How things Move
Ancient Greek philosopher and scientist Aristotle developed the
earliest theory of how things move.
natural motion – motion that could maintain itself without the
aid of an outside agent. (Pushing a rock off the
ledge, falls to the ground)
liquids falling or running downhill, air rising, flames leaping upward
Aristotle believed everything was made of four elements
Aristotle's
Periodic
Table
Fire
Air
Water
“Natural Motion”
(vertical)
Earth
“Violent Motion” (horizontal)
How things Move
Aristotle's
Periodic
Table
water’s natural
resting place
is on top of earth
Fire
Air
“Natural Motion”
(vertical)
• Each element has its own
natural motion, and its own
place that it strives to be
Water
Earth
“Violent Motion” (horizontal)
• Aristotle believed an objects
natural motion was determined
by how much of each element
the object contained (rock sink
in water because it contained
mostly earth, wood floated
because it contained mostly air)
earth moves downward
because Earth’s center
is it’s natural resting place
• Violent Motion – motion that forced objects to behave contrary to an objects
natural motion, meaning an external push or pull was needed
How things Move
• Aristotle
believed that all motion on Earth was either “natural” or “violent”
• Motion not on earth followed a different set of rules
• 5th element – ether (from the Greek word for to kindle or blaze) – had no
weight and was unchangeable, and perfect in every way
• moon, sun, planets and stars were made of ether
• ether’s natural place was in the “heavens”
and it moved in perfect circles
celestial
motion
“perfect circles”
• object’s on earth could not move the way the
star’s did because they did not contain ether
• Aristotle's physics governed science until about
the mid 16th century
• Popular because it reinforced religious beliefs
“………fuse another five elements…” – Wu-Tang Clan
In the seventeenth century Newton developed Calculus which changed
the way we think about motion.
Describing Motion
Displacement
Instantaneous Speed
Average Speed, velocity
Velocity
Acceleration
Quantities which
characterize motion
Displacement – the change from one position, x1 to another position
x2
Greek letter, “delta”,
mathematically means,
“the change in”.
 x  x f  xi
Displacement is a vector quantity –
a vector has both size (aka magnitude)
and direction
If I start at a position of –2 m, and end at
a position of 3 m, what is my displacement?
 x  x f  xi
 3m   2m   3m  2m  5m
Quantities which only have magnitude or size are known as scalars
Quantities that have both magnitude and direction are known as vectors
Speed – answer the question: “How fast?”
Velocity – answers the questions: “How fast and in what direction am I traveling?”
speed –
1 piece of
info
velocity –
2 pieces of
info
Examples of speed:
55 mi/hr, 20 m/s, 300 km/hr
North
Examples of velocity:
West
55 mi/hr, due West
20 m/s, straight up
300 km/hr, 37 degrees East of North
East
South
Average velocity the ratio of the displacement, x, that occurs during a
particular time interval, t.
vavg
 x x f  xi


t
t f  ti
NOTE: We usually call
ti the starting time and
set it equal to 0, ti=0
Average Velocity:
This is a position versus time plot. From here what is the average
velocity from t = 0 to t = 3s?
vavg
 x x f  xi 2m  1m 1 m




t
t f  ti
3s  0 s 3 s
The slope of
this line is Vavg!
The slope of the line
gives us information
on the direction of the
velocity?
+ slope = + displacement
- slope = - displacement
Average speed, sAVG, is a scalar quantity
s AVG
total distance

total time
GO TO HITT QUESTION
Instantaneous Velocity and Speed
If we want to know the velocity of a particle at an instant we
simply obtain the average velocity by shrinking the time interval
 t closer and closer to 0.
x dx
v  lim

t 0 t
dt
total distance
average speed 
total time
start
zoom-in
avg. speed 1 
d3
t3
d3
d1
d1
t1
d4
t4
d4
finish
avg. speed 2 
d2
d5
t5
d5
d2
t2
d6
t6
d6
ad infinitum
instantaneous speed at this location
the trip is built out of an infinitely large number of points just like this one
Example: If a particle’s position is given by x = 4-12t+3t2, what
is its velocity at t = 1s?


d
v
4  12t  3t 2  12  6t
dt
Evaluate this as t = 1 s gives us:
m
v  12  61  12  6  6
s
What does the “-”, negative sign mean?
This tells us the direction of the particle at time t = 1s.
Acceleration: When a particle experiences a change in velocity
is undergoes an acceleration. The average acceleration over a
time interval is defined as:
aavg
v v f  vi


t t f  ti
m
units : 2
s
The instantaneous acceleration is the derivative of velocity with
respect to time:
dv
a
dt
dv d  dx  d 2 x
a
   2
dt dt  dt  dt
The acceleration of a particle at any time is the second derivative
of it’s position with respect to time. NOTE: acceleration is a vector
quantity
Typical accelerations:
Ultracentrifuge
3 x 106 m/s2
Batted baseball
3 x 104 m/s2
Bungee Jump
30 m/s2
Acceleration of gravity on Earth
9.81 m/s2
Emergency stop in a car
8 m/s2
Acceleration of gravity on Moon
1.62 m/s2
Note: Acceleration of gravity on Moon = 1.62 = 0.165  16.5 %
9.81
Acceleration of gravity on Earth
• velocity at any time can be found
from the slope of the x(t) graph
• acceleration at any time can be
found from the slope of the v(t)
graph
slope of v(t)
in most cases the acceleration is constant:
Example: Car skidding,
free falling objects, etc.
When the acceleration is constant, the
average acceleration and instantaneous
acceleration are equal so we have:
a  aavg
v v f  vi v f  vi



t t f  ti
t
Multiplying both sides by t:
at  v f  vi
v f  vi  at
NOTE: Check if this is correct, what is the final
velocity equal to at t = 0. Vi !! YEAH!
Check this out:
Remember that:
vavg
 x x f  xi


t
t f  ti
If we set ti = 0 and tf = t we can rewrite this to be:
vavg 
x f  xi
t vavg
multiplying each side by t and rearrange to xf:
t
 x f  xi
 x f  xi  vavg t
it turns out that for a particle experiencing constant acceleration:
vavg 
vi  v f
2
now plug this into this equation
 vi  v f
x f  xi  
 2

t

now from before:
v f  vi  at
plug this into here
 vi  v f
x f  xi  
 2

 vi  vi  at 
t  x f  xi  
t
2



 2vi t  at 2 
 2vi  at 

x f  xi  
t  xi  
2
 2 


1 2
1 2
x f  xi  vi t  at  x f  xi  vi t  at
2
2
we just call this the distance, d, traveled
1 2
d  vi t  at
2
next we can rearrange this equation: v f  vi  at to solve for,
time, t:
v f  vi now substitute this equation into here
t
a
 v f  vi  1  v f  vi 
1 2
  a

d  vi t  at  vi 
2
 a  2  a 
2
2
2
vi v f  vi 1  v f  2v f vi  vi 


 a
2


a
2 
a

vi v f vi2 v2f v f vi vi2

 


a
a 2a
a
2a
2
2
 vi v f
d

2a 2a
2
2
2
2
2ad  v f  vi  v f  vi  2ad
2
Equation:
Do this for homework!
Equation summary:
1
x f  xi  d  (vi  v f )t
2
1 2
x f  xi  d  v f t  at
2
Equation
v f  vi  at
d  vi t 
1 2
at
2
v2f  vi2  2ad
1
(vi  v f )t
2
1
x f  xi  d  v f t  at 2
2
x f  xi  d 
Missing Quantity
x f  xi , d
vf
t
a
vi
Example: On a dry road, a car with good tires may be able to brake
with a constant deceleration of 4.9s m/s2.
How long does it take the car, initially traveling at 24.6 m/s, take to
stop?
v f  vi  at  v f  vi  at
Given: a = -4.9 m/s2
vi = 24.6 m/s
m
vf = 0 m/s
Find:
t=?
t
v f  vi
a

0  24.6
s 5s
m
 4.92 2
s
Objects that undergo free fall are just a case of a particle under
constant acceleration.
Free - Fallin
Aristotelian physics had a short coming
what is I had a rock with some weight.
And I had a container of water, same
size, shape and weight?
• In fact all falling objects fall at the
same rate, called the acceleration of
gravity (neglecting air resistance)
• Drop different objects their speed
will increase at the same rate!
• Their speed will increase by ~ 10 m/s
(32 ft/s) every second
Free Fall Measurement –important ratios
Time Distance
1
5m
Speed
10 m/s
1
Total Distance
5m
Let 1 unit of distance =
the distance the object falls during the first second.
This turns out to be 4.9 m ~ 5 m
The acceleration is uniform, g = 9.8 m/s/s ~ 10 m/s/s
Free Fall Measurement
Time Distance
1
5m
2
15 m
Speed
10 m/s
20 m/s
1
3
Total Distance
5m
20 m
Free Fall Measurement
Time Distance
1
5m
2
15 m
3
25 m
Speed
10 m/s
20 m/s
30 m/s
1
3
5
Total Distance
5m
20 m
45 m
Free Fall Measurement
Time Distance
1
5m
2
15 m
3
25 m
4
35 m
Speed
10 m/s
20 m/s
30 m/s
40 m/s
1
3
5
7
Total Distance
5m
20 m
45 m
80 m
Free Fall Measurement
Time Distance
1
5m
2
15 m
3
25 m
4
35 m
5
45 m
Speed
10 m/s
20 m/s
30 m/s
40 m/s
50 m/s
1
3
5
7
9
Total Distance
5m
20 m
45 m
80 m
125 m
Free Fall Measurement
Time Distance
1
5m
2
15 m
3
25 m
4
35 m
5
45 m
…
…
t
Speed
10 m/s
20 m/s
30 m/s
40 m/s
50 m/s
7
Total Distance
5m
20 m
45 m
80 m
125 m
9
t2
1
3
5
speed of descent ~ time of fall
distance of fall ~ (time of fall)2
Free Falling Object
time (s)
speed (m/s)
1
10
2
20
3
30
4
40
5
50
Speed (meters/second)
50
40
30
20
10
1
2
3
Time (seconds)
4
5
holy
m
m
m
50  10
40
smokes!!
rise y s
m
s
s 
s  10
that’s g
slope 



2
run
x t
5s  1s
4s
s
speed of descent ~ time of fall
Free Falling Object
140
time (s)
distance (m)
1
5
2
20
3
45
4
80
5
125
120
Distance (meters)
100
80
60
40
20
0
1
2
3
4
5
Time (sec)
distance of fall ~ (time of fall)2