1. objectives
Describe motion in terms of frame of reference, displacement, distance, vector, and scalar.
Understand relationship between displacement and distance, vectors and scalars.
Compare distance traveled to displacement.
Draw diagrams showing displacement and/or distance from a point of origin.

2. 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.

3. Displacement vs. 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

4. Scalar vs. Vector SCALAR VECTOR
A measured quantity that has NO DIRECTION
Examples
Distance, Time, Mass, Volume
VECTOR
A measured quantity that includes DIRECTION
SIGN SHOWS DIRECTION
Example
Displacement

5. Sign Conventions Positive Negative
NORTH
EAST
RIGHT
Negative
SOUTH
WEST
LEFT
A bird flies 5 meters north, then 7 meters south
Distance =
Displacement =
A ball rolls 5 meters north.
Distance =
Displacement =
A cat runs 8 meters west.
Distance =
Displacement =
5 m
12 m
8 m
-2 m
+5 m
-8 m

6. Example A man drives his car 3 miles north, then 4 miles east.
Distance
7 mi
3 mi
North
Displacement
5 mi
Northeast
What distance did he travel?
What is his displacement from his point of origin?

7. 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?
Who is farthest from the house?
Who is closest to the house?
The second man
The first man
The second man

8. Distance vs. Displacementxi 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 xi
∆ denotes change

9. example Distance (m) Displacement (m)
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

10. 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.
12m
Distance:
Displacement:

11. Distance vs. Time Graphs
During what time interval was the object NOT MOVING?
2 – 3 seconds
The interval on the graph where
the distance remains constant!

12. Displacement vs. Time Graphs
During what time interval(s) was the object to the left of the origin?
During what time interval(s) was the object NOT MOVING?
At what distance from the origin does the object stop?
The object’s final position is
at +1 meter (1 meter to the
right of the origin)
When the displacement is
negative, the object has a
position to the left of the origin
1 – 2 and 4 – 5 seconds
Constant displacement means that
the object doesn’t move

13. Objectives Know: Understand: Be able to:
- 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

14. Velocity vs. Speed VELOCITY SPEED
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

15. As an object moves, it often undergoes changes in speed

16. Types of speed and velocity
Initial speed
Final speed
Average speed
Instantaneous speed
Initial velocity
Final velocity
Average velocity
Instantaneous velocity

17. Calculate Average Speed and Average Velocity
The average speed during the course of a motion is often computed using the following formula:
In contrast, the average velocity is often computed using this formula
Does NOT include DIRECTION!
vavg = ∆x ∆t =
tf - ti
xf - xi d t

18. What does this remind you of? What is happening in this
Average Velocity
What does this remind you of?
SLOPE OF A GRAPH!
What is happening in this
graph?
INCREASING
SLOPE
Moving with
CONSTANT
positive velocity
Moving with
INCREASING
velocity
Motionless
Object
CONSTANT
POSITIVE
SLOPE
CONSTANT
ZERO
SLOPE

19. vavg = 3 mi / (8.3 min + 8.9 min + 9.2 min)
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 speed during her walk?
vavg = d / t
vavg = 3 mi / (8.3 min min min)
vavg = 0.11 mi / min

20. 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

21. 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
(∆t )
vavg =
(∆t )
∆x = vavg (∆t ) = (6.02 m/s)(137 s) = 825 m
Answer: 825 m East

22. example What is the coach's average speed and average velocity?
average speed = ( ) yd / 10min = 9.5 yd/min
average velocity = ( ) yd / 10 min = yd/min

23. Instantaneous Speed
Instantaneous Speed - the speed at any given instant in time.

24. 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

25. In conclusion
Speed and velocity are kinematics quantities that have distinctly different definitions. Speed, being a _______quantity, is the rate at which an object covers ___________. The average speed is the _____________ (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a _________quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the ______________ or position change (a vector quantity) per time ratio.
scalar
distance
distance
vector
displacement

26. Using v-t Graphs What can we DO with a v-t graph? Find distance
traveled
Find displacement
Find average
velocity
Area under the graph
Area on TOP = POSITIVE
Area on BOTTOM = NEGATIVE
Area under the graph
Area on TOP and BOTTOM
both considered POSITIVE
How do you use the v-t graph to find DISPLACEMENT?
How do you use the v-t graph to find DISTANCE TRAVELED?
How do you use the v-t graph to find
AVERAGE VELOCITY?

27. objective
Construct and interpret graphs of position vs. time and velocity vs. time.

28. 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)

29. As the slope goes, so goes the velocity
Slow, Positive, Constant Velocity
Fast, Positive, Constant Velocity
Fast, Negative, Constant Velocity
Slow, Negative
Constant Velocity

30. Check Your Understanding
Use the principle of slope to describe the motion of the objects depicted by the two plots below.
slope is negative, increasing
velocity is negative, increasing (faster)
Slope is positive, increasing
velocity is positive, increasing (faster)

32. 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

33. Velocity information on P-t graph
Initial velocity
Final velocity
Average velocity
Instantaneous velocity
+10 m/s

34. 0 m/s
~+22 m/s
+10 m/s
Slope of the tangent line
Initial velocity
Final velocity
Average velocity
Instantaneous velocity

35. 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)

36. 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

37. Practice – determine average velocity
7.3 m/s

38. 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.

39. Acceleration Acceleration aavg = ∆v ∆t = tf - ti vf - vi
change in VELOCITY occuring over TIME
Measured in m/s2
Vector
Acceleration = (change in velocity) / time
aavg = ∆v ∆t =
tf - ti
vf - vi
Questions:
“If an object has a large velocity, does it necessarily have a large acceleration?
If an object has a large acceleration, does it necessarily have a large velocity?”

40. What are the three ways to accelerate your car?
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

41. 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!!!

43. Negative Acceleration Negative Acceleration Positive Acceleration
Negative Velocity
Negative Velocity
Positive Velocity
Positive Velocity
Negative Acceleration
Negative Acceleration
Positive Acceleration
Positive Acceleration
Slowing down
Eventually speeds up in + direction!
Slowing down
Eventually speeds up in – direction!
Speeding up in + direction
Speeding up in - direction

44. +v -v +v -v +a -a -a +a The car will… The car will… The car will…

45. Equations “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”
“Acceleration is a rate
of change in velocity”
“The slope of a v-t
graph tells what the
ACCELERATION IS
DOING!”

46. 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?
aavg = ∆v ∆t =
tf - ti
vf - vi
aavg = (28 m/s – 0) / 3.0 s = 9.3 m/s/s down

47. 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?
aavg = ∆v ∆t ∆v
aavg ∆t
∆t = ∆v
aavg
= (0.0 m/s – 9.0 m/s) / -1.8 m/s2 = 5.0 s

48. 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 meters per second.
What is Car A’s acceleration?
If Car B has the same acceleration, what is its speed at 15 seconds?
vf = vi + at
vf = +3 m/s + (2 m/s2)(15 s)
vf = +33 m/s
vf = vi + at
25 m/s = -5 m/s + a(15 s)
a = +2 m/s2

49. The Meaning of Constant Acceleration
The velocity is changing by a constant amount - in each second of time.

50. 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

51. 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

52. Speeding up or slowing down?
Speeding up means that the magnitude (or numerical value) of the velocity is getting large.

53. Example: describe velocity and acceleration for each v-t diagram
Velocity is positive. increasing, speeding up
Acceleration is constant, positive
Velocity is positive decreasing, slowing down
Acceleration is constant, negative
Velocity is negative decreasing, slowing down
Acceleration is constant, positive
Velocity is negative, increasing, speeding up
Acceleration is constant, negative

54. Example 1 Determine acceleration From 0 s to 4 s: From 4 s to 8 s:

55. 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.
t = second
t = second
t = second
+40 m/s/s
+20 m/s/s
-20 m/s/s

56. 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

57. vavg = vf + vi 2 vavg = ∆x ∆t = vf + vi 2 ∆x ∆t ∆x = ½ (∆t)(vf + vi)
The displacement equals to the area under the velocity vs. time graph

58. 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.

59. area = ½ base x ( height1 + height2)
The shaded area is representative of the displacement
area = base x height
area = ½ base x height
area = ½ base x ( height1 + height2)

60. 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).
90 m
120 m

61. example
Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B).
5 m
45 m

62. 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

63. Recap v = ∆x ∆t = 2 vf + vi Average velocity: a = ∆v ∆t = tf - ti
Acceleration:
Final velocity:
vf = vi + a∆t
Displacement:
∆x = ½ (∆t)(vf + vi)
Displacement:
∆x = vi∆t + ½ a∆t2

64. Another equation for final velocity
vf = vi + a∆t
Displacement:
∆x = ½ (∆t)(vf + vi)
Final velocity:
vf2= vi2 + 2a∆x

65. Constant Non-Zero Acceleration
“Distance or Displacement”
Equation
“Timeless or Shortcut
Equation”
DOES NOT INCLUDE
TIME!

66. In one dimensional motion:
∆x = d

67. 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

68. 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

69. 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 final velocity?
How long does it take the train to reach its final velocity?
vf2 = vi2 + 2ad
vf2 = 0 + 2(1.20 m/s2)(500. m)
vf = 34.6 m/s
vf = vi + at
34.6 m/s = 0 + (1.20 m/s2) t
t = 28.8 s

70. 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 miles per hour2.
If the speed limit is 55 mph, will he get a ticket?
What would his acceleration need to be to not get a ticket?
vf2 = vi2 + 2ad
vf2 = (85. mph)2 + 2(-500. mph2)(2.00 mi)
vf = 72.3 mph *YES*
vf2 = vi2 + 2ad
(55 mph)2 = (85. mph)2 + 2a(2.00 mi)
a = mph2

71. Objective
Apply kinematics equations to calculate distance, time, or velocity under conditions of constant acceleration.

72. 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

73. example
A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75 m?
vf = m/s

74. 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

75. 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

76. Objectives
Relate the motion of a freely falling body to motion with constant acceleration.
Calculate displacement, velocity, and time at various points in the motion of a freely falling object.
Compare the motion of different objects in free fall.

77. 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.

78. “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?

79. (the ‘-’ sign means “downward”)
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.

80. 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

81. Velocity of a dropped object
An object falls from rest. What is its velocity at the end of one second? Two seconds? Three seconds?
a = -g = m/s2 (‘-’ means “downward”)
t = 0
vi = 0 m/s
t = 1 s
v = m/s
v = m/s
t = 2 s
t = 3 s
v = m/s

82. Displacement of a dropped object
An object falls from rest. How far has it fallen at the end of one second? Two seconds? Three seconds?
t = 0
d = 0 m
t = 1 s
d = m
d = m
t = 2 s
t = 3 s
d = m

83. Velocity and distance of a free falling object dropped from rest+

84. Throwing Downward
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 = m/s2
vi = -10 m/s
“a” and “vi” in SAME DIRECTION

85. Building Example (cont.)
What is the object’s velocity as it hits the ground?
vi = -10 m/s
d = -175 m
a = m/s2
vf = ?
How long does it take the object to hit the ground?
t = ?
vf2 = vi2 + 2ad
vf2 = 0 + 2(-9.81 m/s2)(-175 m)
vf = m/s
vf = vi + at
-58.6 m/s = -10 m/s + (-9.81 m/s2) t
t = 4.95 s

86. Throwing Upward
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
t = 2 s
v = m/s
t = 1 s
v = m/s
t = 0 s
vi = +50 m/s
“a” and “vi” in OPPOSITE DIRECTION

87. Cannon (cont.)
What is the object’s velocity as it reaches the top of its flight?
How long does it take the cannonball to reach the top of its flight?
What is the maximum height of the cannonball?
0 m/s
(All objects momentarily STOP at the top of their flight)
vf = vi + at
0 = +50 m/s + (-9.81 m/s2) t
t = 5.1 s
d = vit + ½ at2
d = (+50 m/s)(5.1 s) + ½ (-9.81 m/s2)(5.1 s)2
d = m

88. 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?

89. 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.
Time (s)
Velocity (m/s) 20 1 10 2 3
-10 4
-20
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.

90. 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 = m/s2; ∆y = 0 m;
Find: vf
vf2 = vi2 + 2a∆y
vf2 = vi2 + 0
vf2 = vi2 vi vf
The ball has the same velocity, but in the opposite direction

91. Class work
A ball is thrown straight up into the air at an initial velocity of 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)
19.62
-9.81 1 2 3 4

92. Free Fall by Graphs of a dropped object
A position versus time graph for a free-falling object
A velocity versus time graph for a free-falling object
velocity increases in the negative direction at constant rate – slope is neg. & constant
velocity increase in the negative direction – slope is neg. & increasing

93. Free Fall Graphs of upward object
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)
Downward: velocity increases in negative direction at the same constant rate of m/s/s (slope remains the same), reaches the same speed as it started upward.
velocity
time
Upward: displacement increases, slope is positive, decreasing (velocity is positive, decreasing)
Top: slope = 0 (its velocity is zero)
Downward: displacement decreases, its slope increases in negative direction (velocity is negative, increasing)
position
time