free fall with air drag force

[UrbanXrisis]

If the equation F=-bv^2 describes the drag force of an object...then the differential equation for the object's motion would be:

dv/dt= -g+bv/m

or is it...

dv/dt= g-bv/m

After solving the equation, should I get...
V=mg/b[1-e^(-bt/m)]

Also, does this look like the position v time graph of the object?
http://home.earthlink.net/~urban-xrisis/phy001.gif

[Pyrrhus]

Ok i'll do the analysis.

Forces acting on a body falling.
Down positive (Using resistive force proportional to the speed)

We got a first order DE

mg - bv = m \frac{dv}{dt}

To not get into much detail, this type of DE

\frac{dy}{dt} = ay - b

Has the following solution

y = \frac{b}{a} + ce^{at}

For an initial value, to find C.

y = \frac{b}{a} + [y_{o} - \frac{b}{a}]e^{at}

Thus for our case the solution is

v = \frac{mg}{b} + [v_{o} - \frac{mg}{b}]e^{-\frac{bt}{m}}

If we arrange the terms and v_{o} = 0

v = \frac{mg}{b} - \frac{mg}{b}e^{-\frac{bt}{m}}

v = \frac{mg}{b}(1 - e^{-\frac{bt}{m}})

[UrbanXrisis]

instead of F=-bv^2, what if F=-2v? The question tells me to ignore the effects of gravity in this problem. Would the DE then be dv/dt= 2v/m?

Since f=bv...and b=-2

dv/dt= g-bv/m
dv/dt= g-(-2)v/m
the question asks to ignore gravity...
dv/dt= -(-2)v/m
dv/dt= 2v/m

[Pyrrhus]

If we ignore gravity, then indeed our Newtonian analysis will be

The Object is falling, and the air drag is in the opposite direction
Down positive.

m\frac{dv}{dt} = -bv

\frac{dv}{dt} = \frac{-bv}{m}

Solving this:

\frac{dv}{dt} = \frac{-bv}{m}

\frac{dv}{v} = \frac{-b}{m}dt

\int_{v_{o}}^{v} \frac{dv}{v} = \int_{0}^{t} \frac{-b}{m}dt

\ln |v|]_{v_{o}}^{v} = \frac{-b}{m}t]_{0}^{t}

For v_{o} = 0

ln |v| = \frac{-b}{m}t

v = e^{\frac{-b}{m}t}

[UrbanXrisis]

Also, does this look like the position v time graph of the object?
http://home.earthlink.net/~urban-xrisis/phy001.gif

Does this look correct?

[UrbanXrisis]

If we ignore gravity, then indeed our Newtonian analysis will be

The Object is falling, and the air drag is in the opposite direction
Down positive.

m\frac{dv}{dt} = -bv

\frac{dv}{dt} = \frac{-bv}{m}

Solving this:

\frac{dv}{dt} = \frac{-bv}{m}

\frac{dv}{v} = \frac{-b}{m}dt

\int_{v_{o}}^{v} \frac{dv}{v} = \int_{0}^{t} \frac{-b}{m}dt

However, the retarding force is not F=bv, it is F=-2v as stated above. Correct me if I’m wrong, but isn’t \frac{dv}{dt} = \frac{-bv}{m} for when F=bv? Should it be \frac{dv}{dt} = \frac{2v}{m} ?

[Pyrrhus]

Sure, just substitute b with 2, and for the graphic, It looks rather odd to me, but maybe someone else can find it the physical meaning.

[UrbanXrisis]

my graph represents the object accelerating and to at point where it has constant velocity because of the drag force

or should it look something like this...
http://home.earthlink.net/~urban-xrisis/phy002.gif

[Pyrrhus]

the graphic looks better now, and i added the extra solve steps for neglecting gravity.

[blad]

what is b in this equation