One-Dimensional Motion with and without air resistance.
One of the first applications you ever considered in Calculus 1 was the motion of an object in free
fall.
The equations of motion are : position
velocity
acceleration
y(t) = -16t2 + v0t +y0
v(t) = -32t +v0
a(t) = -32
We are now going to consider the case where we add in air resistance.
We'll consider the case in which the air resistance force is proportional to the velocity.
dv
Starting with Newton's Second Law of Motion : m
mg v . Where β is called the drag
dt
coefficient and depends on the physical properties of the body.
dv

v.
m
g
dt
We can write this in the form the first order linear DE:
dv
dt

v
m
g

m
dt
The integrating factor is : e
We obtain

m
dt
e
e
d e
 
t
m
v
 
t
m
g e
 
t
m
dt
Then :
e
 
t
m
v
g e
 
t
m
dt
mg 
e

 
t
m
c
mg
v( t )
 
t
m
c e

t
Applying the initial condition v(0) = v0 we have finally v
mg

g
m

v 0 e
m
Recall in the English system the weight of an object is mg so to formulate this in terms weight and
 g t
not mass we would have v
w
w


v 0 e
w
Let’s Compare the velocity in the cases with and without air resistance.
Let vf represent the velocity of an object in free fall and vr represent the velocity of an object with
air resistance.
g 32
v 0 64
 5
m 1.563
w 50
y 0 80
v f( t )
v r( t )
g t
v0
w
w


 g t
v 0 e
w
50
v f( t )
0
1
2
3
4
5
v r( t )
50
t
Notice with air resistance the velocity appears to approach a limiting value whereas in the free fall
case the velocity continues to decrease. This limiting value is called the terminal velocity and we
can see
 g t
lim
t

g
32
w
w


v 0 e
w
w

.
v0
64
m 1.563
w 50
y 0 80
 5
v f( t )
g t
v r( t )
v0
 g t
w
w


v f( t )
v 0 e
0
w
2
4
v r( t )
w

100
200
t
View animation VELOCITY and note as  decreases vf approaches vr and the terminal velocity
– > -
To prove this formally show using some algebraic manipulation and L'Hoptital's Rule
t
mg
g
m

lim

0 
v 0 e
m
=
g t
v0
Let’s Consider The position functions. Using the same conventions as before we have:
2
y f( t )
16t v 0t y 0 .
Using
dy r
dt
 g t
v r( t )
w
w


v 0 e
w
we can separate variables and integrate to obtain:
 g t
y r( t )
w
t

w  w
 g 
v 0 e
w
C.
Applying the initial condition we obtain C = y0 +
w  w
 g 
v0 .
 g t
w
t

Finally we obtain: y r( t )
w  w
 g 
v 0 e
w
y0
w  w
 g 
v0 .
POSIT ION
20 0
y f( t )
10 0
y r( t )
0
2
4
10 0
t
View animation POSITION and Again as  decreases and you will see the position function for
the air resistance case approach the position function for the case with no air resistance
Let’s consider the two together
g 32
v 0 64
 5
m 1.563
w 50
y 0 80
y f( t )
2
16t
v 0t
y0
 g t
y r( t )
w
t

w  w
 g 
v 0 e
w
y0
w  w
 g 
v0
In the first graph let's consider what happens in the first 2 seconds Because of air resistance yr
reaches its maximum height and starts falling before yf.
POSIT ION
15 0
y f( t )
y r( t ) 10 0
50
0
1
t
2
VELOCITY
10 0
v f( t )
50
v r( t )
0
1
2
50
t
View animation FIRST 2 SECS
In this second graph We consider the entire trajectories. Note that without air resistance the
particle rises more quickly then as it starts to fall it speeds up catches and passes the particle
with air resistance.
The reason being of course that with air resistance the particle reaches its terminal velocity. Note
also that as the particle reaches terminal velocity the plot of vr approaches a constant value and
the position function is becoming linear.
POSIT ION
20 0
y f( t )
10 0
y r( t )
0
2
4
10 0
t
VELOCITY
v f( t )
50
v r( t )
0
2
4
t
View animation ENTIRE TRAJECTORY
In certain problems such as releasing an object or parachute problems since the motion is one
direction we are more concerned with speed and distance rather than position and velocity. In
these cases we take down as the positive direction.
dv
The DE takes the form m
mg v
dt
t
and the solution is v
mg
mg


v 0 e
m
. ( note :Typically in these cases v 0 is 0.)
g 32
v0 0

5
m 1.563
w 50
y 0 80
v f( t ) g t
v0
 g t
v r( t )
w
w


v 0 e
w
100
v f( t )
v r( t )
50
w

0
1
2
3
4
5
t
For distance we then have:
Using the same conventions as before we have:
y f( t )
Using
2
16t
dy r
dt
v 0 t
y 0.
 g t
v r( t )
w
w


v 0 e
w
we can separate variables and integrate to obtain:
 g t
y r( t )
w
t

w  w
 g 
v 0 e
w
C.
Applying the initial condition we obtain C = y0 -
w  w
 g 
v0 .
 g t
w
t

Finally we obtain: y r( t )
w  w
 g 
v 0 e
w
y0
w  w
 g 
v0 .
g 32
v0 0

5
m 1.563
w 50
y0 0
y f( t )
2
16t
v 0 t
y0
 g t
y r( t )
w
t

w  w
 g 
v 0 e
w
y0
6
8
w  w
 g 
v0
100
80
y f( t )
60
y r( t )
40
20
0
2
4
t
Here we have taken the release point y(0) to be 0.
Note in 2 secs without air resistance the object has already fallen 100 ft whereas with air
resistance the object has only fallen about 20 ft.
Summary
Velocity
Metric System
dv
m
dt
mg v
t
v
mg

g
m

v 0 e
m
English System
g
g 
v
w
w
w


dv
dt
 g t
v
v 0 e
w
Speed
dv
m
mg v
dt
t
v
mg
mg


g
  v
w
dv
g
dt
v
w
w


v 0 e
m
.
 g t
v 0 e
w
Position
Metric System
y r( t )
mg 
t

m  mg


 t
v 0 e
m
y0
m  mg


v0
English System
 g t
y r( t )
w
t

w  w
 g 
v 0 e
w
w  w
 g 
y0
Distance
Metric System
 t
y r( t )
mg 
t

m  mg


v 0 e
m
y0
m  mg


v0
English System
 g t
y r( t )
w
t

w  w
 g 
v 0 e
w
y0
w  w
 g 
v0
v0