kinmatics

1. Kinematics in One Dimension

2. Kinematics deals with the concepts that
are needed to describe motion.
Dynamics deals with the effect that forces
have on motion.
Together, kinematics and dynamics form
the branch of physics known as Mechanics.

3. 2.1 Displacement
position
initial

o
x

position
final

x

nt
displaceme



 o
x
x
x




4. 2.1 Displacement
m
0
.
2

o
x

m
0
.
7

x

m
0
.
5

x

m
0
.
5
m
2.0
m
7.0 




 o
x
x
x




5. m
0
.
2

x

m
0
.
7

o
x

m
0
.
5


x

m
0
.
5
m
7.0
m
2.0 





 o
x
x
x




6. m
0
.
2


o
x

m
0
.
7

x

m
0
.
5

x

  m
0
.
7
m
.0
2
m
5.0 





 o
x
x
x




7. Average speed is the distance traveled divided by the time
required to cover the distance.
time
Elapsed
Distance
speed
Average 
SI units for speed: meters per second (m/s)

8. Example 1 Distance Run by a Jogger
How far does a jogger run in 1.5 hours (5400 s) if his
average speed is 2.22 m/s?
time
Elapsed
Distance
speed
Average 
  
   m
12000
s
5400
s
m
22
.
2
time
Elapsed
speed
Average
Distance




9. Average velocity is the displacement divided by the elapsed
time.
time
Elapsed
nt
Displaceme
velocity
Average 
t
t
t o
o






x
x
x
v





10. Example 2 The World’s Fastest Jet-Engine Car
Andy Green in the car ThrustSSC set a world record of
341.1 m/s in 1997. To establish such a record, the driver
makes two runs through the course, one in each direction,
to nullify wind effects. From the data, determine the average
velocity for each run.

11. s
m
5
.
339
s
4.740
m
1609







t
x
v


s
m
7
.
342
s
4.695
m
1609







t
x
v



12. Position-Time Graphs
• We can use a
postion-time graph to
illustrate the motion of
an object.
• Position is on the y-
axis
• Time is on the x-axis

13. Plotting a Distance-Time Graph
• Axis
– Distance (position) on
y-axis (vertical)
– Time on x-axis
(horizontal)
• Slope is the velocity
– Steeper slope = faster
– No slope (horizontal
line) = staying still

14. Where and When
• We can use a position
time graph to tell us
where an object is at any
moment in time.
• Where was the car at 4
s?
• 30 m
• How long did it take the
car to travel 20 m?
• 3.0 s

15. Interpret this graph…

16. Describing in Words

17. Describing in Words
• Describe the motion of
the object.
• When is the object
moving in the positive
direction?
• Negative direction?
• When is the object
stopped?
• When is the object
moving the fastest?
• The slowest?

18. The notion of acceleration emerges when a change in
velocity is combined with the time during which the
change occurs.

19. t
t
t o
o






v
v
v
a




DEFINITION OF AVERAGE ACCELERATION

20. Example 3 Acceleration and Increasing Velocity
Determine the average acceleration of the plane.
s
m
0

o
v

h
km
260

v

s
0

o
t s
29

t
s
h
km
0
.
9
s
0
s
29
h
km
0
h
km
260








o
o
t
t
v
v
a




22. Example 3 Acceleration and Decreasing
Velocity
2
s
m
0
.
5
s
9
s
12
s
m
28
s
m
13








o
o
t
t
v
v
a




23. Equations of Kinematics for Constant Acceleration
 t
v
v
x o 
 2
1
2
2
1
at
t
v
x o 

at
v
v o 

ax
v
v o 2
2
2



24. Five kinematic variables:
1. displacement, x
2. acceleration (constant), a
3. final velocity (at time t), v
4. initial velocity, vo
5. elapsed time, t

25. Kinematics for Constant Acceleration
The speedboat has a constant
acceleration of +2.0 m/s². If the initial
velocity of the boat is +6.0 m/s, find the
boat’s displacement after 8.0 seconds.

26. Kinematics for Constant Acceleration
     
m
110
s
0
.
8
s
m
0
.
2
s
0
.
8
s
m
0
.
6
2
2
2
1
2
2
1





 at
t
v
x o
2

27. Kinematics for Constant Acceleration
A jet is taking off from the deck of an aircraft
carrier, as the Figure shows. Starting from rest,
the jet is catapulted with a constant
acceleration of +31 m/s² along a straight line
and reaches a velocity of +62 m/s. Find the
displacement of the jet.

28. Kinematics for Constant Acceleration
Example 6 Catapulting a Jet
Find its displacement.
s
m
0

o
v
??

x
2
s
m
31


a
s
m
62


v

29. Kinematics for Constant Acceleration
   
  m
62
s
m
31
2
s
m
0
s
m
62
2 2
2
2
2
2






a
v
v
x o

30. Applications of the Equations of Kinematics
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables.
4. Verify that the information contains values for at least three
of the five kinematic variables. Select the appropriate equation.
5. When the motion is divided into segments, remember that
the final velocity of one segment is the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a
kinematics problem.

31. Applications of the Equations of Kinematics
Example 8 An Accelerating Spacecraft
A spacecraft is traveling with a velocity of +3250 m/s. Suddenly
the retrorockets are fired, and the spacecraft begins to slow down
with an acceleration whose magnitude is 10.0 m/s2. What is
the velocity of the spacecraft when the displacement of the craft
is +215 km, relative to the point where the retrorockets began
firing?

32. Applications of the Equations of Kinematics
Example 8 An Accelerating Spacecraft
A spacecraft is traveling with a velocity of +3250 m/s. Suddenly
the retrorockets are fired, and the spacecraft begins to slow down
with an acceleration whose magnitude is 10.0 m/s2. What is
the velocity of the spacecraft when the displacement of the craft
is +215 km, relative to the point where the retrorockets began
firing?
x a v vo t
+215000 m -10.0 m/s2 ? +3250 m/s

33. Applications of the Equations of Kinematics
Example 8 An Accelerating Spacecraft
x a v vo t
+215000 m -10.0 m/s2 ? +3250 m/s
V=?
V=?
V=?

34. Applications of the Equations of Kinematics
ax
v
v o 2
2
2


x a v vo t
+215000 m -10.0 m/s2 ? +3250 m/s
ax
v
v o 2
2


    
s
m
5
.
2502
m
215000
s
m
0
.
10
2
s
m
3250 2
2






v
v

35. Applications of the Equations of Kinematics
ax
v
v o 2
2
2

 ax
v
v o 2
2


    
s
m
5
.
2502
m
215000
s
m
0
.
10
2
s
m
3250 2
2






v
v

36. Applications of the Equations of Kinematics
Example 8 An Accelerating Spacecraft
A spacecraft is traveling with a velocity of +3250 m/s. Suddenly
the retrorockets are fired, and the spacecraft begins to slow down
with an acceleration whose magnitude is 10.0 m/s2. What is
the velocity of the spacecraft when the displacement of the craft
is +215 km, relative to the point where the retrorockets began
firing?

38. FREE FALL

40. Freely Falling Bodies
In the absence of air resistance, it
is found that all bodies at the same
location above the Earth fall
vertically with the same
acceleration.
This idealized motion is called
free-fall and the acceleration of a
freely falling body is called the
acceleration due to gravity.

41. Falling Objects
● Aristotle believed that
heavier objects would fall
faster, but never tested it.
● Galileo Galilei is famous
for being the first in
recorded history to test
whether heavier objects
actually fall faster.
● Conclusion: all objects fall
at about the same speed.
http://commons.wikimedia.org/
wiki/File:Leaning_Tower_of_Pi
sa.jpg

42. Freely Falling Bodies
A free-falling object is an object
which is falling under the sole
influence of gravity
1. A free-falling objects do not
encounter air resistance.
2. All free-falling objects (on Earth)
accelerate downwards at a rate
of…
near the Earth’s Surface.
2
2
s
ft
2
.
32
or
s
m
80
.
9

g

43. 2
s
m
80
.
9

g

46. Same acceleration for
different masses and forces
applied
Why doesn’t the
elephant fall faster?

49. Freely Falling Bodies
Example 10 A Falling Stone
A stone is dropped from the top of a tall
building and falls freely from rest. What is the
position y and velocity of the stone after 1.0
s, 2. 0 s, and 3.00s of free fall?

50. Freely Falling Bodies
y a v vo t
? -9.80 m/s2 ? 0 m/s 1.00 s

51. y a v vo t
? -9.80 m/s2 ? 0 m/s 1.00 s
     
m
9
.
4
s
00
.
1
s
m
80
.
9
s
00
.
3
s
m
0
2
2
2
1
2
2
1







y
y
at
t
v
y o

52. Freely Falling Bodies
Example 12 How
High Does it Go?
The referee
tosses the coin up
with an initial
speed of 5.00m/s.
In the absence if
air resistance,
how high does the
coin go above its
point of release?

53. Freely Falling Bodies
y a v vi t
? -9.80 m/s2 0 m/s +5.00
m/s

54. y a v vi t
? -9.80 m/s2 0 m/s +5.00
m/s
ay
v
v i
f 2
2
2


a
v
v
y o
2
2
2


   
  m
28
.
1
s
m
80
.
9
2
s
m
00
.
5
s
m
0
2 2
2
2
2
2






a
v
v
y o

55. Freely Falling Bodies
What is the
time the
coin is in the
air before
returning to its
release point?

56. Freely Falling Bodies
Conceptual Example 14 Acceleration Versus
Velocity
 There are Three Parts To The Motion Of The Coin.
 On the way up, the coin has a vector velocity that
is directed upward and has decreasing
magnitude.
 At the top of its path, the coin momentarily has
zero velocity.
 On the way down, the coin has downward-pointing
velocity with an increasing magnitude.
 In the absence of air resistance, does
the acceleration of the coin, like the
velocity, change from one part to
another?

57. vᵢ= 5 m/s
The motion of an object that is
thrown upward and eventually returns to
earth has a symmetry that is useful to
keep in mind from the point of view of
problem solving.
The calculations just completed
indicate that a time symmetry exists
in free-fall motion, in the sense that the
time required for the object to reach
maximum height equals the time for it to
return to its starting point.
At any displacement y above
the point of release, the coin’s speed
during the upward trip EQUALS
the speed at the same point
during the downward trip.

58. Freely Falling Bodies
What is the
time the
coin is in the
air before
returning to its
release point?

60. Position-Time Graphs
• We can use a
postion-time graph to
illustrate the motion of
an object.
• Position is on the y-
axis
• Time is on the x-axis

61. Plotting a Distance-Time Graph
• Axis
– Distance (position) on
y-axis (vertical)
– Time on x-axis
(horizontal)
• Slope is the velocity
– Steeper slope = faster
– No slope (horizontal
line) = staying still

62. Where and When
• We can use a position
time graph to tell us
where an object is at any
moment in time.
• Where was the car at 4
s?
• 30 m
• How long did it take the
car to travel 20 m?
• 3.2 s

63. Interpret this graph…

64. Describing in Words

65. Describing in Words
• Describe the motion of
the object.
• When is the object
moving in the positive
direction?
• Negative direction.
• When is the object
stopped?
• When is the object
moving the fastest?
• The slowest?

66. Accelerated Motion
• In a position/displacement
time graph a straight line
denotes constant velocity.
• In a position/displacement
time graph a curved line
denotes changing velocity
(acceleration).
• The instantaneous velocity
is a line tangent to the
curve.

67. Accelerated Motion
• In a velocity time graph a
line with no slope means
constant velocity and no
acceleration.
• In a velocity time graph a
sloping line means a
changing velocity and the
object is accelerating.

68. Velocity
• Velocity changes when an object…
– Speeds Up
– Slows Down
– Change direction

69. Velocity-Time Graphs
• Velocity is placed on
the vertical or y-axis.
• Time is place on the
horizontal or x-axis.
• We can interpret the
motion of an object
using a velocity-time
graph.

70. Constant Velocity
• Objects with a
constant velocity have
no acceleration
• This is graphed as a
flat line on a velocity
time graph.

71. Changing Velocity
• Objects with a
changing velocity are
undergoing
acceleration.
• Acceleration is
represented on a
velocity time graph as
a sloped line.

72. Positive and Negative Velocity
• The first set of
graphs show an
object traveling
in a positive
direction.
• The second set
of graphs show
an object
traveling in a
negative
direction.

73. Speeding Up and Slowing Down
•The graphs on the left represent an object
speeding up.
•The graphs on the right represent an object
that is slowing down.

74. Two Stage Rocket
• Between which
time does the
rocket have the
greatest
acceleration?
• At which point
does the velocity
of the rocket
change.

75. Displacement from a Velocity-Time
Graph
• The shaded region under
a velocity time graph
represents the
displacement of the
object.
• The method used to find
the area under a line on a
velocity-time graph
depends on whether the
section bounded by the
line and the axes is a
rectangle, a triangle

76. s
m
4
s
2
m
8
Slope 






t
x

78. 2
s
m
6
s
2
s
m
12
Slope 






t
v