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# Graphical analysis of motion

## on Aug 15, 2011

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lecture

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• An object at zero velocity is at rest. The object will not change position.
• Positive velocity indicates direction towards the RIGHT or EAST
• Straight line upward indicates positive constantvelocity
• Negative velocity indicates an object moving toward the LEFT or WEST
• A straight downward slope indicates constant negative velocity to which an object moves constantly leftward
• Positive velocity indicates object moves towards RIGHT, Positive acceleration indicates object speeds up
• Increasing positive slope shows acceleration.. A change in velocity speeding up

## Graphical analysis of motionPresentation Transcript

• Tabulating Time, Displacement and Velocity at a given Acceleration
• A car starting from rest, accelerating at 2m/s2 to the right? Describe its position every second from 0-5 sec.
Since the starts at rest. vo= 0m/s, deriving vf=at
Using d=vavet
t=0, v=0, d=0 at 1
At 1 sec
vf=2m/s2(1s) = 2m/s
d=2m/s(1s)=2m
• Graphical Analysis of Motion
• Position –Time Graph
• Case 1: Zero Velocity
50
1
2
3
4
5
40
30
Position
20
10
0
Time
Object at 30 meters with zero velocity
Object at rest and in the same position for a long time
• 30
55
10
15

20
25
35
40

45
50
t=1
t=3
t=4
t=5
t=2
Case 2: Positive, Constant velocity
Consider a car moving with a constant velocity of 10 m/s to the right
• 50
1
2
3
4
5
40
30
Position
20
10
0
Time
Straight line sloping upward
Case 2: Positive, Constant velocity
• 30
55
10
15

20
25
35
40

45
50
t=4
t=1
t=2
t=3
t=5
t=0
60
Case 3: Negative, Constant velocity
Consider a car moving with a constant velocity of 10 m/s to the left
• 50
1
2
3
4
5
40
30
Position
20
10
0
Time
Straight line sloping downward
Case 3: Negative, Constant velocity
• t=4
t=1
t=0
t=3
t=2
40
2.5
0
10

22.50
Case 4: Positive Velocity, Positive Acceleration
Consider a car starts from rest and accelerates at 5m/s2
• 50
1
2
3
4
5
40
30
Position
20
10
0
Time
Case 4: Positive Velocity, Positive Acceleration
Increasing, positive slope
The car is accelerating or speeding up to the right
• t=0
t=2
t=1
t=3
t=4
t=5
0
40
62.25

22.50
60
52.25
Case 5: Positive Velocity, Negative Acceleration
Consider a car moving at 25m/s and accelerates at -5m/s2
• 1
2
3
4
5
50
40
Position
30
20
10
60
0
Time
Case 5: Positive Velocity, Negative Acceleration
Decreasing Positive slope
The car is decelerating or slowing down to the right
• t=5
t=1
t=2
t=3
t=4
t=0
0
40
62.25

22.50
60
52.25
Case 6: Negative Velocity, Negative Acceleration
Consider a car at position 62.25m starting from rest and moving leftward with an acceleration of -5m/s2
• Increasing Negative slope
Accelerating or speeding up to the left
1
2
3
4
5
50
40
Position
30
20
60
10
0
Time
Case 6: Negative Velocity, Negative Acceleration
• t=5
t=2
t=3
t=4
t=1
t=0
62.5
40
2.5
0
10

22.50
Case 7: Negative Velocity, Positive Acceleration
Consider a car at position 62.25m starting at -25m/s but accelerating at +5m/s2
• 1
2
3
4
5
50
40
Position
30
20
60
10
0
Time
Case 7: Negative Velocity, Positive Acceleration
Decreasing, negative slope
The car is decelerating or slowing down to the left
• Velocity – Time Graph
• Case 1: Zero Velocity
Object at rest in same position
V+
0
V-
• V+
0
V-
Case 2: Positive Constant Velocity
Object is moving at constant velocity towards the right
• V+
0
V-
Case 3: Negative constant Velocity
Object is at constant negative velocity indicating motions towards the LEFT
• V+
0
V-
Case 4: Positive velocity, Positive acceleration
Indicates the object speeds up towards the right
• V+
0
V-
Case 5: Positive velocity, Negative acceleration
Indicates the object slows down towards the right
• V+
0
V-
Case 6: Negative velocity, Negative acceleration
Indicates the object speeds up towards the left
• V+
0
V-
Case 4: Negative velocity, Positive acceleration
Indicates the object slows down towards the left
• Fill in the blanks by describing position time graphs and velocity time graphs.
In a v-t graph an object that moves with a positive velocity and a negative acceleration; the object slows down to the ________________.

In a v-t graph an object that has a constant negative velocity is represented by a straight line parallel but _______ the x-axis.

In a p-t graph an object that is moving at a negative acceleration, negative velocity has an increasing negative slope which __________towards the left.

• A car that moves to the right and speeding up both has ________ velocity and acceleration. An object that is in zero velocity is at 20 meter; its position is constant and is represented by a line which is ____________ to the x-axis.
• Case 4: sloping upward above x
Case 5: sloping down above x
Case 6: sloping down below x-axis
Case7: sloping up below x-axis
• One dimensional vertical motion
Free fall: a state where an object which moves under the sole influence of GRAVITY. (acceleration due to gravity)
• Two important characteristics of a free falling object:
Gravity = 9.8m/s2
Downward direction
• Mass and Acceleration
Acceleration is directly proportional to the force and inversely proportional to the mass.
• When will an object fall faster?
An object will fall faster if there is an appreciable amount of air resistance.
Air resistance is the friction present in the air. The larger the velocity of an object the larger is the air resistance.
• Quantities involved in free fall
vo= velocity initialvf= final velocityg= gravity (9.8 m/s2)t=timey= displacement
• Equations of one-dimensional vertical motion
vf=v0 + gty= v0t + 1/2 gt2vf2=vo2 + 2gy
• Case 1 : Dropped (vo=0)Objects that are dropped has an initial velocity of ZERO.Example:A ball is dropped from a height of 50m Compute for the time of flight to reach the ground.
50 meters
• From y= v0t + 1/2 gt2 and vo=0
t= 2𝑦𝑔
Same as velocity before impact from vf2=vo2 + 2gy
vf=2𝑔𝑦

• Solving for time and velocity of an object dropped at 50 meters above the ground.
t= 2𝑦𝑔 t= 2(50𝑚)9.8𝑚/𝑠2 = 3.2 s
vf=2𝑔𝑦=29.8𝑚𝑠250𝑚
= 31.3 m/s

• Creating the time-velocity-displacement table of the motion, from 0s to 3.2 sec intervals of 0.8 s we use the following formulas :
vf=v0+ gty= v0t + 1/2 gt2
50 meters