Question

@anemonix
#4

Answer 1

a(t)=ge^(-bt)
V(t)=Integral[a(t)dt] = g*Int[e^(-bt) dt]
u = -bt
du= - b*dt
g*Int[e^(-bt)dt]=(-g/b)*Int[e^(-bt)*(-bdt)]=(-g/b)Int(e^u du)

Answer 2

V(t) = (-g/b)*e^u + C
V(t) = (-g/b)e^(-bt) + C.
Initial position and velocity are both 0, this means, when t=0, V(t)=0, and S(t)=0.

Answer 3

V(0) = (-g/b)*e^(-b*0) + C
0 = (-g/b)*1 + C
C = g/b, so:
V(t) = (-g/b)e^(-bt) + (g/b)

Answer 4

S(t) = Int V(t) dt

Answer 5

Oh! That is also the same as \[\int\limits_{1}^{t} g e^(-bt)\] -bt is the exponent of e for Velocity right?

Answer 6

I am not sure, but I think it's the same as the integral from 0 to t, not 1 to t, we can evaluate and check, but this is a typical initial condition problem from calc. And yes, -bt is the exponent. I can write this out here
http://www.twiddla.com/935664