Recent content by Gaidzahg

Well, since v0 = v(0), then wouldn't v0 = e^C? I guess that makes sense to me now. Thank you. On another note, where can I learn how to use the hashtags and other code you used to make the equations look nice?

Okay, so then after integration we get t = -m/b ln v + C t - C = -m/b ln v -(t - C)b/m = ln v (C-t)b/m = ln v e^(C-t)b/m = v (e^Cb/m) * (e^-tb/m) = v does that look right?

Exactly. What I'm trying to understand is how is the equation derived so that the v0 appears in it. You are correct that the answer I came up with gives a nonsensical answer.

Maybe I'm not understanding you, or you are not understanding me. There is no solution given at time t = 0. The whole point of the problem is deriving the equation that shows the velocity as a function of time.

The book only says that the initial velocity at time zero is v-naught (v0). The books solution is literally just v=v0e^-(b/m)t

F = ma -bv = ma -bv/m = a -bv/m = dv/dt dt = -mdv/bv ∫dt = -m/b ∫dv/v t = -m/b ln v -(b/m)t = ln v e^-(b/m)t = v