objectives
1. Describe motion in terms of frame of
reference, displacement, distance, vector, and
scalar.
2. Understand relationship between displacement
and distance, vectors and scalars.
3. Compare distance traveled to displacement.
4. Draw diagrams showing displacement and/or
distance from a point of origin.
Description of motion depends on the frame
of reference
• A frame of reference can be thought of as any spot
you are doing your measurement from as long as it
is not accelerating. The reference point is arbitrary,
but once chosen, it must be used throughout the
problem.
Displacement vs. Distance
• DISTANCE
– The complete length of the path
traveled by a moving object
• DISPLACEMENT
– the length of the straight-line path from
a moving object’s origin to its final
position
Scalar vs. Vector
• SCALAR
– A measured quantity that has NO
DIRECTION
– Examples
• Distance, Time, Mass, Volume
• VECTOR
– A measured quantity that includes
DIRECTION
– SIGN SHOWS DIRECTION
– Example
• Displacement
Sign Conventions
Positive
Negative
NORTH
EAST
RIGHT
SOUTH
WEST
LEFT
A bird fliesAA5ball
cat
meters
rolls
runsnorth,
58 meters
meters
then
north.
west.
7 meters south
Distance = 512
8m
mm
Displacement = -2
+5m
-8
m
Example
• A man drives his car 3 miles north, then
4 miles east.
4 mi
East
3 mi
North
Distance
7 mi
Displacement
5 mi
Northeast
What distance did he travel?
What is his displacement from his point of origin?
Example
• Three men leave the same house on foot. The first man
walks 30 feet north, then 40 feet west. The second man
walks 90 feet south, then 88 feet north. The third man
walks 10 feet east, then 50 feet west.
• Which man has traveled the greatest distance?
The second man
• Who is farthest from the house?
The first man
• Who is closest to the house?
The second man
Distance vs. Displacement
xi
xf
• There could be many distances between xf and xi many
be many, distance depends on the path.
• There is only one displacement between xf and xi.
displacement refers to shortest distance between the xf
and xi and direction from xi to xf
Displacement = change in position = final position – initial position
∆x
=
xf
∆ denotes change
-
xi
example
• Use the diagram to determine the resulting displacement and the
distance traveled by the skier during these three minutes.
Distance (m) Displacement (m)
AB
180
+180
BC
140
-140
CD
100
+100
TOTAL
420
+140
Example
• Consider the motion depicted in the
diagram below. A physics teacher walks 4
meters East, 2 meters South, 4 meters
West, and finally 2 meters North.
Distance:
12m
Displacement: 0
Distance vs. Time Graphs
During what time interval was the object NOT MOVING?
Distance vs. Time
18
2 – 3 seconds
The interval
on the graph where
14
the distance
remains constant!
12
Distance (m)
16
10
8
6
4
2
0
0
1
2
3
4
Time (s)
5
6
7
Displacement vs. Time Graphs
During
During
At
what
what
what
time
distance
time
interval(s)
interval(s)
fromwas
thewas
the
origin
object
thedoes
object
tothe
the
NOT
object
left MOVING?
of stop?
the origin?
Displacement vs. Time
6
5
Displacement (m)
4
3
2
The object’s final position is
at +1 meter
(1 meter to the
When
the displacement
is
of the
origin)
the
object
has a
1 – 2 and negative,
4 – 5right
seconds
position to the
left that
of the origin
Constant displacement
means
the object doesn’t move
1
0
-1 0
1
2
3
4
-2
-3
-4
Time (s)
5
6
7
Objectives
• Know:
– - Definitions of velocity and speed.
– - Equation for average velocity/speed.
• Understand:
– - Relationship between speed, distance and time.
– - Relationship between vectors and scalars.
• Be able to:
– - Use velocity/speed equation to find unknown.
– - Draw and interpret d/t and v/t graphs.
Homework – castle leaning – must show work on a
separate sheet of paper
Velocity vs. Speed
• VELOCITY
– change in DISPLACEMENT occurring over time
– Includes both MAGNITUDE and DIRECTION
• VECTOR
• The direction of the velocity vector is simply the
same as the direction that an object is moving.
• SPEED
– change in DISTANCE occurring over time
– Inclues ONLY MAGNITUDE
• SCALAR
As an object moves, it often
undergoes changes in speed
Types of speed and velocity
•
•
•
•
Initial speed
Final speed
Average speed
Instantaneous speed
•
•
•
•
Initial velocity
Final velocity
Average velocity
Instantaneous velocity
Calculate Average Speed
and Average Velocity
• The average speed during the course of a motion is
often computed using the following formula:
Does NOT include DIRECTION!
• In contrast, the average velocity is often computed using
this formula
∆x
x f - xi
vavg = ∆t = t - t
f
i
= d
t
Average Velocity
d
v
t
What does this remind you of?
SLOPE OF AWhat
GRAPH!
is happening in this
Position vs. Time
Position
Position
graph?
INCREASING
SLOPE
CONSTANT
CONSTANT
POSITIVE
ZERO
SLOPESLOPE
Time
Moving with
Motionless
INCREASING
CONSTANT
Object
positive
velocity
velocity
Example
• Sally gets up one morning and
decides to take a three mile walk.
She completes the first mile in 8.3
minutes, the second mile in 8.9
minutes, and the third mile in 9.2
minutes.
– What is her average
during her
vavg = d / speed
t
walk?
v = 3 mi / (8.3 min + 8.9 min + 9.2 min)
avg
vavg = 0.11 mi / min
Example
• Tom gets on his bike at 12:00 in the
afternoon and begins riding west. At
12:30 he has ridden 8 miles.
– What was his average velocity during
his ride?
vavg = d / t
vavg = 8 mi / 30 min
vavg = 0.27 mi / min WEST
Example – finding displacement
• During a race on level ground, Andre runs with
an average velocity of 6.02 m/s to the east.
What displacement does Andre cover in 137 s?
∆x
(∆t ) vavg = ∆t (∆t )
∆x = vavg (∆t ) = (6.02 m/s)(137 s) = 825 m
Answer: 825 m East
example
• What is the coach's average speed and average
velocity?
average speed = (35 + 20 + 40) yd / 10min = 9.5 yd/min
average velocity = (-35 + 20 - 40) yd / 10 min = - 5.5 yd/min
Instantaneous Speed
• Instantaneous Speed - the speed at any given instant
in time.
Average Speed versus
Instantaneous Speed
Average Speed - the average of all instantaneous
speeds; found simply by a distance/time ratio.
During your trip, there may have been times that you were
stopped and other times that your speedometer was
reading 50 miles per hour. Yet, on average, you were
moving with a speed of
In conclusion
• Speed and velocity are kinematics quantities
that have distinctly different definitions. Speed,
scalar
being a _______quantity,
is the rate at which an
distance
object covers ___________.
The average speed
is the _____________
(a scalar quantity) per
distance
time ratio. Speed is ignorant of direction. On the
vector
other hand, velocity is a _________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.
Using v-t Graphs
Velocity (m/s)
Velocity vs. Time
35
30
25
20
15
10
5
0
-5 0
-10
-15
-20
What can we DO
with a v-t graph?
Find distance
traveled
1
2
3
4
5
6
7
8
9
10
11
Time (s)
Area
under
under
the
graph to find
How doArea
you use
thethe
v-tgraph
graph
DISTANCE
AVERAGE
DISPLACEMENT?
TRAVELED?
VELOCITY?
Area
AreaononTOP
TOPand
= POSITIVE
BOTTOM
Area
both
onconsidered
BOTTOMPOSITIVE
= NEGATIVE
Find
displacement
Find average
velocity
objective
1. Construct and interpret graphs of
position vs. time and velocity vs. time.
Graphical Interpretation of
Velocity: position vs. time graph
• The Meaning of Shape for a p-t Graph
Straight line Constant Velocity
(positive)
Curved line –
Changing Velocity
(increasing)
As the slope goes, so goes the velocity
Slow, Positive,
Constant Velocity
Fast, Positive,
Constant Velocity
Fast, Negative,
Constant Velocity
Slow, Negative
Constant Velocity
Check Your Understanding
• Use the principle of slope to describe the motion
of the objects depicted by the two plots below.
Slope is positive, increasing
slope is negative, increasing
velocity is positive,
increasing (faster)
velocity is negative,
increasing (faster)
The slope of the line on a position-time graph
is equal to the velocity of the object
Slope = constant = +10 m/s
Velocity = constant = +10 m/s
Slope is changing –
increasing in positive
direction
velocity is changing –
increasing in positive
direction
Velocity information on P-t graph
•
•
•
•
Initial velocity
Final velocity
Average velocity
Instantaneous velocity
•
•
•
•
+10 m/s
+10 m/s
+10 m/s
+10 m/s
•Initial velocity
•Final velocity
•Average velocity
•Instantaneous velocity
•
•
•
•
0 m/s
~+22 m/s
+10 m/s
Slope of the tangent
line
As the slope goes, so goes the velocity
Slope is negative,
increasing (steeper)
Slope is negative,
decreasing (flatter)
velocity is negative,
increasing (faster)
velocity is negative,
decreasing (slower)
example
Describe the velocity of the object between 0-5 s and
between 5-10 s.
The velocity is 5 m/s between 0-5 seconds
The velocity is zero between 5-10 seconds
Practice – determine average velocity
7.3 m/s
objectives
1. Describe motion in terms of changing
velocity.
2. Compare graphical representations of
accelerated and nonaccelerated motions.
3. Apply kinematics equations to calculate
distance, time, or velocity under conditions
of constant acceleration.
Acceleration
• Acceleration change in VELOCITY occuring over TIME
Measured in m/s2
Vector
Acceleration = (change in velocity) / time
∆v
vf - vi
aavg = ∆t = t - t
f
i
Questions:
1. “If an object has a large velocity, does it necessarily have
a large acceleration?
2. If an object has a large acceleration, does it necessarily
have a large velocity?”
• Anytime an object's velocity is changing,
the object is said to be accelerating; it has
an acceleration.
• What are the three ways to accelerate
your car?
Gas pedal, break, steering wheel
The Direction of the Acceleration Vector
• The direction of the acceleration vector depends on
whether the object is speeding up or slowing down
• If an object is speeding up, then its acceleration is in the
same direction of its motion.
– moving in the + direction, the acceleration + direction
– moving in the - direction, the acceleration - direction
• If an object is slowing down, then its acceleration is in
the opposite direction of its motion.
– moving in the + direction, the acceleration - direction
– moving in
• The direction of velocity and acceleration do not have to
be the same!!!
Negative
Positive Velocity
Velocity
Negative
Positive Acceleration
Acceleration
Speeding
Speeding
Slowing
up
up in
indown
+- direction
direction
Eventually speeds up in +
– direction!
+v
-v
+v
-v
+a
-a
-a
+a
The car will…
The car will…
The car will…
The car will…
Equations
aavg
v

t
“Acceleration is a rate
of change in velocity”
“The slope of a v-t
graph tells what the
ACCELERATION IS
DOING!”
v f  vi  at
“An object’s velocity
at any point in time can
be found by considering:
- its starting velocity
- its acceleration
- the amount of time over
which it accelerates”
Example – average acceleration
• What is the acceleration of an amusement
park ride that falls from rest to a speed of
28 m/s in 3.0 s?
∆v
vf - vi
aavg = ∆t = t - t
f
i
aavg = (28 m/s – 0) / 3.0 s = 9.3 m/s/s down
Example – finding time of
accelertion
• 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?
∆v
aavg = ∆t
∆v
aavg ∆t
∆t = ∆v = (0.0 m/s – 9.0 m/s) / -1.8 m/s2 = 5.0 s
aavg
Example
• Two cars start at the same point . Car A
starts with a velocity of -5 meters per
second while Car B starts with a velocity
of +3 meters per second. At the end of 15
seconds, Car A has a velocity of +25
vf = vi + at
meters per second.
vf = +3 m/s + (2 m/s2)(15 s)
vf =acceleration?
+33 m/s
– What is Car A’s
vf = vi + at
25 m/s = -5 m/s + a(15 s)
– If Car B has the
acceleration,
what is its
2
a =same
+2 m/s
speed at 15 seconds?
The Meaning of Constant
Acceleration
The velocity is changing by a
constant amount - in each
second of time.
Constant motion vs. accelerated motion
ticker tape
constant velocity: displacement is changing by a
constant amount - in each second of time.
constant acceleration: displacement is increasing in
each second of time. Velocity is changing by a constant
amount - in each second of time
Describing Motion with Velocity vs. Time
Graphs
The slope value of any straight line on a velocity-time graph
is the acceleration of the object
Velocity: Constant
Slope = 0
Acceleration = 0
Velocity: increasing
Slope: +, constant
Acceleration: +, constant
Speeding up or slowing down?
• Speeding up means that the magnitude (or
numerical value) of the velocity is getting large.
Example: describe velocity and acceleration for
each v-t diagram
Velocity is
positive.
increasing,
speeding up
Velocity is
positive
decreasing,
slowing down
Velocity is
negative
decreasing,
slowing down
Velocity is
negative,
increasing,
speeding up
Acceleration is
constant,
positive
Acceleration is
constant,
negative
Acceleration is
constant,
positive
Acceleration is
constant,
negative
Example 1
Determine acceleration
From 0 s to 4 s: 0 m/s2
From 4 s to 8 s: 2 m/s2
•
Example 2
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.
+40 m/s/s
1. t = 0 - 1 second
+20 m/s/s
2. t = 1 - 4 second
3. t = 4 - 12 second -20 m/s/s
objectives
• Understand Relationships for all kinematics
equations.
• Understand Relationship between slope of
kinematics graphs and quantities they represent.
• Use any kinematics equation to find unknown.
• Draw and interpret all kinematics graphs.
Homework – castle learning
vavg =
vf + v i
2
vf + vi
2
∆x
vavg =
∆t
∆x
= ∆t
∆x = ½ (∆t)(vf + vi)
The displacement equals to the
area under the velocity vs. time
graph
Determining the Area on a v-t
Graph
For velocity versus time graphs, the
area bound by the line and the axes
represents the displacement.
The shaded area is representative of the
displacement
area = base x height
area = ½ base x height
area = ½ base x ( height1 + height2)
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).
120 m
90 m
example
• Determine the displacement of the object
during the first second (Practice A) and
during the first 3 seconds (Practice B).
5m
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).
25 m
40 m
Recap
∆x
vf + v i
• Average velocity: v = ∆t =
2
• Acceleration:
∆v
vf - vi
a = ∆t = t - t
f
i
• Final velocity:
vf = vi + a∆t
• Displacement:
∆x = ½ (∆t)(vf + vi)
• Displacement:
∆x = vi∆t + ½ a∆t2
Another equation for final velocity
• Final velocity:
vf = vi + a∆t
• Displacement:
∆x = ½ (∆t)(vf + vi)
• Final velocity:
vf2= vi2 + 2a∆x
Constant Non-Zero
Acceleration
1 2
d  vi t  at
2
“Distance or Displacement”
Equation
v  v  2ad
“Timeless or Shortcut
Equation”
2
f
2
i
DOES NOT INCLUDE
TIME!
In one
dimensional
motion:
∆x = d
Example
• A bicyclist accelerates from 5.0 m/s to a
velocity of 16 m/s in 8 s. Assuming uniform
acceleration, what displacement does the
bicyclist travel during this time interval?
∆x = ½ (vf + vi)(∆t)
84 m
Example
• A racing car reaches a speed of 42 m/s. It then begins a
uniform negative acceleration, using its parachute and
breaking system, and comes to rest 5.5 s later. Find how
far the car moves before stopping.
120 m
Example
• A train starts from rest and leaves
Greenburg station and travels for 500.
meters with an acceleration of 1.20 meters
per second2.
– What is the train’s
vf2 = final
vi2 + velocity?
2ad
vf2 = 0 + 2(1.20 m/s2)(500. m)
vf = 34.6 m/s
– How long does it take the train to reach its final
velocity?
vf = vi + at
34.6 m/s = 0 + (1.20 m/s2) t
t = 28.8 s
Example
• A driver traveling at 85. miles per hour sees a
police car hiding in the trees 2.00 miles ahead.
He applies his brakes, decelerating at -500.
miles per hour2.
2 =mph,
– If the speed limit isv 55
v 2 + will
2adhe get a ticket?
vf2 = (85.
f
mph)2
i
+ 2(-500. mph2)(2.00 mi)
vf = 72.3 mph *YES*
– What would his acceleration
need to be to not get a
vf2 = vi2 + 2ad
ticket?
(55 mph)2 = (85. mph)2 + 2a(2.00 mi)
a = -1050 mph2
Objective
• Apply kinematics equations to calculate
distance, time, or velocity under conditions
of constant acceleration.
example
• A barge moving with a speed of 1.00 m/s increases
speed uniformly, so that in 30.0 s it has traveled 60.2 m.
What is the magnitude of the barge’s acceleration?
a  6.71  102 m/s2
example
• A person pushing a stroller starts from rest, uniformly
accelerating at a rate of 0.500 m/s2. What is the velocity
of the stroller after it has traveled 4.75 m?
vf = +2.18 m/s
example
• An aircraft has a landing speed of 302 km/h. The landing
area of an aircraft carrier is 195 m long. What is the
minimum uniform accelerating required for a safe
landing?
a = -18 m/s2
example
• A plane starting at rest at one end of a runway
undergoes a uniform acceleration of 4.8 m/s2 for 15 s
before takeoff. What is its speed at takeoff? How long
must the runway be for the plane to be able to take off?
vf = 72 m/s
∆x = 540 m
Objectives
1. Relate the motion of a freely falling
body to motion with constant
acceleration.
2. Calculate displacement, velocity, and
time at various points in the motion of a
freely falling object.
3. Compare the motion of different objects
in free fall.
Two important motion characteristics that
are true of free-falling objects
• Free-falling objects do not
encounter air resistance.
• All free-falling objects (on Earth)
accelerate downwards at a
constant rate.
“g” - The “Magic” Number
• “g” means ACCELERATION DUE TO GRAVITY
• Each planet, star, moon, or other large object
has its own value for “g”
Examples
• “g” is 1.62 m/s2 on the Moon
• “g” is 26 m/s2 on Jupiter
• “g” is 9.81 m/s2 on Earth
If the feather and elephant experiment were performed on
the moon, would they still fall at the same rate, just like
on Earth?
a = -g = -9.81m/s2 for all free fall objects.
(the ‘-’ sign means “downward”)
•
•
•
•
Dropped from rest
Throwing downward
Throwing upward
Throwing side ways
• Since free fall motion has constant
acceleration, we can apply all
kinematics equations on free fall
motion.
Free fall – motion with constant acceleration
a = -g
To solve free fall motion problems, we can use kinematics
equations with constant acceleration.
a = -g
∆x = ∆y
Velocity of a dropped object
• An object falls from rest. What is its
velocity at the end of one second? Two
= -9.81seconds?
m/s2 (‘-’ means “downward”)
v f  viseconds?
 at a = -gThree
t=0
vi = 0 m/s
t=1s
v = -9.81 m/s
t=2s
v = -19.62 m/s
t=3s
v = -29.43 m/s
Displacement of a dropped object
• An object falls from rest. How far has it
fallen at the end of one1 second? Two
d seconds?
vi t  at 2
seconds? Three
2
t=0
d=0m
t=1s
d = -4.905 m
t=2s
d = -19.62 m
t=3s
d = -44.145 m
Velocity and distance of a free falling object
dropped from rest
+
Throwing Downward
av=i =-9.81
-10 m/s 2
• An object is thrown downward from
the top of a 175 meter building with
an initial speed of 10 m/s.
– What acceleration does it
experience?
• -9.81 m/s2
– What is the object’s initial
velocity?
• -10 m/s
“a” and “vi” in SAME DIRECTION
Building Example (cont.)
• What is the object’s velocity as it hits the
ground?
•
•
•
•
vi = -10 m/s
d = -175 m
a = -9.81 m/s2
vf = ?
vf2 = vi2 + 2ad
vf2 = 0 + 2(-9.81 m/s2)(-175 m)
vf = -58.6 m/s
• How long does it take the object to hit the
vf = vi + at
ground?
-58.6 m/s = -10 m/s + (-9.81 m/s2) t
• t=?
t = 4.95 s
Throwing Upward
t=2s
v = +30.38 m/s
t=1s
v = +40.19 m/s
t=0s
vi = +50 m/s
v f  vi  at
• A cannon fires a shot
directly upward with an
initial velocity of 50 m/s.
– What acceleration does
the cannonball
experience?
• -9.81 m/s2
– What is the cannonball’s
initial velocity?
• +50 m/s
“a” and “vi” in OPPOSITE
DIRECTION
Cannon (cont.)
• What is the object’s velocity as it reaches the
top of its flight?
0 m/s
(All objects momentarily STOP at the top of their flight)
• How long does it take the cannonball to reach
the top of its flight?
vf = vi + at
0 = +50 m/s + (-9.81 m/s2) t
t = 5.1 s
• What is the maximum height of the
cannonball?
d = v t + ½ at2
i
d = (+50 m/s)(5.1 s) + ½ (-9.81 m/s2)(5.1 s)2
d = +127.4 m
Question
•
If a ball thrown into the air, and caught at
the same point you it was released. Is
the speed of the ball equal to, more than,
or less than the initial speed when it is
caught?
Acceleration is the rate at which an object changes its
velocity.
• To accelerate at -10 m/s/s
means to change the
velocity by -10 m/s each
second.
• If the velocity and time for
a free-falling object being
tossed upward from a
position with speed of 20
m/s were tabulated, then
one would note the
pattern.
Time (s)
Velocity (m/s)
0
20
1
10
2
0
3
-10
4
-20
Question
•
If a ball thrown into the air, and caught at the
same point you it was released. Is the speed of
the ball equal to, more than, or less than the
initial speed when it is caught?
given: vi, a = -9.81 m/s2; ∆y = 0 m;
Find: vf
vf2 = vi2 + 2a∆y
vf2 = vi2 + 0
vi
vf
vf2 = vi2
The ball has the same velocity,
but in the opposite direction
Class work
• A ball is thrown straight up into the air at an initial
velocity of 39.24 m/s. Fill in the table showing the ball’s
position, velocity, and acceleration each second for the
first 4.00 s of its motion.
t (s)
Y (m)
v (m/s)
a (m/s2)
0
0
19.62
-9.81
1
2
3
4
Free Fall by Graphs of a dropped object
A position versus time graph
for a free-falling object
velocity increase in the
negative direction – slope
is neg. & increasing
A velocity versus time graph
for a free-falling object
velocity increases in the
negative direction at
constant rate – slope is neg.
& constant
Free Fall Graphs of upward object
velocity
Upward: velocity is big, positive, decreasing,
slope is constant (a = -9.8 m/s/s).
Top, velocity is zero. Slope remains the same
(acceleration is still -9.8 m/s/s)
position
time
Downward: velocity increases in negative
direction at the same constant rate of -9.81
m/s/s (slope remains the same), reaches the
same speed as it started upward.
Upward: displacement increases, slope is
positive, decreasing (velocity is positive,
decreasing)
Top: slope = 0 (its velocity is zero)
time
Downward: displacement decreases, its slope
increases in negative direction (velocity is
negative, increasing)