# Dimension kinematics by MikeJenny

VIEWS: 249 PAGES: 181

• pg 1
```									               9/3 Do now
• Give at least three words to describe
motion.

• All do now count as extra credit for the
next quiz.
• If you are absent or late for the do now,
you will not get the credit.
• You have 5 min to finish the Do now
Homework – Due Tue. 9/7
• Posted on school website – go to school
website, click on teacher, click on
LaBarbera, click on regents physics
homework.
One Dimensional Kinematics -
Chapter Outline
1. Describing Motion with Words
2. Describing Motion with Diagrams
3. Describing Motion with Position vs. Time
Graphs
4. Describing Motion with Velocity vs. Time
Graphs
5. Free Fall and the Acceleration of Gravity
6. Describing Motion with Equations
Lesson 1 : Describing Motion with Words
objectives
1. Introduction to the Language of
Kinematics
2. Differentiate Scalars and Vectors
3. Describe motion in terms of
•   frame of reference,
•   Displacement and distance,
•   speed and velocity
•   Acceleration
Introduction to the Language of
Kinematics
• A typical physics course concerns itself
with a variety of broad topics. One such
topic is mechanics - the study of the
motion of objects.
• Kinematics is the science of describing
the motion of objects using words,
diagrams, numbers, graphs, and
equations. Kinematics is a branch of
mechanics.
Motion takes place over time and
depends upon the frame of reference
• http://www.teachersdomain.org/resource/ls
ps07.sci.phys.fund.frameref/
• http://www.phy.ntnu.edu.tw/ntnujava/index
.php?topic=140
• motion is relative
• A frame of reference is a background
against which you can measure changes
in position.
• Unless stated otherwise, we choose the
Earth as our frame of reference.
Scalars and Vectors
• The motion of objects can be described by
words. Words and phrases such as going fast,
stopped, slowing down, speeding up, and
turning provide a sufficient vocabulary for
describing the motion of objects. In physics, we
use these words and many more. We will be
expanding upon this vocabulary list with words
such as distance, displacement, speed, velocity,
and acceleration.
• Scalars are quantities which are fully described
by a magnitude (or numerical value) alone.
• Vectors are quantities which are fully described
by both a magnitude and a direction.
example
Quantity                  Category

a. 5 m                      Scalar

b. 30 m/sec, East           vector

c. 5 mi., North            vector

d. 20 degrees Celsius       Scalar

e. 256 bytes               Scalar

f. 4000 Calories           Scalar
Distance and Displacement
• Distance and displacement are two quantities
which may seem to mean the same thing yet
have distinctly different definitions and
meanings.
• Distance is a scalar quantity which refers to
"how much ground an object has covered"
during its motion.
• Displacement is a vector quantity which refers
to "how far out of place an object is"; it is the
object's overall change in position.
Distance and displacement

A

B
example
•   A car is driven from Buffalo
to Albany and on to New
York City, as shown in the
diagram.
•   Compared to the magnitude
of the car's total
displacement, the distance
driven is

1. shorter
2. longer
3. the same
example
• A physics teacher walks 4 meters East, 2
meters South, 4 meters West, and finally 2
meters North.

Distance:           12 m
Displacement:       0
Example
•    A student walk east 3.0 meters, then walk west 4.0
meters,
1.   What is the student’s displacement?                N
2.   What is the student’s distance traveled?
1. Use scales to draw the diagram: 1 cm = 1 m
4mW

3mE
resultant

1. The displacement is 1 meter west - vector

2. Distance traveled is 7.0 meters - scalar
example
• The diagram below shows the position of a cross-country
skier at various times. At each of the indicated times, the
skier turns around and reverses the direction of travel. In
other words, the skier moves from A to B to C to D.
• Use the diagram to determine the resulting
displacement and the distance traveled by the skier
during these three minutes.

The skier covers a distance of (180 m+140 m+100 m) = 420 m
and has a displacement of 140 m, rightward.
example
• A football coach paces back and forth along the sidelines.
The diagram below shows several of coach's positions at
various times. At each marked position, the coach makes a
"U-turn" and moves in the opposite direction. In other words,
the coach moves from position A to B to C to D.
• What is the coach's resulting displacement and distance of
travel? Click the button to see the answer.

The coach covers a distance of (35 yds+20 yds+40 yds) = 95 yds
and has a displacement of 55 yards, left.
1. What is the displacement of the cross-country
team if they begin at the school, run 10 miles
and finish back at the school?
The displacement of the runners is 0 miles.

2. What is the distance and the displacement of
the race car drivers in the Indy 500?

The displacement of the cars is 0 miles. The
successful cars have covered a distance of 500
miles
1st period
chairs from all the desks. Thank you!
Do now
• What is a scalar?
• What is a vector?

You have 5 min.
objectives
• Differentiate between speed and velocity
• Calculate speed and velocity
Speed and Velocity
• Speed is a scalar quantity which refers to "how fast an
object is moving." An object with no movement at all has a
zero speed.
• Velocity is a vector quantity which refers to "the rate at
which an object changes its position.“
• Velocity is a vector quantity.
• For example: It would not be enough to say that an object
has a velocity of 55 mi/hr. One must include direction
information in order to fully describe the velocity of the
object. You must describe an object's velocity as being 55
mi/hr, east.
• This is one of the essential differences between
speed and velocity. Speed is a scalar quantity and does
not keep track of direction; velocity is a vector quantity
and is direction aware.
Direction of the velocity
• The direction of the velocity vector is simply the same as
the direction which an object is moving. It would not
matter whether the object is speeding up or slowing
down.
Calculating Average Speed and
Average Velocity
example
• While on vacation, Lisa Carr traveled a
total distance of 440 miles. Her trip took 8
hours. What was her average speed?
example
• During a race on level ground, André runs
with an average velocity of 6.02 m/s to the
east. What distance does André cover in
137 s?
v=d/t
6.02 m/s = d / 137 s
d = 825 m

Note: the calculator answer is 824.74 m, but both the
values for velocity and time have three significant
figures, so the displacement must be reported as 825 m
Average Speed versus
Instantaneous Speed
• Instantaneous Speed - the speed
at any given instant in time.

Average Speed - the average of all
instantaneous speeds; found simply
by a distance/time ratio.
Constant speed vs. changing
speed
example
• Now let's consider the motion of that physics teacher
again. The physics teacher walks 4 meters East, 2
meters South, 4 meters West, and finally 2 meters North.
The entire motion lasted for 24 seconds. Determine the
average speed and the average velocity.

her average speed is 0.50 m/s. However, her average
velocity is 0 m/s.
example
• The diagram below shows the position of a cross-country
skier at various times. At each of the indicated times, the
skier turns around and reverses the direction of travel. In
other words, the skier moves from A to B to C to D.
Determine the average speed and the average velocity
of the skier during these three minutes.

The skier’s average speed is (420 m) / (3 min) = 140 m/min
The skier’s average velocity is (140 m, right) / (3 min) = 46.7 m/min, right
example
• A football coach paces back and forth along the sidelines.
The diagram below shows several of coach's positions at
various times. At each marked position, the coach makes a
"U-turn" and moves in the opposite direction. In other words,
the coach moves from position A to B to C to D. What is the
coach's average speed and average velocity?

The coach’s average speed is (95 yd) / (10 min) = 9.5 yd/min
The coach’s average velocity is (55 yd, left) / (10 min) = 5.5 yd/min, left
In conclusion
• speed and velocity are kinematics quantities
which have distinctly different definitions. Speed,
being a scalar quantity, is the rate at which an
object covers distance. The average speed is
the distance (a scalar quantity) per time ratio.
Speed is ignorant of direction. On the other
hand, velocity is a vector quantity; it is direction-
aware. Velocity is the rate at which the position
changes. The average velocity is the
displacement or position change (a vector
quantity) per time ratio.
homework
• Online School website

Practice

•Text book page 44 – practice 2A #4-6
4. Simpson drives his car with an average velocity of 48.0 km/h to the
east. How long will it take him to drive 144 km on a straight highway?

Given: v = 48.0 km/h east
d = 144 km east
unknown: t = ?

v=d/t
48.0 km/h = 144 km / t
t = 144 km/ (48.0 km/h) = 3 hr = 3.00 hr

Note: since the significant figures of given values is 3, so
the displacement must be reported as 3 significant figures
which is 3.00 hr
9/8 do now
• Text book
– page 45 – conceptual challenge
• Homework is due You may come post session to finish
the homework.
• Lab safety letter is due

1. Displacement = 0; average velocity = 0;
average speed = d / t = (2 x 1.75 m + 2 x 2.25 m) / 23 s = 0.35 m/s

2. No, the velocities are not the same because they are traveling in
different directions.
Objective - Section review
• Text book page 47 – 1, 2, 4, 5, 6

• castle learning

• Finish the lab

• Homework is due
6. A bus travels 280 km south along a straight path with an average velocity of
88 km/h to the south. The bus stops for 24 min, then it travels 210 km south
with an average velocity of 75 km/h to the south.
a. How long does the total trip last?
b. what is the average velocity for the total trip?

Given: v1 = 88 km/h; d1 = 280 km south                   unknown: ttotal = ?
t2 = 24 min = (24/60) hr = 0.4 hr
v3 = 75 km/h; d3 = 210 km south
v=d/t
88 km/h = 280 km / t1
t1 = 280 km/ (88 km/h) = 3.18 hr
t3 = 210 km/ (75 km/h) = 2.80 hr
ttotal = t1 + t2 + t3 = 3.18 hr + 0.4 hr + 2.80 hr = 6.38 hr = 6.4 hr
Note: since the significant figures of given values is 2, so
the displacement must be reported as 2 significant figures
which is 6.4 hr
6. A bus travels 280 km south along a straight path with an average velocity of
88 km/h to the south. The bus stops for 24 min, then it travels 210 km
south with an average velocity of 75 km/h to the south.
a. How long does the total trip last?
b. what is the average velocity for the total trip?

Given: v1 = 88 km/h; d1 = 280 km south                  unknown: vavg = ?
t2 = 24 min = (24/60) hr = 0.4 hr
v3 = 75 km/h; d3 = 210 km south
v = d / t (total displacement / total time)
v = (280 km S. + 210 km S.) / 6.4 hr = 76.56 km/hr south
V = 77 km/h south

Note: since the significant figures of given values is 2, so
the displacement must be reported as 2 significant figures
which is 77 km/h south
Lab 1: intro to physics
objective: to acquire training in scientific
methods of observation and hypothesis
• Make your own group with 3 or less
people in a group.
• Go to each of 7 stations
– Follow directions and describe your
observations
– Try to give a reason for your observation
9/10 do now
•   A bus travels 300 km south along a
straight path with an average velocity of
100 km/h to the south. The bus stops for 30
min, then it travels 150 km south with an
average velocity of 75 km/h to the south.
[you need to convert the min. to hr.]
•   How long does the total trip last?
•   [write equation, substitute number and
objective
• Finish the lab
• Homework correction
• Define and calculate acceleration
Do now Class work
• Practice packet: page 1-6
1. Describing Motion Verbally with Distance
and Displacement
2. Describing Motion Verbally with Speed
and Velocity
9/13 do now
• Write in your own words the definition of
acceleration.
objective
• Define acceleration
• Calculate acceleration
Acceleration
• Acceleration is a vector quantity which is defined as the
rate at which an object changes its velocity. An object is
accelerating if it is changing its velocity.
• Acceleration has to do with changing how fast an object
is moving. If an object is not changing its velocity, then
the object is not accelerating.

The velocity is changing by a
constant amount - 10 m/s - in each
second of time. Anytime an object's
velocity is changing, the object is
said to be accelerating; it has an
acceleration.
The Meaning of Constant Acceleration
• Sometimes an accelerating object will change its velocity
by the same amount each second. This is referred to as a
constant acceleration since the velocity is changing by a
constant amount each second. The data tables below
depict motions of objects with a constant acceleration and
a changing acceleration. Note that each object has a
changing velocity.
Calculating the Average
Acceleration

• Acceleration values are expressed in units
of velocity/time. The units on acceleration
are velocity units divided by time units -
Typical acceleration units is (m/s)/s or
m/s2
The Direction of the Acceleration Vector
• The direction of the acceleration vector depends
on two things:
– whether the object is speeding up or slowing down
– whether the object is moving in the + or - direction
• The general RULE OF THUMB is: If an object is
slowing down, then its acceleration is in the
opposite direction of its motion.
• This RULE OF THUMB can be applied to
determine whether the sign of the acceleration
of an object is positive or negative.
• Consider the two data tables below.

Object is speeding up       Object is slowing
in the positive             down in the negative
direction, its              direction, its
acceleration is in the      acceleration is in the
same direction of its       opposite direction of its
motion – positive.          motion – positive.
• Consider the two data tables below.

Object is speeding up
Object is slowing down
in the negative
in the positive direction,
direction, its
its acceleration is in the
acceleration is in the
opposite direction of its
same direction of its
motion – negative.
motion – negative.
• Use the equation for acceleration to determine the
acceleration for the following two motions.

Use a = (v - v ) / t and pick any   Use a = (v -v ) / t and pick any two
points.   f i
two points.f i
a = (0 m/s - 8 m/s) / (4 s)
a = (8 m/s - 0 m/s) / (4 s)
a = (-8 m/s) / (4 s)
a = (8 m/s) / (4 s)
a = -2 m/s/s
a = 2 m/s/s
Calculating average acceleration
•     A shuttle bus slows to a stop with an average acceleration of
-1.8 m/s2. How long does it take the bus to slow from 9.0 m/s
to 0.0 m/s?
Given: aavg = -1.8 m/s2 ; vi = 9.0 m/s; vf = 0 m/s

unknown: t = ?

a = ∆v / t
-1.8 m/s2 = (0 – 9.0 m/s) / t
t = (0 – 9.0 m/s) / (-1.8 m/s2) = 5.0 s
example
• Text book – page 49 – practice 2B - #5
Given: aavg = 4.7 x 10-3 m/s2 ; t = 5.0 min = 5 x 60 s = 300 s

unknown:
a. ∆v = ?     b. If vi = 1.7 m/s    vf = ?

a.     a = ∆v / t
4.7 x 10-3 m/s2 = ∆v / 300 s
∆v = (4.7 x 10-3 m/s2) x 300 s = 1.4 m/s

b.     ∆v = vf - vi
1.4 m/s = vf – 1.7 m/s
vf = 1.4 m/s + 1.7 m/s = 3.1 m/s
9/14 do now
•   Practice packet: Acceleration - page 7

Homework
castle learning: acceleration1
Objective – review acceleration
•    The direction of acceleration
•    Practice packet: Acceleration - page 8
•    Text book – page 50 - Conceptual challenge:
1, 2, 3
1. No; the ball could change direction at the very
top.
2. Slowing down
3. Jennifer’s acceleration could be positive if she is
moving in the negative direction.
9/15 do now
•  Observe when a ball is thrown upward,
describe
1. Its direction of velocity, give the reason
2. Its direction of acceleration, give the
Lesson 2 Objective : Describing
Motion with Diagrams
1. Ticker Tape Diagrams
2. Vector Diagrams
Ticker Tape Diagrams
• A common way of analyzing the motion of objects in
physics labs is to perform a ticker tape analysis. A long
tape is attached to a moving object and threaded
through a device that places a tick upon the tape at
regular intervals of time - say every 0.10 second. As the
object moves, it drags the tape through the "ticker," thus
leaving a trail of dots. The trail of dots provides a history
of the object's motion and therefore a representation of
the object's motion.
• The distance between dots on a ticker tape
represents the object's position change during
that time interval.
• The analysis of a ticker tape diagram will
also reveal if the object is moving with a
constant velocity or accelerating.
Vector Diagrams
• Vector diagrams are diagrams which depict the
direction and relative magnitude of a vector quantity
by a vector arrow. Vector diagrams can be used to
describe the velocity of a moving object during its
motion.
• In a vector diagram, the magnitude of a
vector quantity is represented by the size
of the vector arrow.
• Vector diagrams can be used to represent any vector
quantity. In future studies, vector diagrams will be used
to represent a variety of physical quantities such as
acceleration, force, and momentum. It will become a
very important representation of an object's motion as
we proceed further in our studies of the physics of
motion.
• This animation depicts some information about the car's
motion. The velocity and acceleration of the car are
depicted by vector arrows. The direction of these arrows
are representative of the direction of the velocity and
acceleration vectors.
• Purpose: question: How can I find my speed while I am in
motion?
• Material: spark timer, spark recording tape, ruler
• Procedure: (write down what you did to answer the
question)
• Data: (set up a data table)
• conclusion:
A complete lab write-up includes a Title, a Purpose, the material, a
Procedure, a Data section, and a Conclusion. The Data section should
include collected data for several trials; column headings should be
labeled and units shown. One sample calculation should be shown. Speed
values for all the trials (except those which are obvious ) should be averaged;
should be indicated as such in the Data section (a is a wholly scientific means
of doing so). The Conclusion should respond to the question raised in the
Purpose of the lab.
9/15 do now
•  Observe when a ball is thrown upward,
describe
1. Its direction of velocity, give the reason
2. Its direction of acceleration, give the
9/16 Do now - practice
• Practice packet –
– Page 12 #5, #6

Homework
Hand out – due tomorrow

Quiz tomorrow – you may have your homework
for reference only.
I have hall duty today – no post session.
Brief review
• What are the words we learned to describe motion?

Distance, displacement, speed, velocity, acceleration
Scalar quantity:   Distance, speed
Vector quantity:   Displacement, velocity, acceleration

Ticker tape diagram:
Vector diagram:
Lesson 3 objective : Describing
Motion with Position vs. Time
Graphs

1. The Meaning of Shape for a p-t Graph
2. The Meaning of Slope for a p-t Graph
3. Determining the Slope on a p-t Graph
The Meaning of Shape for a p-t Graph
• To begin, consider a car moving with a constant,
rightward (+) velocity - say of +10 m/s.

If the position-time data for such a car were graphed,
then the resulting graph would look like this:

Note that a motion described as
a constant, positive velocity
results in a line of constant
and positive slope when
plotted as a position-time
graph.
• Now consider a car moving with a rightward (+), changing
velocity - that is, a car that is moving rightward but
speeding up or accelerating.

If the position-time data for such a car were graphed, then
the resulting graph would look like this:

Note that a motion described
as a changing, positive
velocity results in a line of
changing and positive
slope when plotted as a
position-time graph.
• The position vs. time graphs for the two types of motion -
constant velocity and changing velocity (acceleration) -
are depicted as follows.

Constant Velocity                  Positive Velocity
Positive Velocity                 Changing Velocity
(acceleration)
The Importance of Slope
• The slope of the line on a position-time graph
reveals useful information about the velocity of
the object. It is often said, "As the slope goes, so
goes the velocity." Whatever characteristics the
velocity has, the slope will exhibit the same (and
vice versa).
– If the velocity is constant, then the slope is constant
(i.e., a straight line).
– If the velocity is changing, then the slope is changing
(i.e., a curved line).
– If the velocity is positive, then the slope is positive (i.e.,
moving upwards and to the right).
• "As the slope goes, so goes the velocity."

Constant, Positive,     Constant, Positive,
Slow Velocity           Fast Velocity

Constant, Positive,   Constant, Positive,
smaller slope         bigger slope
Constant, negative       Constant, negative
Slow Velocity            Fast Velocity

Constant, negative,    Constant, negative,
smaller slope          bigger slope
Negative (-) Velocity      Leftward (-) Velocity
Slow to Fast               Fast to Slow

Negative slope,         Negative slope,
increasing              decreasing
• Use the principle of slope to describe the motion of the
objects depicted by the two plots below. In your
description, be sure to include such information as the
direction of the velocity vector (i.e., positive or negative),
whether there is a constant velocity or an acceleration,
and whether the object is moving slow, fast, from slow to
fast or from fast to slow. Be complete in your description.

positive slope, increasing        Negative slope, increasing

Positive velocity slow to fast.   negative velocity slow to fast.
The Meaning of Slope for a p-t Graph
• The shape of the line on the graph
(straight, curving, steeply sloped, mildly
sloped, etc.) is descriptive of the object's
motion. The slope of a position vs. time
object's velocity.
• For example, a small, negative, constant
slope means a slow, negative, constant
velocity; a changing slope (curved line)
means a changing velocity.
• Consider a car moving with a constant velocity of +10
m/s for 5 seconds. The diagram below depicts such a
motion.

Note that during the first 5
seconds, the line on the graph
slopes up 10 m for every 1 second
along the horizontal (time) axis.
That is, the slope of the line is +10
meter/1 second. In this case, the
slope of the line (10 m/s) is
equal to the velocity of the car.
• Consider a car moving at a constant velocity of +5 m/s
for 5 seconds, abruptly stopping, and then remaining at
rest (v = 0 m/s) for 5 seconds.

For the first five seconds the line on
the graph slopes up 5 meters for
every 1 second along the horizontal
(time) axis. the line on the position vs.
time graph has a slope of +5 meters/1
second for the first five seconds.
Thus, the slope of the line on the
graph equals the velocity of the car.
During the last 5 seconds (5 to 10
seconds), the line slopes up 0 meters.
That is, the slope of the line is 0 m/s -
the same as the velocity during this
time interval.
an important principle
• The slope of the line on a position-time graph is equal to
the velocity of the object
Determining the Slope on a p-t Graph
• The slope of the line on a position versus
time graph is equal to the velocity of the
object.
– If the object is moving with a velocity of +4 m/s,
then the slope of the line will be +4 m/s.
– If the object is moving with a velocity of -8 m/s,
then the slope of the line will be -8 m/s.
– If the object has a velocity of 0 m/s, then the
slope of the line will be 0 m/s.
• Let's begin by considering the position versus time graph

To calculate the slope we must use the slope equation.

1.   Pick two points on the line and determine their coordinates.
2.   Determine the difference in y-coordinates of these two points (rise).
3.   Determine the difference in x-coordinates for these two points (run).
4.   Divide the difference in y-coordinates by the difference in x-
coordinates (rise/run or slope)
• To determine the slope for the graph:
9/17 do now
• Describe the velocity in the following graphs using
words: positive or negative, constant or increasing or
decreasing or zero:
A                     B                  C
p                      p                     p

t                     t                   t

Positive, constant       Positive, increasing           zero
objectives
• Using Position-time graph to describe
motion
• Quiz
• Finish and hand in all your homework –
including Castlelearning, school website,
handout
Example – calculate slope

• Slope = (y2 – y1) / (x2 – x1)
• Slope = (26.0 m – 2.0 m) / (0.0 s – 8.0 s) = -3.0 m/s
• Determine the velocity of the object as portrayed
by the graph below.

• Slope = (y2 – y1) / (x2 – x1)
• Slope = (25 m – 5 m) / (5 s – 0 s) = 4 m/s
Class work
• Describing Motion with Position-Time
Graphs
• Text book – page 47 #3
• Review for the quiz
9/17 do now
• Describe the velocity in the following graphs using
words: positive, constant, increasing, decreasing, zero:
A                     B                  C
p                      p                     p

t                     t                   t

Positive, constant       Positive, increasing           zero
objective
• Quiz
• Finish and hand in all your homework
9/20 do now
• Use a new sheet of paper
p

As the slope goes, so goes
the velocity
t

Describe the velocity of the object as portrayed by the graph
using words such as (positive or negative), (constant or zero
or increasing or decreasing).
Lesson 4 objective: Describing Motion
with Velocity vs. Time Graphs:

1.   The Meaning of Shape for a v-t Graph
2.   The Meaning of Slope for a v-t Graph
3.   Relating the Shape to the Motion
4.   Determining the Slope on a v-t Graph
5.   Determining the Area on a v-t Graph
The Meaning of Shape for a v-t Graph
• Consider a car moving with a constant, positive
velocity - say of +10 m/s. A car moving with a constant
velocity is a car with zero acceleration.

Note that a motion described as
a constant, positive velocity
results in a line of zero slope (a
horizontal line has zero slope)
when plotted as a velocity-time
graph. Furthermore, only
positive velocity values are
plotted, corresponding to a
motion with positive velocity.
• Now consider a car moving with a positive, changing
velocity - that is, a car that is moving rightward but
speeding up or accelerating. Since the car is moving in the
positive direction and speeding up, the car is said to have a
positive acceleration.

Note that a motion described as a
changing, positive velocity results
in a sloped line when plotted as a
velocity-time graph. The slope of
the line is positive, corresponding
to the positive acceleration.
Furthermore, only positive velocity
values are plotted, corresponding
to a motion with positive velocity.
• The velocity vs. time graphs for the two types of motion -
constant velocity and changing velocity (acceleration) -
can be summarized as follows.

Positive Velocity                 Positive Velocity
Zero Acceleration               Positive Acceleration
The Importance of Slope
• The slope of the line on a velocity-time
graph reveals useful information about the
acceleration of the object.
– If the acceleration is zero, then the slope is
zero (i.e., a horizontal line).
– If the acceleration is positive, then the slope is
positive (i.e., an upward sloping line).
– If the acceleration is negative, then the slope
is negative (i.e., a downward sloping line).
• The slope of a velocity-time graph reveals information
about an object's acceleration. But how can one tell
whether the object is moving in the positive direction (i.e.,
positive velocity) or in the negative direction (i.e., negative
velocity)? And how can one tell if the object is speeding up
or slowing down?
• The answers to these questions hinge on one's ability to

A positive velocity means the       A negative velocity means the
object is moving in the             object is moving in the
positive direction;                 negative direction;
The slope of a velocity-time graph reveals information about an
object's acceleration. But how can one tell if the object is
speeding up or slowing down?

Speeding up means that           Slowing down means that
the magnitude (or                the magnitude (or
numerical value) of the          numerical value) of the
velocity is getting large.       velocity is getting smaller
•    Consider the graph at the right. The object whose motion
is represented by this graph is ... (include all that are true):
1.   moving in the positive direction.
2.   moving with a constant velocity.
3.   moving with a negative velocity.
4.   slowing down.
5.   changing directions.
6.   speeding up.
7.   moving with a positive acceleration.
8.   moving with a constant acceleration.
The Meaning of Slope for a v-t Graph
• The shape of a velocity versus time graph reveals
pertinent information about an object's acceleration.
– If the acceleration is zero, then the velocity-time
graph is a horizontal line (i.e., the slope is zero).
– If the acceleration is positive, then the line is an
upward sloping line (i.e., the slope is positive).
– If the acceleration is negative, then the velocity-time
graph is a downward sloping line (i.e., the slope is
negative).
– If the acceleration is great, then the line slopes up
steeply (i.e., the slope is great).
• The shape of the line on the graph (horizontal, sloped,
steeply sloped, mildly sloped, etc.) is descriptive of the
object's motion.
• Consider a car moving with a constant velocity of +10 m/s.
A car moving with a constant velocity has an acceleration
of 0 m/s/s.

Time (s)      Velocity (m/s)

0               10
1               10
2               10
3               10
4               10
5               10

Note that the line on the graph is horizontal. That is the
slope of the line is 0 m/s/s. In this case, it is obvious that the
slope of the line (0 m/s/s) is the same as the acceleration (0
m/s/s) of the car.
• Consider a car moving with a constant velocity of +10 m/s.
A car moving with a constant velocity has an acceleration
of 0 m/s/s.

Time (s)      Velocity (m/s)

0               10
1               10
2               10
3               10
4               10
5               10

Note that the line on the graph is horizontal. That is the
slope of the line is 0 m/s/s. In this case, it is obvious that the
slope of the line (0 m/s/s) is the same as the acceleration (0
m/s/s) of the car.
• Consider a car moving with a constant velocity of +10 m/s.
A car moving with a constant velocity has an acceleration
of 0 m/s/s.

Time (s)     Velocity (m/s)

0               0
1              10
2              20
3              30
4              40
5              50

Note that the line on the graph is diagonal - that is, it has a
slope. The slope of the line can be calculated as 10 m/s/s.
The slope of the line (10 m/s/s) is the same as the
acceleration (10 m/s/s) of the car.
• Consider the motion of a car which first       Time (s)   Velocity (m/s)
travels with a constant velocity (a=0 m/s/s)
0             2
of 2 m/s for four seconds and then
accelerates at a rate of +2 m/s/s for four        1             2
seconds. That is, in the first four seconds,      2             2
the car is not changing its velocity (the         3             2
velocity remains at 2 m/s) and then the car
increases its velocity by 2 m/s per second        4             2
over the next four seconds.                       5             4
6             6
7             8
8            10
Velocity (m/s)

The slope of the line on a
velocity-time graph is equal
to the acceleration of the
object
Time (s)
•    The velocity-time graph for a two-stage rocket is
shown below. Use the graph and your understanding
of slope calculations to determine the acceleration of
the rocket during the listed time intervals.
1.   t = 0 - 1 second 40 m/s2
2.   t = 1 - 4 second 20 m/s2
3.   t = 4 - 12 second -20 m/s2
Relating the Shape to the Motion
• The shape of a velocity vs. time graph reveals pertinent
– If the acceleration is zero, then the velocity-time
graph is a horizontal line - having a slope of zero.
– If the acceleration is positive, then the line is an
upward sloping line - having a positive slope.
– If the acceleration is negative, then the velocity-time
graph is a downward sloping line - having a negative
slope.
– If the acceleration is great, then the line slopes up
steeply - having a large slope.
• The shape of the line on the graph (horizontal, sloped,
steeply sloped, mildly sloped, etc.) is descriptive of the
object's motion.
9/21 do now
Use words like constant, positive, zero, negative,
increasing, decreasing to describe velocity and acceleration
of the object in graph A and graph B

A                          B

position               t

Velocity: Constant, positive         Velocity: Constant, positive
Acceleration: Zero (zero slope)      Acceleration: Zero
objectives
1. Relating the Shape to the Motion
2. Determining the Slope on a v-t Graph
3. Determining the Area on a v-t Graph
Use words like constant, positive, zero, negative,
increasing, decreasing to describe velocity and acceleration
of the object.

Constant, negative velocity,
Zero acceleration (zero slope)
Use words like constant, positive, zero, negative,
increasing, decreasing to describe velocity and acceleration
of the object.

increasing, positive velocity,
Constant, positive acceleration
(constant, positive slope)
Use words like constant, positive, zero, negative,
increasing, decreasing to describe velocity and acceleration
of the object.

positive velocity, slowing down
Acceleration is in the direction
opposite of its motion – negative,
constant (negative, constant
slope) is
Use words like constant, positive, zero, negative, slowing
down, speeding up to describe velocity and acceleration of
the object.

negative velocity, slowing
down
Positive, constant acceleration
(positive, constant slope)
Use words like constant, positive, zero, negative,
increasing, decreasing to describe velocity and acceleration
of the object.

negative velocity, speeding up.
Negative, constant acceleration.
(negative, constant slope)
• Describe the motion depicted by the following velocity-time
graphs. In your descriptions, use words such as positive,
negative, speeding up or slowing down, constant, zero to
describe velocity and acceleration during the various time
intervals (e.g., intervals A, B, and C).
A: positive, constant velocity,
zero acceleration

B: positive velocity, slowing down, constant
negative acceleration

C: positive, constant velocity,
zero acceleration
A: positive, decreasing velocity, negative,
constant acceleration

B: zero velocity, zero acceleration

C: negative velocity, speeding up, negative
constant acceleration
A: positive, constant velocity, zero
acceleration

B: positive velocity, slowing down, negative,
constant acceleration

C: negative velocity, speeding up, negative
constant acceleration
Determining the Slope on a v-t Graph
• The slope of the line on a velocity versus
time graph is equal to the acceleration of
the object.
– If the object is moving with an acceleration of
+4 m/s/s (i.e., changing its velocity by 4 m/s
per second), then the slope of the line will be
+4 m/s/s.
– If the object is moving with an acceleration of
-8 m/s/s, then the slope of the line will be -8
m/s/s.
– If the object has a velocity of 0 m/s, then the
slope of the line will be 0 m/s.
• Let's begin by considering the velocity versus time graph

To calculate the slope we must use the slope equation.

1.   Pick two points on the line and determine their coordinates.
2.   Determine the difference in y-coordinates of these two points (rise).
3.   Determine the difference in x-coordinates for these two points (run).
4.   Divide the difference in y-coordinates by the difference in x-
coordinates (rise/run or slope)
• To determine the slope for the graph:
• Consider the velocity-time graph below.
Determine the acceleration (i.e., slope) of the
object as portrayed by the graph.

The acceleration (i.e.,
slope) is 4 m/s/s.
Determining the Area on a v-t Graph
• A plot of velocity-time can be used to
determine the acceleration of an object
(the slope).
• A plot of velocity versus time can also be
used to determine the displacement of an
object.
• For velocity versus time graphs, the area
bound by the line and the axes
represents the displacement.
9/21 do now
Use words like constant, positive, zero, negative,
increasing, decreasing to describe velocity and acceleration
of the object in graph A and graph B

A                          B

position               t

Velocity: Constant, positive         Velocity: Constant, positive
Acceleration: Zero (zero slope)      Acceleration: Zero
objective
Determining the Area on a v-t Graph
Determining the Area on a v-t Graph
• A plot of velocity-time can be used to
determine the acceleration of an object
(the slope).
• A plot of velocity versus time can also be
used to determine the displacement of an
object.
• For velocity versus time graphs, the area
bound by the line and the axes
represents the displacement.
• The diagram below shows three different velocity-time
graphs; the shaded regions between the line and the
time-axis represents the displacement during the stated
time interval.

180 m                 80 m            125 m – 20 m
= 105 m
example
• Determine the displacement (i.e., the area) of the object
during the first 4 seconds (Practice A) and from 3 to 6
seconds (Practice B).

The area of under the line            The area of under the line
represents the displacement.          represents the displacement.
A = (4 s) • (30 m/s) = 120 m          A = (3 s) • (30 m/s) = 90 m
That is, the object was displaced     That is, the object was displaced
120 m during the first 4 seconds of   90 m during the interval of time
motion                                between 3 and 6 seconds.
example
• Determine the displacement of the object during the first
second (Practice A) and during the first 3 seconds
(Practice B).

The area of under the line          The area of under the line
represents the displacement.        represents the displacement.
Area = ½ • b • h                    Area = ½ • b • h
Area = ½ • (1 s) • (10 m/s) = 5 m   Area = ½ • (3 s) • (30 m/s) = 45 m
example
• Determine the displacement of the object during the time
interval from 2 to 3 seconds (Practice A) and during the
first 2 seconds (Practice B).

The area of a trapezoid represents       The area of a trapezoid represents
the displacement                         the displacement
Area = ½ • b1 • h1 – ½ b2 • h2 (b1 = 3   Area = ½ • b1 • h1 – ½ b2 • h2 (b1 = 3 s,
s, b2 = 2s, h1 = 30 m/s, h2 = 20 m/s.)   b2 = 2s, h1 = 30 m/s, h2 = 10 m/s.)
Area = 45 m – 20 m = 25 m                Area = 45 m – 5 m = 40 m
That is, the object was displaced 25     That is, the object was displaced 40 m
m during the time interval from 2 to 3   during the time interval from 1 to 2
seconds.                                 seconds.
9/22 Do now
A. What does the slope in position-time graph
represent?

B. What does the slope in velocity-time graph
represent?

C. What does the area bound by the line and the axes
in velocity-time graph represent?
9/23 do now
• Describe the object’s velocity in the diagram A
and B. choose one word for each blank.
p                                    v

t                                   t

A                               B

Velocity: ____________ (positive or    Velocity: ____________ (positive or
negative), _____________(constant or   negative), _____________(constant or
zero or increasing or decreasing)      zero or increasing or decreasing)
objectives
• Position-time graph lab
• Review graphs

Homework – castle learning:
Lab 3 : Position-Time Graphs Lab
Direction:
Open up logger pro program – open file folder on top – click physics with vernier (physics
with computer) – click 01a
Question:
How can the following types of motion be described with a position-time graph? (moving in the
positive direction versus moving in the negative direction; moving fast versus moving slow; moving
with a constant speed versus moving with a gradually changing speed; speeding up versus slowing
down; etc.)
Purpose:
To contrast the shape and slope of the position-time graphs for the following types of motion:
– moving in the + direction versus moving in the - direction
– moving fast versus moving slow
– a constant speed motion versus a gradually changing speed
– a speeding up motion versus a slowing down motion (use incline)
– combinations of the above (use incline)
Material:
– Computer, Logger Pro program, motion sensor, computer interface,
Report requirement:
A complete lab write-up includes Title, Purpose, Materials, Data section, and
Conclusion/Discussion.
– The Data section should include one graph for each contrasting set of two motions; axes
should be labeled; labels or color coding or some other method should be used to distinguish
between the two motions.
– The Conclusion/Discussion section should provide a thorough discussion of the differences in
the position-time graphs for the variety of motions under study.
Class work
1. Describing Motion with Velocity-Time
Graphs
2. Describing Motion Graphically
3. Interpreting Velocity-Time Graphs
4. Graphing Summary
5. Kinematics Graphing - Mathematical
Analysis
6. Text book – page 59 - #5, 6
9/24 do now
• The graph below shows the velocity of a race car moving
along a straight line as a function of time. What is the
magnitude of the displacement of the car from t = 3.0
seconds to t = 4.0 seconds?
objectives
• Graph quiz
• Practice packet – finish pages up to page
23.
9/27 do now
• The velocity-time graph represents the motion of a 3-
kilogram cart along a straight line. The cart starts at t = 0
and initially moves north. What is the magnitude of the
displacement of the cart
1. during t = 0 and t = 5 seconds?

2. during t = 5 and t = 7 seconds?

3. During t = 0 and t = 7 seconds?
• The velocity-time graph represents the motion of a 3-
kilogram cart along a straight line. The cart starts at t = 0
and initially moves north. What is the acceleration of the
cart
1. during t = 0 and t = 3 seconds?

2. during t = 3 and t = 7 seconds?
• The velocity-time graph represents the motion of a 3-
kilogram cart along a straight line. The cart starts at t = 0
and initially moves north. What is the velocity of the cart

1. At t = 3 seconds?

2. during t = 6 seconds?
•   The graph represents the relationship between the
displacement of an object and its time travel along a
straight line.

a. What is the magnitude of the object's total
displacement during t = 0 to t = 8.0 seconds?
b. What is the average speed of the object during t = 0
to t = 4.0 seconds?
c. What is the magnitude of acceleration of the object
during t = 0 to t = 2.0 seconds?
9/27 do now
•    A locomotive starts at rest and accelerates at 0.12
meters per second squared to a speed of 2.4 meters
per second in 20. seconds. Describe the locomotive’s
acceleration and velocity.

1.   Acceleration: _______________(constant,
increasing, decreasing, zero)
2.   velocity: _______________(constant, increasing,
decreasing, zero)
objectives
• Free fall

homework

• Castle learning – free fall
Lesson 5 objective : relate the
motion of a Free Falling body to
motion with constant Acceleration
1.   Introduction to Free Fall
2.   The Acceleration of Gravity
3.   Representing Free Fall by Graphs
4.   How Fast? and How Far?
5.   The Big Misconception
Homework: castle learning
Introduction to Free Fall
• A free-falling object is an object which is falling
under the sole influence of gravity. Any object
which is being acted upon only be the force of
gravity is said to be in a state of free fall.
• There are two important motion characteristics
which are true of free-falling objects:
– Free-falling objects do not encounter air
resistance.
– All free-falling objects (on Earth) accelerate
downwards at a rate of 9.81 m/s/s.
The dot diagram at the right depicts the acceleration
of a free-falling object.
The fact that the distance which the object travels
every interval of time is increasing is a sure sign that
the ball is speeding up as it falls downward
• A free-falling object is an object which is falling under
the sole influence of gravity. A free-falling object has
an acceleration of 9.81 m/s/s, downward (on Earth).

• It is known as the acceleration of gravity - the
acceleration for any object moving under the sole
influence of gravity.

• The symbol g is used to represent acceleration of
gravity.

• g = 9.81 m/s/s (m/s2), downward
• Acceleration is the rate at which an object changes its velocity. It is
the ratio of velocity change to time between any two points in an object's
path. To accelerate at 9.8 m/s/s means to change the velocity by 9.8
m/s each second.

If the velocity and time             Time (s)          Velocity (m/s)
for a free-falling object
being dropped from a
position of rest were                    0                     0
tabulated, then one
would note the                           1                   - 9.8
following pattern.                       2                  - 19.6
3                  - 29.4
4                  - 39.2
5                  - 49.0
For an object falling
freely from rest, it’s
speed increases by
9.81 m/s every
second.
Representing Free Fall by Graphs
• We can describe the motion of objects through the use
of graphs - position versus time and velocity vs. time
graphs.
• A free falling object is accelerating.

Position vs. time graph
Position (m)

Position (m)
or

Time (s)                                  Time (s)

Downward is set to be                      Downward is set to be
positive.                                  negative.
Representing Free Fall by Graphs
• A free falling object is accelerating.

Velocity vs. time graph
velocity (m/s)

Velocity (m/s)
Time (s)

or
Time (s)

Downward is set to be                     Downward is set to be
positive.                                 negative.
How Fast? and How Far?
• Free-falling objects are in a state of acceleration.
Specifically, they are accelerating at a rate of 9.81 m/s/s.
• The velocity of a free-falling object is changing by 9.8 m/s
every second.
• If dropped from a position of rest, the object will be
traveling
– At 1 s – v = 9.81 m/s
– At 2 s – v = 19,6 m/s
– At 3 s – v = 29.4 m/s.
• Thus, the velocity of a free-falling object which has been
dropped from a position of rest is dependent upon the
time for which it has fallen. The formula for determining
the velocity of a falling object after a time of t seconds is
• vf = g * t where g is the acceleration of gravity.
• g = 9.81 m/s/s
vf = g * t
• where g is the acceleration of gravity. The
value for g on Earth is 9.8 m/s/s.
• The above equation can be used to
calculate the velocity of the object after
any given amount of time when dropped
from rest.
• Example Calculations:
– At t = 6 s
vf = (9.81 m/s2) * (6 s) = 58.8 m/s
– At t = 8 s
vf = (9.81 m/s2) * (8 s) = 78.4 m/s
• The distance which a free-falling object has fallen from a
position of rest is also dependent upon the time of fall.
This distance can be computed by use of the formula; d
= ½ ·g·t2 where g is the acceleration of gravity (9.81
m/s/s on Earth).
• Example
• t=1s
d = ½ ·(9.81 m/s2)·(1 s)2 = 4.9 m
• t=2s
d = ½ ·(9.81 m/s2)·(2 s)2 = 19.6 m
• t=5s
d = ½ ·(9.81 m/s2)·(5 s)2 = 122.5 m
• Try to remember
these distances:
• 1s–5m
• 2 s – 20 m
• 3 s – 45 m
• 4 s – 80 m
• 5 s – 123 m
The Big Misconception
Acceleration is
• The acceleration of a free-         the same at all
falling object (on earth) is 9.81   points.
m/s/s. This value (known as the
acceleration of gravity) is the
same for all free-falling objects
regardless of how long they
have been falling, or whether
they were initially dropped from
rest or thrown up into the air.
The Big Misconception
• “Doesn't a more massive object accelerate at a
greater rate than a less massive object?"
The answer is absolutely not! That is, absolutely not if we
are considering the specific type of falling motion
known as free-fall. Free-fall is the motion of objects which
move under the sole influence of gravity; free-falling objects
do not encounter air resistance.
More massive objects will only fall faster if there is an
appreciable amount of air resistance present.
practice
• Free fall work sheet
• Text book – page 65 - #1, 3, 4, 6a, b.
9/28/10 do now
• Sketch a graph best represents the relationship between
mass and acceleration due to gravity for objects near the
surface of the Earth.

Homework: castlelearning: kinematics 1, kinematics 2
Lesson 6 objective : Describing Motion with
Equations
1.    The Kinematics Equations
2.    Kinematics Equations and Problem-Solving
3.    Kinematics Equations and Free Fall
4.    Sample Problems and Solutions
5.    Kinematics Equations and Graphs
• The kinematics equations are a set of four
equations which can be utilized to predict
unknown information about an object's motion if
other information is known.

d = vit + ½ a·t2
vf = vi + a·t
vf2 = vi2 + 2a·d
v +v
d = vavg·t = i f ·t
2

d: displacement   t: time
a: acceleration   vi: initial velocity   vf: final velocity
Kinematics Equations and Problem-
Solving
• We are using the equations to determine unknown
information about an object's motion. The strategy of
finding the unknown involves the following steps:
1. Construct an informative diagram of the physical
situation
2. Identify and list the given information in variable form.
3. Identify and list the unknown information in variable
form.
4. Identify and list the equation which will be used to
determine unknown information from known information.
5. Substitute known values into the equation and use
appropriate algebraic steps to solve for the unknown
information.
mathematically correct.
Example A
• Ima Hurryin is approaching a stoplight moving with a
velocity of +30.0 m/s. The light turns yellow, and Ima applies
the brakes and skids to a stop. If Ima's acceleration is -8.00
m/s2, then determine the displacement of the car during the
skidding process.
Given:         unknown:
vi = +30.0 m/s
vf = 0 m/s
a = - 8.00 m/s2       d = ??

vf2 = vf2 + 2·a·d
(0 m/s)2 = (30 m/s)2 + 2(- 8.00 m/s2)·d
d = 56.3 m The value seems reasonable. It takes a car about 4 seconds
to come to a stop. Its average speed is 15 m/s, so the
distance should be close to 60 m.
Example B
• Ben Rushin is waiting at a stoplight. When it finally turns
green, Ben accelerated from rest at a rate of a 6.00 m/s2
for a time of 4.10 seconds. Determine the displacement
of Ben's car during this time period
Diagram:              Given:         unknown:

vi = 0 m/s
t = 4.10 s        d = ??
a = 6.00 m/s2

d = vit + ½ ·a·t2
d = (0 m/s)(4.10 s) + ½ (6.00 m/s2)·(4.10 s)2
The value seems reasonable. Its speed at the end
d = 50.4 m of 4.10 s is about 24 m/s, so its average is about
12 m/s, its distance should be around 48 m.
Kinematics Equations and Free Fall
• A free-falling object is an object which is falling under the
sole influence of gravity. Such an object will
experience a downward acceleration of 9.81 m/s/s.
• Whether the object is falling downward or rising upward
towards its peak, if it is under the sole influence of
gravity, then its acceleration value is 9.8 m/s/s.
• Like any moving object, the motion of an object in free
fall can be described by four kinematics equations.

d = vit + ½ a·t2
vf = vi + a·t
vf2 = vi2 + 2a·d
vi+vf
d = vavg·t =         ·t
2
• There are a few conceptual characteristics of free fall motion:
1. An object in free fall experiences an acceleration of -9.81
2. If an object is merely dropped (as opposed to being thrown)
from an elevated height, then the initial velocity of the object is
0 m/s.
3. If an object is projected upwards in a perfectly vertical direction,
then it will slow down as it rises upward. The instant at which
it reaches the peak of its trajectory, its velocity is 0 m/s. This
value can be used as one of the motion parameters in the
kinematic equations; for example, the final velocity (vf) after
traveling to the peak would be assigned a value of 0 m/s.
4. If an object is projected upwards in a perfectly vertical direction,
then the velocity at which it is projected is equal in magnitude
and opposite in sign to the velocity which it has when it returns
to the same height. That is, a ball projected vertically with an
upward velocity of +30 m/s will have a downward velocity of -
30 m/s when it returns to the same height.
Example A
• Luke Autbeloe drops a pile of roof shingles from the top of a
roof located 8.52 meters above the ground. Determine the
time required for the shingles to reach the ground.
Diagram:              Given:         unknown:

vi = 0.0 m/s
d = -8.52 m        t = ??
a = - 9.81 m/s2

d = vit + ½ ·a·t2

-8.52 m = (0 m/s)(t) + ½ (-9.81 m/s2)·t2
t = 1.32 s The value seems reasonable enough. It falls about
5 m at the end of 1 s, about 20 m at the end of 2 s.
Example B
• Rex Things throws his mother's crystal vase vertically
upwards with an initial velocity of 26.2 m/s. Determine the
height to which the vase will rise above its initial height.
Diagram:                   Given:             unknown:

vi = 26.2 m/s
vf = 0 m/s
d = ??
a = -9.81 m/s2

vf2 = vf2 + 2·a·d
(0 m/s)2 = (26.2 m/s)2 + 2(- 9.81 m/s2)·d
d = 35 m The value seems reasonable. It takes the vase less than 3
seconds to reach the top. The distance should be less than 45
meters and more than 20 m..
example
• A spark timer is used to record the position of a lab cart
accelerating uniformly from rest. Each 0.10 second, the
timer marks a dot on a recording tape to indicate the
position of the cart at that instant, as shown.

• The linear measurement between t = 0 second to t =
0.30 is 5.4 cm.
• Calculate the magnitude of the acceleration of the cart
during that time interval.
Given: t = 0.30 s, d = 5.4 cm, vi = 0
Find: a = ?          d = vit + ½ at2
a = 120 cm/s2
9/29 do now
• Which combination of graphs above best
describes free-fall motion? [Neglect air
resistance.]
1 D kinematics review
•   Lesson 1 : Describing Motion with        •   Lesson 4 : Describing Motion with
Words                                        Velocity vs. Time Graphs
Introduction to the Language of Kinematics   The Meaning of Shape for a v-t Graph
Scalars and Vectors                          The Meaning of Slope for a v-t Graph
Distance and Displacement                    Relating the Shape to the Motion
Speed and Velocity                           Determining the Slope on a v-t Graph
Acceleration                                 Determining the Area on a v-t Graph
• Lesson 2 : Describing Motion with          • Lesson 5 : Free Fall and the
Diagrams                                     Acceleration of Gravity
Introduction to Diagrams                     Introduction to Free Fall
Ticker Tape Diagrams                         The Acceleration of Gravity
Vector Diagrams                              Representing Free Fall by Graphs
• Lesson 3 : Describing Motion with          How Fast? and How Far?
Position vs. Time Graphs                 The Big Misconception
The Meaning of Shape for a p-t Graph         • Lesson 6 : Describing Motion with
The Meaning of Slope for a p-t Graph             Equations
Determining the Slope on a p-t Graph         The Kinematics Equations
Kinematics Equations and Problem-Solving
Kinematics Equations and Free Fall
Sample Problems and Solutions
Kinematics Equations and Graphs
9/30 do now
•    A roller coaster, traveling with an initial speed of 15
meters per second, decelerates uniformly at -7.0 meters
per second2 to a full stop. Approximately how far does
the roller coaster travel during its deceleration? [show
work]
vi      vf     vavg     a       t   d

15 m/s   0 m/s          -7 m/s2       ?

vf2 = vf2 + 2·a·d
0 = (15 m/s)2 + 2 (-7 m/s2)·d
-(15 m/s)2 = 2 (-7 m/s2)·d
d = 16 m
example
• A 1.0-kilogram ball is dropped from the roof of a
building 40. meters tall. What is the approximate time
of fall? [Neglect air resistance.] [show work]

vi    vf   vavg      a      t     d
0 m/s               -9.81 m/s2 ?   -40 m

d = vit + ½ ·a·t2

t = 2.86 s
objectives
• Review
• Test is tomorrow – do your homework –
castle learning assignments!
Homework note book
• Text book
– P65 - #1, 2, 4, 5
– P53 – #2 (18.8 m); #4 (no; the plane needs 1
km to land.)
– P55 - #3 (-7.5 m/s; 19 m); #4 (2.5 s; 32 m)
– P58 - #3 (16 m/s, 7 s)
– P64 – #4 (3.7 m; 0.76 s)
objectives
• Lab – determine the acceleration
• Review
• Test is tomorrow – do your homework –
castle learning assignments!
10/1 do now
• A ball is thrown straight downward with a speed of 0.50 meter per
second from a height of 4.0 meters. What is the speed of the ball 0.70
second after it is released? [Neglect friction.]

vi        vf     vavg         a          t      d
-0.50 m/s     ?               -9.81 m/s2 0.7 s    -4.0 m

vf = vi + a·t

vf = -7.4m/s

Speed is 7.4m/s
• The velocity-time graph represents the motion of a 3-
kilogram cart along a straight line. The cart starts at t = 0
and initially moves north. What is the magnitude of the
displacement of the cart between t = 3 and t = 7
seconds? [Show work]

0m
example
•    The graph represents the relationship between the
displacement of an object and its time travel along a
straight line.
1.   What is the magnitude of the object's total displacement
during t = 0 seconds and t = 8.0 seconds?         0m
2.   What is the average speed of the object during t = 0
seconds and t = 4.0 seconds? [show work] 2 m/s
3.   What is the magnitude of acceleration of the object
during t = 0 seconds and t = 2.0 seconds?

0m
objectives
• Lab
• Chapter 1 test
• Homework note book is due
Lab 4 : determine acceleration
Question:
How can you find the acceleration of a moving cart on an incline?
Purpose:
To determine the acceleration of a moving cart on an incline.
Material:
spark timer, data recording tape, ruler, incline, tapes, graph paper
Report requirement:
A complete lab write-up includes Title, Purpose, Materials, Data section, and Conclusion/Discussion.
The Data section should include
–      Data table:
• collected data for three trials;
• column headings should be labeled and units shown.
• displacement values for all the trials (except those which are bvious ) should be averaged;
• One sample calculation of acceleration should be shown.
• Final average of the acceleration should be shown.
– Displacement vs. time graph:
• Title of the graph
• Axises of displacement and time with appropriat scale and unit
• Plot the data points on the graph
• Draw the best fit line
– Velocity vs. time graph:
• Title of the graph
• Axises of velocity and time with appropriate scale and unit
• Find velocity at a time instant by using kinematics equations v(t) = 2 · Vaverage (0 – t)
• Plot the data points on the graph
• Draw the best fit line
• Determine the acceleration of the object
• The Conclusion should respond to the question raised in the Purpose of the lab.
– Procedure
– Compare acceleration determined in velocity time graph with the acceleration value determined in data table. Discuss your
result.
2. Time of Free-Fall
A long string of metal washers is constructed such that the first
one is 30 cm from the bottom end, the second is 130 cm from
the bottom end, and the third is 2.74 m from the bottom end.
When the string is released, the washers will sound out at
constant time intervals.
1. Units of Measure

A meter stick, yard stick, 1 kilogram
mass and 1 pound weight are laid out
on a demonstration table for inspection
and comparison of relative size.
Mechanics
Runner's Speed
• 5. Take a stopwatch and a meter stick to a
running track or sidewalk. Use the meter stick to
measure 20 m. Mark the distance with pieces of
masking tape. Measure the time it takes to walk
20 m. Calculate your average speed. Measure
the time it takes you to jog, or run, the same
distance. What is your average speed?
Materials: stopwatch; Meter stick; Masking tape.

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