### Chapter 2 Notes - Beaumont High School

Chapter 2
Kinematics in One Dimension
Objectives
• We will compare and contrast distance to
displacement, and speed to velocity
• We will be able to solve problems using
distance, displacement, speed and
velocity
Kinematics deals with the concepts that
are needed to describe motion.
These concepts are:
Displacement
Velocity
Acceleration
Time
Dynamics deals with the effect that forces
have on motion.
Together, kinematics and dynamics form
the branch of physics known as Mechanics.
Right now we are focused on Kinematics,
how things move. Not what is pushing
them to move (dynamics)
2.1 Displacement
Displacement is the shortest distance
from the initial to the final position

x o  initial
position

x  final position
  
 x  x  x o  displaceme
nt
2.1 Displacement
Problem: What is the displacement?

x o  2 .0 m

 x  5 .0 m

x  7 .0 m
  
 x  x  x o  7.0 m  2.0 m  5 . 0 m
2.1 Displacement
Problem: What is the displacement?

x  2 .0 m

 x   5 .0 m

x o  7 .0 m
r
r
r
 x  x  x o  2.0 m  7.0 m   5 .0 m
*Notice*
r
r
r
 x  x  x o  2.0 m  7.0 m   5 .0 m
• In the last problem the answer is -5.0m.
What does that mean?
• It means that the object traveled in the
negative direction 5.0m. IT DOES NOT
MEAN THE OBJECT WAS WALKING
BACKWARDS OR GOING BACK IN
TIME!
2.1 Displacement

x o   2 .0 m

x  5 .0 m

 x  7 .0 m
  
 x  x  x o  5.0 m    2 .0  m  7 . 0 m
*Distance is not Displacement*
• Distance refers to the
total amount of land
covered
(For Example: you walk
around a track and you
have covered 400 m.)
• Displacement is the final
point – the initial point
(For Example: you walk
around a track and your
displacement is 0 m.)
2.1.1. The branch of physics that deals with motion is called
mechanics. Kinematics is the portion of mechanics that
describes motion without any reference to which of the
following concepts?
a) forces
b) accelerations
c) velocities
d) displacements
e) time
2.1.2. A particle travels along a curved path between two points
A and B as shown. Complete the following statement: The
displacement of the particle does not depend on
a) the location of A.
b) the location of B.
c) the direction of A from B.
d) the distance traveled from A to B.
e) the shortest distance between A and B.
2.1.3. For which one of the following situations will the path
length equal the magnitude of the displacement?
a) An Olympic athlete is running around an oval track.
b) A roller coaster car travels up and down two hills.
c) A truck travels 4 miles west; and then, it stops and travels 2
miles west.
d) A ball rises and falls after being thrown straight up from the
earth's surface.
e) A ball on the end of a string is moving in a vertical circle.
2.2 Speed and Velocity
Average speed is the distance traveled divided by
the time required to cover the distance.
Average
speed 
Distance
Elapsed
time
SI units for speed: meters per second (m/s)
Try to let go of miles per hour!
2.2 Speed and Velocity
Example 1 Distance Run by a Jogger
How far does a jogger run in 1.5 hours (5400 s) if his
average speed is 2.22 m/s?
Average
speed 
Distance
Elapsed
Distance
  Average
speed
time
 Elapsed
  2 . 22 m s  5400 s   12000 m
time

2.2.1. A turtle and a rabbit are to have a race. The turtle’s
average speed is 0.9 m/s. The rabbit’s average speed is 9
m/s. The distance from the starting line to the finish line is
1500 m. The rabbit decides to let the turtle run before he
starts running to give the turtle a head start. What,
approximately, is the maximum time the rabbit can wait
before starting to run and still win the race?
a)
b)
c)
d)
e)
15 minutes
18 minutes
20 minutes
22 minutes
25 minutes
2.2 Speed and Velocity
Average velocity is the displacement divided by the
elapsed time.
Average
velocity

Displaceme
Elapsed

v
 
x  xo
t  to

nt
time

x
t
Remember – Its ALWAYS FINAL minus INITIAL.
Even if the number turns out to be NEGATIVE
2.2 Speed and Velocity
Example 2 The World’s Fastest JetEngine Car
Andy Green in the car
ThrustSSC set a world
record of 341.1 m/s in 1997.
To establish such a record,
the driver makes two runs
through the course, one in
each direction, to nullify
wind effects. From the
data, determine the average
velocity for each run.
2.2 Speed and Velocity

v

v

x
t

x
t

 1609 m
  339 . 5 m s
4.740 s

 1609 m
4.695 s
  342 . 7 m s
2.2.2. Which
one of the following quantities is defined
as the distance traveled divided by the elapsed
time for the travel?
a)
b)
c)
d)
e)
average speed
average velocity
average acceleration
instantaneous velocity
instantaneous acceleration
2.2.3. Which
one of the following quantities is defined
as an object’s displacement divided by the elapsed
time for the displacement?
a)
b)
c)
d)
e)
average speed
average velocity
average acceleration
instantaneous velocity
instantaneous acceleration
2.3 Acceleration
Acceleration occurs when there is a change in
velocity during a specific time period
2.3.1. Which one of the following situations does the object
have no acceleration?
a) A ball at the end of a string is whirled in a horizontal
circle at a constant speed.
b) Seeing a red traffic light ahead, the driver of a minivan
steps on the brake. As a result, the minivan slows from
15 m/s to stop before reaching the light.
c) A boulder starts from rest and rolls down a mountain.
d) An elevator in a tall skyscraper moves upward at a
constant speed of 3 m/s.
2.3 Acceleration
DEFINITION OF AVERAGE ACCELERATION

a 
 
v  vo
t  to


v
t
Distance divided by time2
2.3 Acceleration
Example 3 Acceleration and Increasing Velocity
Determine the average acceleration of the plane.

vo  0 m s

a 
 
v  vo
t  to

v  260 km h

to  0 s
260 km h  0 km h
29 s  0 s
t  29 s
  9 .0
km h
s
2.3 Acceleration
2.3.4. A sports car starts from rest. After 10.0 s, the speed of
the car is 25.0 m/s. What is the magnitude of the car’s
acceleration?
a) 2.50 m/s2
b) 5.00 m/s2
c) 10.0 m/s2
d) 25.0 m/s2
e) 250 m/s2
2.3 Acceleration
Example 3 – Acceleration and Decreasing Velocity
Solve for acceleration

a 
 
v  vo
t  to

13 m s  28 m s
12 s  9 s
  5 .0 m s
2
2.3 Acceleration
Common Usage
• If an object is slowing
down it is still
“accelerating” because
the velocity is changing.
• However, most people
refer to that as
“decelerating”
2.3.2. In which one of the following situations does the car have
an acceleration that is directed due north?
a) A car travels northward with a constant speed of 24 m/s.
b) A car is traveling southward as its speed increases from
24 m/s to 33 m/s.
c) A car is traveling southward as its speed decreases from 24
m/s to 18 m/s.
d) A car is traveling northward as its speed decreases from 24
m/s to 18 m/s.
e) A car travels southward with a constant speed of 24 m/s.
2.3.3. A postal truck driver driving due east gently steps on her
brake as she approaches an intersection to reduce the speed
of the truck. What is the direction of the truck’s
acceleration, if any?
a) There is no acceleration in this situation.
b) due north
c) due east
d) due south
e) due west
Question
• How many “Accelerators” does a car
have?
•
•
•
•
3
Gas pedal
Brake
Steering Wheel – A change in direction is
a change in velocity
2.3.4. The drawing shows the position of a rolling ball at one second
intervals. Which one of the following phrases best describes the
motion of this ball?
a) constant position
b) constant velocity
c) increasing velocity
d) constant acceleration
e) decreasing velocity
2.3.5. A police cruiser is parked by the side of the road when a
speeding car passes. The cruiser follows the speeding car.
Consider the following diagrams where the dots represent
the cruiser’s position at 0.5-s intervals. Which diagram(s)
are possible representations of the cruiser’s motion?
a) A only
b) B, D, or E only
c) C only
d) E only
e) A or C only
Some Minutia
• So far we have analyzed
average velocity, speed and
acceleration.
• Instantaneous speed,
velocity or acceleration is the
speed, velocity or
acceleration of an object at a
specific time.
For example – Speedometer
gives us instantaneous
speed.
Easy Rule of Thumb or Hands – HA!
• Velocity – Right Hand, Acceleration – Left
• Arms together – object speeding up
• Arms separate – object slowing down
2.4.1. Complete the following statement: For an object
moving at constant acceleration, the distance traveled
a) increases for each second that the object moves.
b) is the same regardless of the time that the object moves.
c) is the same for each second that the object moves.
d) cannot be determined, even if the elapsed time is known.
e) decreases for each second that the object moves.
2.4.2. Complete the following statement: For an object
moving with a negative velocity and a positive
acceleration, the distance traveled
a) increases for each second that the object moves.
b) is the same regardless of the time that the object moves.
c) is the same for each second that the object moves.
d) cannot be determined, even if the elapsed time is known.
e) decreases for each second that the object moves.
2.4.3. At one particular moment, a subway train is moving
with a positive velocity and negative acceleration.
Which of the following phrases best describes the
motion of this train? Assume the front of the train is
pointing in the positive x direction.
a) The train is moving forward as it slows down.
b) The train is moving in reverse as it slows down.
c) The train is moving faster as it moves forward.
d) The train is moving faster as it moves in reverse.
e) There is no way to determine whether the train is moving
forward or in reverse.
2.4 Equations of Kinematics for Constant Acceleration

v
 
x  xo
t  to

a 
 
v  vo
t  to
It is customary to dispense with the use of boldface
Symbols overdrawn with arrows for the displacement,
velocity, and acceleration vectors. We will, however,
continue to convey the directions with a plus or minus
sign.
v
x  xo
t  to
a
v  vo
t  to
Objectives
We will be able to differentiate and use
the 4 equations of kinematics to solve
kinematic problems
2.4 Equations of Kinematics for Constant Acceleration
Equations of Kinematics for Constant Acceleration
v  v o  at
x
1
2
v o  v  t
v  v  2 ax
2
2
o
x  x o  v ot 
1
2
at
2
2.4 Equations of Kinematics for Constant Acceleration
How far does the boat travel?
x  x o  v ot 
1
2
at
2
 0 m  + 6 .0 m s 8 .0 s  
  110 m
1
2
2 .0 m
s
2
8 .0 s 
2
2.4 Equations of Kinematics for Constant Acceleration
Example 6 Catapulting a Jet
Find the displacement of the jet
vo  0 m s
x  ??
a   31 m s
v   62 m s
2
2.4 Equations of Kinematics for Constant Acceleration
v v
2
x
2a
2
o

62 m s 

2
 0 m s 
2 31 m s
2

2
  62 m
2.5 Applications of the Equations of Kinematics
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables.
4. Verify that the information contains values for at least three
of the five kinematic variables. Select the appropriate equation.
5. When the motion is divided into segments, remember that
the final velocity of one segment is the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a
kinematics problem.
2.5 Applications of the Equations of Kinematics
Example 8 An Accelerating Spacecraft
A spacecraft is traveling with a velocity of +3250 m/s. Suddenly
the retrorockets are fired, and the spacecraft begins to slow down
with an acceleration whose magnitude is 10.0 m/s2. What is
the velocity of the spacecraft when the displacement of the craft
is +215 km, relative to the point where the retrorockets began
firing?
x
a
v
vo
+215000 m
-10.0 m/s2
?
+3250 m/s
t
2.5 Applications of the Equations of Kinematics
2.5 Applications of the Equations of Kinematics
x
a
v
vo
+215000 m
-10.0 m/s2
?
+3250 m/s
v  v  2 ax
2
v
2
o
3250
  2500 m s
v

m s   2 10 . 0 m s
2
t
v  2 ax
2
o
2
215000
m
2.5.1. Starting from rest, two objects accelerate with the
same constant acceleration. Object A accelerates for
three times as much time as object B, however. Which
one of the following statements is true concerning these
objects at the end of their respective periods of
acceleration?
a) Object A will travel three times as far as object B.
b) Object A will travel nine times as far as object B.
c) Object A will travel eight times as far as object B.
d) Object A will be moving 1.5 times faster than object B.
e) Object A will be moving nine times faster than object B.
2.6 Freely Falling Bodies
In the absence of air resistance, it is found that ALL
bodies at the same location above the Earth fall
vertically with the same acceleration.
This idealized motion is called free-fall and the
acceleration of a freely falling body is called the
acceleration due to gravity.
g  9 . 80 m s
2
or
32 . 2 ft s
2
2.6 Freely Falling Bodies
g  9 . 80 m s
2
2.6 Freely Falling Bodies
Example 10 A Falling Stone
A stone is dropped from the top of a tall building.
After 3.00s of free fall, what is the displacement y
of the stone?
2.6 Freely Falling Bodies
y
a
?
-9.80 m/s2
v
vo
t
0 m/s
3.00 s
2.6 Freely Falling Bodies
y
a
?
-9.80 m/s2
y  y o  v ot 
1
2
at
vo
t
0 m/s
3.00 s
2
 0 m   0 m s 3 .00 s  
  44 .1 m
v
1
2
 9 .80 m
s
2
3 .00
s
2
2.6 Freely Falling Bodies
Example 12 How High Does it
Go?
The referee tosses the coin
up with an initial speed of
5.00m/s. In the absence if
air resistance, how high
does the coin go above
its point of release?
2.6 Freely Falling Bodies
y
a
v
vo
?
-9.80 m/s2
0 m/s
+5.00
m/s
t
2.6 Freely Falling Bodies
y
a
v
vo
?
-9.80 m/s2
0 m/s
+5.00
m/s
v v
2
v  v  2 ay
2
2
o
v v
2
y
2a
y
t
2
o
2a
2
o

0 m s 

2
 5 . 00 m s 
2  9 . 80 m s
2

2
 1 . 28 m
2.6 Freely Falling Bodies
Conceptual Example 14 Acceleration Versus Velocity
The following picture shows the
path of the coin from the previous
problem with velocity vectors.
On this picture. Draw acceleration
vectors.
2.6 Freely Falling Bodies
Conceptual Example 15 Taking Advantage of Symmetry
Does the pellet in part b strike the ground beneath the
cliff with a smaller, greater, or the same speed as the
pellet in part a?
2.6 Freely Falling Bodies
Conceptual Example 15 Taking Advantage of Symmetry
Same speed as part A.
In the absence of air resistance, an object fired in the air
at a certain velocity, vo, will have the same speed, vo,
when it returns to the same height at which it as fired.
Another side note
• In free fall problems
if an object is moving
up, at the top of its
path the velocity is
always 0 m/s.
However the
acceleration is still 9.8m/s2.
2.6.1. A rock is released from rest from a hot air balloon that
is at rest with respect to the ground a few meters below.
If we ignore air resistance as the rock falls, which one of
the following statements is true?
a) The rock will take longer than one second to reach the
ground.
b) The instantaneous speed of the rock just before it reaches
the ground will be 9.8 m/s.
c) The rock is considered a freely falling body after it is
released.
d) As the rock falls, its acceleration is 9.8 m/s2, directed
upward.
e) After the ball is released it falls at a constant speed of 9.8
m/s.
2.6.2. Ping-pong ball A is filled with sand. Ping-pong ball B
is identical to A, except that it is empty inside. Ball A is
somewhat heavier than ball B because of the sand inside.
Both balls are simultaneously dropped from rest from the
top of a building. Which of these two balls has the
greater acceleration due to gravity, if any, as they fall?
a)
b)
c)
d)
ball A
ball B
Both ball A and ball B have zero acceleration.
Both ball A and ball B have the same acceleration.
2.6.3. A ball is thrown vertically upward from the surface of
the earth. The ball rises to some maximum height and
falls back toward the surface of the earth. Which one of
the following statements concerning this situation is true
if air resistance is neglected?
a) As the ball rises, its acceleration vector points upward.
b) The ball is a freely falling body for the duration of its
flight.
c) The acceleration of the ball is zero when the ball is at its
highest point.
d) The speed of the ball is negative while the ball falls back
toward the earth.
e) The velocity and acceleration of the ball always point in
the same direction.
Objectives
We will be able to describe the motion of
an object when given a position, velocity
or acceleration graph
When given one of the graphs, we will be
able to draw the other 2 graphs.
2.7 Graphical Analysis of Velocity and Acceleration – Most
important part of chapter!
• What is the definition of slope?
• Rise / run or Δy/Δx
2.7 Graphical Analysis of Velocity and Acceleration – Most
important part of chapter!
• What is the slope of the graph below?
2.7 Graphical Analysis of Velocity and Acceleration
Slope 
x
t

8m
2s
 4 m s
Notice…
• On the graph, the y-axis is position and
the x-axis is time.
• Slope is Δy/Δx or position over time
• The slope of a position-time graph IS THE
VELOCITY!!
2.7 Graphical Analysis of Velocity and Acceleration
What is the velocity for each part of the graph?
More Mystery Walk 1
 Now Sketch a velocity vs time graph for
Mystery Walk 1.
Velocity vs Time Graph Walk 1
Remember! Velocity is Speed in a particular
direction. Our walker reversed direction to
Mystery Walk 2
 Based on the graph, describe what the walker
did.
Mystery Walk 2
 He waited for 4 seconds before starting to walk
slowly at constant velocity
Mystery Walk 2
 Now, sketch a velocity time graph of Mystery
Walk 2.
Mystery Walk 2 Velocity vs Time Graph
Mystery Walk 3
 Based on the graph describe what the walker
did during her walk.
Mystery Walk 3
 She walked backward very slowly. After 5
seconds, she ran forward for 5 more seconds.
Mystery Walk 3
 Sketch a Velocity vs Time graph of Mystery
Walk 3.
Velocity vs Time for Mystery Walk 3
Which v vs t Graph
Corresponds to the Given d vs t Graph?
?
Which v vs t Graph
Corresponds to the Given d vs t Graph?
?
2.7.1. A dog is initially walking due east. He stops, noticing a cat
behind him. He runs due west and stops when the cat disappears
into some bushes. He starts walking due east again. Then, a
motorcycle passes him and he runs due east after it. The dog gets
tired and stops running. Which of the following graphs correctly
represent the position versus time of the dog?
2.7 Graphical Analysis of Velocity and Acceleration
•Another Important point
•The area UNDER a velocity-time graph is
equal to the DISPLACEMENT of the object
2.7 Graphical Analysis of Velocity and Acceleration
•Notice is the graph below the velocity is not
constant because the slope is not constant.
•In the graph below the velocity is changing,
therefore the object is ACCELERATING!
2.7 Graphical Analysis of Velocity and Acceleration
2.7 Graphical Analysis of Velocity and Acceleration
•Another Important point
•The area UNDER a acceleration-time graph is
equal to the velocity of the object
Which v vs t Graph
Corresponds to the Given d vs t Graph?
?
Which v vs t Graph
Corresponds to the Given d vs t Graph?
?
2.7.3. Consider the graph
the position versus time
graph shown. Which
curve on the graph best
represents a constantly
accelerating car?
a)
A
b) B
c) C
d) D
e) None of the curves
represent a constantly
accelerating car.
2.7.3. The graph shows the velocity of an object versus the
elapsed time. During which interval on the graph does the
object’s acceleration decrease with time?
a) A
b) B
c) C
d) D
e) E
2.7.4. Complete the following statement: the instantaneous
acceleration of an object can be determined by determining
the slope of
a) the object’s velocity versus elapsed time graph.
b) the object’s displacement versus elapsed time graph.
c) the object’s distance versus elapsed time graph.
d) the object’s acceleration versus elapsed time graph.
2.7 Graphical Analysis of Velocity and Acceleration
Slope 
v
t

 12 m s
2s
 6 m s
2
Notice…
• On the graph, the y-axis is velocity and the
x-axis is time.
• Slope is Δy/Δx or velocity over time
• The slope of a velocity-time graph IS THE
ACCELERATION!!
x
All 3 Graphs
t
v
t
a
t
Graphical Comparison
Given the displacementtime graph (a)
The velocity-time graph
is found by measuring
the slope of the positiontime graph at every
instant.
The acceleration-time
graph is found by
measuring the slope of
the velocity-time graph at
every instant.
Section 2.4
Graph Practice- On Board
Try making all three graphs for the following scenario:
1. Schmedrick starts out north of home. At time zero he’s
driving a cement mixer south very fast at a constant speed.
2. He accidentally runs over an innocent moose crossing
the road, so he slows to a stop to check on the poor moose.
3. He pauses for a while until he determines the moose is
squashed flat and deader than a doornail.
4. Fleeing the scene of the crime, Schmedrick takes off
again in the same direction, speeding up quickly.
5. When his conscience gets the better of him, he slows,
turns around, and returns to the crash site.