- #1

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For a particle moving downward and taking positive upwards, I have

mdv/dt=-mg+bv

^{2}

The terminal velocity comes out at v

_{l}=-√(mg/b) (negative as it obviously has its terminal velocity downwards).

Substituting this in gives

dv/dt=-g(1-(v/v

_{l})

^{2})

Now make the substitution z=v/v

_{l}so dv/dt=v

_{l}dz/dt. Then we obtain

v

_{l}dz/dt=-g(1-z

^{2})

Noting that 1/1-z

^{2}=0.5[(1/1+z)+(1/1-z)] and separating variables we get

∫[(1/1+z)+(1/1-z)]dz=-2g/v

_{l}∫dt

Integrating each side then gives (the initial conditions are v=0 at t=0 so the constant of integration is zero)

ln(1+z/1-z)=-2gt/v

_{l}.

Next let k=v

_{l}/2g so that

ln(1+z/1-z)=-t/k

e

^{-t/k}=1+z/1-z

e

^{-t/k}-ze

^{-t/k}=1+z

z(1+e

^{-t/k})=e

^{-t/k}-1

z=(e

^{-t/k}-1)/(e

^{-t/k}+1)

Therefore

v=v

_{l}[(e

^{-t/k}-1)/(e

^{-t/k}+1)]

Now, the correct answer should be

v=v

_{l}[(1-e

^{-t/k})/(1+e

^{-t/k})]

i.e

v=-v

_{l}[(e

^{-t/k}-1)/(e

^{-t/k}+1)]

which must be right because as t→∞, v→v

_{l}which by the definition of v

_{l}is expected.

I've spent a lot of time trying to work out where I'm going wrong and it's driving me crazy. Thanks in advance :)