- #1
Decypher
- 2
- 0
Example problem from a Classical Mechanics book.
Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.
F = m(dv/dt) = -mg-kmv
dv/(kv + g) = -dt
(1/k)ln(kv + g) = -t + c
kv + g = e^(-kt+kc)
v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt)
I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
I know a property of exponents allows you to separate the exponents powers when they are being added, then each one becomes the power of an exponent which is where e^-kt comes from, but I am not sure about where the other half of the exponent goes.
Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.
F = m(dv/dt) = -mg-kmv
dv/(kv + g) = -dt
(1/k)ln(kv + g) = -t + c
kv + g = e^(-kt+kc)
v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt)
I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
I know a property of exponents allows you to separate the exponents powers when they are being added, then each one becomes the power of an exponent which is where e^-kt comes from, but I am not sure about where the other half of the exponent goes.