Kinematics
AP Physics 1
Defining the important variables
Kinematics is a way of describing the motion of objects
without describing the causes. You can describe an
object’s motion:
In words
Mathematically
Pictorially Graphically
No matter HOW we describe the motion, there are several KEY VARIABLES
that we use.
Symbol
Variable
Units
t
Time
s
a
Acceleration
m/s/s
x or y
Displacement
m
vo
Initial velocity
m/s
v
Final velocity
m/s
g or ag
Acceleration due to
gravity
m/s/s
Reference Frames
Any measurement of position, displacement,
velocity or acceleration must be made with
respect to a defined reference frame—this is first
step in problem solution.
Possible reference frames:

•Window with up = + or –
•Un-stretched net with up = + or •Stretched net with up = + or –
•Ground = not sufficient
information
Module 2-1
Average Speed and Velocity
distance traveled
average speed 
time elapsed
displaceme nt
average velocity (v ) 
time elapsed
x2  x1
x
v 

t 2  t1
t
Average velocity is a vector, so it has magnitude
and direction. In one dimension we use + or minus
sign to indicate direction.
Module 2-6
Instantaneous Velocity
instantaneous velocity is defined as the average velocity
over an infinitesimally short time interval.
v 
Module 2-9
lim
t  0
x
t
The 3 Kinematic equations
There are 3 major
kinematic equations
than can be used to
describe the motion in
DETAIL. All are used
when the acceleration
is CONSTANT.
v f  vo  at
2
1
x  xo  vot 
at
2
2
2
v f  vo  2a( x f  xo )
Kinematic #1
v f  vo
v
a

t
t
v f  vo  at
v f  vo  at
Kinematic #1
Example: A boat moves slowly out of a marina (so as to not
leave a wake) with a speed of 1.50 m/s. As soon as it
passes the breakwater, leaving the marina, it throttles up
and accelerates at 2.40 m/s/s.
a) How fast is the boat moving after accelerating for 5 seconds?
What do I
know?
vo= 1.50 m/s
a = 2.40 m/s/s
t=5s
What do I
want?
v=?
v  vo  at
v  (1.50)  (2.40)(5)
v  13.5 m/s
Kinematic #2
2
1
x  xo  voxt  at
2
b) How far did the boat travel during that time?
x  xo  voxt  1 at 2
2
2
1
x  0  (1.5)(5)  (2.40)(5 )
2
x  37.5 m
Kinematic #3
v  v  2a( x  xo )
2
2
o
Example: You are driving through town at 12 m/s when suddenly a ball rolls
out in front of your car. You apply the brakes and begin decelerating at
3.5 m/s/s.
How far do you travel before coming to a complete stop?
What do I
know?
vo= 12 m/s
a = -3.5 m/s/s
V = 0 m/s
What do I
want?
x=?
v 2  vo2  2a( x  xo )
0  12 2  2(3.5)( x  0)
 144  7 x
x  20.57 m
Common Problems Students Have
I don’t know which equation to choose!!!
Equation
v  vo  at
Missing Variable
x
x  xo  voxt  1 at 2
2
v
v  v  2a( x  xo )
t
2
2
o
Up and Down Motion
For object that is thrown upward
and returns to starting position:
1
y  y  v t  gt
2
0
2
0
• assumes up is positive
•velocity changes sign (direction)
but acceleration does not
v  v0  gt
•Velocity at top is zero
•time up = time down
•Velocity returning to starting
position = velocity when it was
released but opposite sign
Module 4-3
Acceleration due to Gravity
Module 4-4
Kinematics for the VERTICAL Direction
All 3 kinematics can be used to analyze one
dimensional motion in either the X direction OR the
y direction.
v  vo  at  v y  voy  gt
2
2
1
1
x  xo  voxt 
at  y  yo  voyt 
gt
2
2
2
2
2
2
v  vox  2a ( x  xo )  v y  voy  2 g ( y  yo )
“g” or ag – The Acceleration due to gravity
The acceleration due to gravity is a special constant that exists
in a VACUUM, meaning without air resistance. If an object is
in FREE FALL, gravity will CHANGE an objects velocity by
9.8 m/s every second.
g  ag  9.8 m / s
The acceleration due to gravity:
•ALWAYS ACTS DOWNWARD
•IS ALWAYS CONSTANT near the
surface of Earth
2
Examples
A stone is dropped at rest from the top of a cliff. It is
observed to hit the ground 5.78 s later. How high
is the cliff?
What do I
know?
v = 0 m/s
oy
g = -9.8 m/s2
yo=0 m
t = 5.78 s
What do I
want?
y=?
Which variable is NOT given and
NOT asked for?
Final Velocity!
y  yo  voyt  1 gt 2
2
y  (0)(5.78)  4.9(5.78) 2
y  -163.7 m
H =163.7m
Examples
A pitcher throws a fastball with a velocity of 43.5 m/s. It is
determined that during the windup and delivery the ball covers
a displacement of 2.5 meters. This is from the point behind the
body when the ball is at rest to the point of release. Calculate
the acceleration during his throwing motion.
What do I
know?
vo= 0 m/s
x = 2.5 m
v = 43.5 m/s
What do I
want?
a=?
Which variable is NOT given and
NOT asked for?
TIME
v  v  2a( x  xo )
2
2
o
43.5  0  2a(2.5  0)
a  378.5 m/s/s
2
2
Examples
How long does it take a car at rest to cross a 35.0 m
intersection after the light turns green, if the acceleration
of the car is a constant 2.00 m/s/s?
What do I
know?
vo= 0 m/s
x = 35 m
a = 2.00 m/s/s
What do I
want?
Which variable is NOT given and
NOT asked for?
Final Velocity
t=?
x  xo  voxt  1 at 2
2
35  0  (0)  1 (2)t 2
2
t  5.92 s
Graphical Analysis of Motion
AP Physics 1
Slope – A basic graph model
A basic model for understanding graphs in physics is SLOPE.
Using the model - Look at the formula for velocity. slope  Rise  v  x
Run
t
Who gets to play the role of the slope? Velocity
Who gets to play the role of the y-axis or the rise? Displacement
Who get to play the role of the x-axis or the run? Time
What does all the mean? It means that if your are given a Displacement
vs. Time graph, to find the velocity of an object during specific time
intervals simply find the slope.
Displacement vs. Time graph
What is the velocity of the object from 0
seconds to 3 seconds?
The velocity is the slope!
Example
It is very important that you are able to look at a graph
and explain it's motion in great detail. These graphs
can be very conceptual.
Look at the time interval t = 0 to t = 9
seconds. What does the slope do?
It increases, the velocity is increasing
Look at the time interval t = 9 to t = 11
seconds. What does the slope do?
No slope. The velocity is ZERO.
Look at the time interval t = 11 to t = 15
seconds. What does the slope do?
The slope is constant and positive. The
object is moving forwards at a constant
velocity.
Look at the time interval t = 15 to t = 17
seconds. What does the slope do?
The slope is constant and negative.
The object is moving backwards at a
constant velocity.
Slope – A basic graph model
Let’s look at another model
Who gets to play the role of the slope? Acceleration
Who gets to play the role of the y-axis or the rise? Velocity
Who get to play the role of the x-axis or the run? Time
What does all the mean? It means that if your are given a Velocity vs.
Time graph. To find the acceleration of an object during specific time
intervals simply find the slope.
Velocity vs. Time Graph
What is the acceleration from 0 to 6s?
What is the acceleration from 6 to 9s?
You could say one of two things here:
The object has a ZERO acceleration
The object has a CONSTANT velocity
What is the acceleration from 14 to 15s?
A negative acceleration is sometimes called DECELERATION. In other
words, the object is slowing down. An object can also have a negative
acceleration if it is falling. In that case the object is speeding up. CONFUSING?
Be careful and make sure you understand WHY the negative sign is there.
Area – the “other” basic graph model
Another basic model for understanding graphs in
physics is AREA.
Let's try to algebraically make our formulas
look like the one above. We'll start with our
formula for velocity.
Who gets to play the role of the base?
Time
Who gets to play the role of the height?
Velocity
What kind of graph is this?
A Velocity vs. Time graph ( velocity = yaxis & time = x-axis)
Who gets to play the role of the
Area?
Displacement
Example
What is the displacement during the
time interval t = 0 to t = 5 seconds?
That happens to be the area!
What is the displacement during the
time interval t = 8 to t = 12 seconds?
Once again...we have to find the area.
During this time period we have a
triangle AND a square. We must find the
area of each section then ADD them
together.
Asquare  BH  4(35)  140m
1
1
BH  (4)(60)  120m
2
2
Atotal  120m  140m
Atriangle 
Displaceme nt  260m
Area – the “other” basic graph model
Let's use our new model again, but for our equation for
acceleration.
What does this mean?
Who gets to play the role of the base? Time
Who gets to play the role of the height? Acceleration
What kind of graph is this?
An Acceleration vs. Time graph ( acceleration = y-axis & time = x-axis)
Who gets to play the role of the Area?
The Velocity
Acceleration vs. Time Graph
What is the velocity during the time
interval t = 3 and t = 6 seconds? Find
the Area!
A  Bh  v  ta
v  (3)(6)  18m / s
Summary
There are 3 types of MOTION graphs
 Displacement(position) vs. Time
 Velocity vs. Time
 Acceleration vs. Time
There are 2 basic graph models
 Slope
 Area
Summary
v (m/s)
a (m/s/s)
x (m)
area = x
t (s)
t (s)
area = v
t (s)
Comparing and Sketching graphs
One of the more difficult applications of graphs in physics is when given
a certain type of graph and asked to draw a different type of graph
List 2 adjectives to describe the SLOPE or VELOCITY
1. The slope is CONSTANT
2. The slope is POSITIVE
x (m)
t (s)
How could you translate what
the SLOPE is doing on the
graph ABOVE to the Y axis on
the graph to the right?
v (m/s)
t (s)
Example
v (m/s)
x (m)
t (s)
t (s)
1st line
• The slope is constant
• The slope is “-”
2nd line
• The slope is “0”
3rd line
• The slope is “+”
• The slope is constant
Example – Graph Matching
What is the SLOPE(a) doing?
a (m/s/s)
The slope is increasing
v (m/s)
t (s)
a (m/s/s)
t (s)
t (s)
a (m/s/s)
t (s)