034

You're right...should start as dx = mvdv/(-mg + kv2)

In my book it says v dv = a ds --> a = v(dv/ds)... so I have vdv/(-g + kv2) = dx ∫vdv/(-g + kv2) [from 0 to v] = ∫dx [from 0 to x] (1/2k) ln(-g+kv2) [from 0 to v] = x --> (1/2k) ln(g-kv2/g) = x Solving for v and skipping a few steps I got v = √(g/k(1-e2kx))

Was I not correct in saying that a = v (dv/ds) ??

But I need the velocity in terms of the distance fallen, x. Integrating the right side, ∫dt , would just give me 't' am I right?

If I change a to d2s/dt2 is this the right path?

Great...just realized that

Okay so I have mv(dv/dx) = mg - kv2 , and divide both sides by the (mg - kv2) Then, (mv/(mg - kv2))dv = dx ∫(mv/mg - mv/kv2) dv = ∫dx Just need a little help integrating if I'm on the right track I guess...

A ball of mass m is released from rest at x=0. Air resistance is expressed as R = kv2 , where k is a positive constant and v denotes velocity. Derive an expression for the speed in terms of the distance x that it has fallen. Identify the terminal v. I know I have to start with ma = mg -...