AP Physics C

Here is the situation! Suppose you drop a marble into a graduated

cylinder full of glycerin. Obviously there will be some drag force acting on the marble. Initially the marble will have some acceleration but after a short time the marble will be moving downward at constant speed.

Physicists have a name for this “constant speed” the marble reaches…it is “Terminal Velocity”. The acceleration at this “Terminal Velocity” is zero.

Think about this: As the marble moves downward through the glycerin, the acceleration of the marble decreases from some downward value to zero at terminal velocity.

Let the drag force provided by the glycerin be defined by:

kvFd Such that the drag force increases proportionally with the velocity.

kv

mg

Now write Newton’s Second Law equation for the marble.

makvmgF

makvmgF

kv

mg

Remember that the key to “terminal

velocity” is a = 0 so…

0 kvmg

Now solve for velocity… in this case terminal velocity k

mgvT

We will now develop an expression for velocity as a function of time, v(t), to find the instantaneous velocity at any given point in time, tLet’s begin with our Newton’s second law equation

Then, solve for “a”

Remembering that

makvmg

vm

kga

dtdva this equation

becomes…

vm

kg

dt

dv The equation in this form is

called a “differential equation”

Starting with the differential equation…

We will need to find an anti-derivative, but first we need to get the equation in the right form…In other words, all of the terms with the variable (v) on one side of the equation and all others on the other.

Time for a little algebra…

vm

kg

dt

dv

dtvm

kgdv

dtvg

dv

mk

dtdvvgvg

dv

mk

mk

1

Working from the equation …

tv

mk

dtdvvg 00

1

It is time to find the anti-derivative and set the limits

tv

mk

dtdvvg 00

1

If we evaluate the right side of the equation it becomes…

tdt ttt

00

tv

mk

dtdvvg 00

1

We want to use a familiar “anti-derivative” form to finish so we will use

uduu

ln1

Because it most closely matches the form of our equation.

To evaluate the left side of the equation

We need to use a method of integration (finding the anti-derivative) called “u” substitution…here is how it works.

dvvg

v

mk 0

1

Let vgu mk so,

then

dvdu mk

We need to rearrange the “du” equation so that we can substitute for “dv”

dudv km next we substitute these

expressions into our anti-derivative

dvvg

v

mk 0

1Substituting the expressions for “u” and “du” into this equation causes it to become…

duu

duu

km

km

1

)(1

We can factor the “-k/m” outside of the integral (anti-derivative) because it is a constant.

Note: We have temporarily stopped using the limits for the integral during the substitution process.

So now our equation with the substitutions is in the general form shown earlier can be evaluated as follows:

uduu

km

km ln

1

Now we need to substitute our expression for “u” back into the equation, so…

v

mk

km vg

0ln Note:

We re-established the limit

v

mk

km vg

0ln So we have

this...

Now it is time to evaluate the limits, so…

g

vg

gvg

mk

km

mk

mk

km

ln

0lnln

tg

vg mk

km

ln

Okay, let’s put the right and left sides back together to get:

Remember that our goal is to solve for velocity, v, as a function of time. More algebra…move the “–k/m” to the other side and then get rid of the “ln”

tg

vgmkm

k

ln Next step: get rid of the “ln” Remember…

So…xe x ln

tkmg

tmk

tmk

tg

vg

mk

mk

mk

mk

mk

etv

gevg

eg

vg

ee

1)(

lnFrom the previous page…

A little algebra (magic)…

More algebra…

solving for velocity as a function of time we get…