Physics 3310 Chapter 2 Notes Mr. Kim 1 Chapter 2 Notes By: Mr. Kim Physics 3310-xx • Definitions: • Kinematics – Study of Motion • Kinetic Energy - Energy associated with motion • • Motion in physics is broken down into 3 categories • 1.) Translational Motion - motion such that an object moves from one position to another along a straight line. • 2.) Rotational Motion - motion such that an object moves from one position to another along a circular path. • 3.) Vibrational Motion - motion such that an object moves back and forth in some type of periodicity. Kinematics in 1-D. •Chapter 2. Describing Motion:
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Physics 3310 Chapter 2 Notes
Mr. Kim 1
Chapter 2 NotesBy: Mr. Kim
Physics 3310-xx
• Definitions:
• Kinematics – Study of Motion• Kinetic Energy - Energy associated with motion•
• Motion in physics is broken down into 3 categories• 1.) Translational Motion - motion such that an object
moves from one position to another along a straight line.• 2.) Rotational Motion - motion such that an object moves
from one position to another along a circular path.• 3.) Vibrational Motion - motion such that an object moves
back and forth in some type of periodicity.
Kinematics in 1-D.
•Chapter 2. Describing Motion:
Physics 3310 Chapter 2 Notes
Mr. Kim 2
Example: Diatomic Molecule Moving Through Space.•Chapter 2. Describing Motion:
Kinematics in 1-D.
•Chapter 2. Describing Motion:
•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.
Physics 3310 Chapter 2 Notes
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Displacement•Chapter 2. Describing Motion:
Displacement: Change in Position.•Chapter 2. Describing Motion:
position initial ix
position final fx
ntdisplaceme if xxx
Physics 3310 Chapter 2 Notes
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Displacement: Change in Position.•Chapter 2. Describing Motion:
m 0.3ix
m 0.9fx
m 0.6x
m 0.6m 3.0m 9.0 if xxx
• Example 1:
Displacement: Change in Position.•Chapter 2. Describing Motion:
m 0.4fx
m 0.12ix
m 0.8x
m 0.8m .021m 4.0 if xxx
• Example 2:
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Displacement: Change in Position.•Chapter 2. Describing Motion:
m 0.4fx
m 0.12x
m 0.12m .08m 4.0 if xxx
m 0.8ix
• Example 3:
Displacement: Change in Position.•Chapter 2. Describing Motion:
m 0.3ix
m 0.11x
m 0.11m .0)3(m 8.0 if xxx
m 0.8fx
• Example 4:
Physics 3310 Chapter 2 Notes
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Speed and Velocity•Chapter 2. Describing Motion:
•Average speed is the total distance traveled divided by the time required to cover the distance.
timeElapsed
Distance speed Average
SI units for speed: meters per second (m/s)
Speed and Velocity•Chapter 2. Describing Motion:
• 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?
timeElapsed
Distance speed Average
m 12000s 5400sm 22.2
timeElapsedspeed Average Distance
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Speed and Velocity•Chapter 2. Describing Motion:
•Average velocity is the displacement divided by the elapsed time.
timeElapsed
ntDisplaceme velocityAverage
ttt if
if
xxx
v
Speed and Velocity•Chapter 2. Describing Motion:
• 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.
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Speed and Velocity•Chapter 2. Describing Motion:
• Example 2 The World’s Fastest Jet-Engine Car
sm5.339s 4.740
m 1609
t
xv
sm7.342s 4.695
m 1609
t
xv
Speed and Velocity•Chapter 2. Describing Motion:
• The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.
tt
xv
0lim
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Acceleration•Chapter 2. Describing Motion:
• The notion of acceleration emerges when a change in velocity is combined with the time during which the change occurs.
Acceleration: Rate of Change in Velocity•Chapter 2. Describing Motion:
• Average acceleration is change in velocity divided by the time during which the change in velocity occurs.
• Determine the average acceleration of a dragster that decreases its velocity of 28 m/s when time is 9.0 seconds to 13.0 m/s when the time is 12.0 seconds.
sm0.28iv
sm0.12fv
s 0.9it s 0.12ft
Given:
2sm0.5s 9s 12
sm28sm13
o
o
tt
vva
Acceleration•Chapter 2. Describing Motion:
• Notice that the acceleration is a negative value.
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POSITIVE ACCELERATION
• Speeding up in a positive direction
• Slowing down in a negative direction
• Speeding up in a negative direction
• Slowing down in a positive direction
NEGATIVE ACCELERATION
Comparison of Positive & Negative Acceleration
•It is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement, velocity, and acceleration vectors. However, we will continue to conveythe directions with a plus or minus sign.
i
i
ttf
f
xx
v
i
i
tt
xxv
f
f
i
i
ttf
f
vv
a
i
i
tt
vva
f
f
o
o
tt
vva
o
o
tt
xxv
Physics 3310 Chapter 2 Notes
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•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
4 Kinematic Equations for Constant Acceleration•Kinematic Equations
• Equation #4• The average of initial and final velocity times by time is displacement.
o
o
tt
xxv
If we let the object be at the origin when the clock starts at zero seconds…
0ox
0ot
BUT!!!
t
xv tvx
vvv o 21 tvvx o 2
1
WHERE
THUS
THEN
Physics 3310 Chapter 2 Notes
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4 Kinematic Equations for Constant Acceleration•Kinematic Equations
• Equation #1• The average acceleration is change in velocity divided by time.
We let the object be at the origin when the clock starts at zero seconds…
0ox
0ot
BUT!!!
Becomes
THUS
THEN
o
o
tt
vva
t
vva o
ovvat atvv o
4 Kinematic Equations for Constant Acceleration•Kinematic Equations
• Equation #2• Combine Equation #1 and Equation #4.
We let the object be at the origin when the clock starts at zero seconds…
BUT!!!
Becomes
THUS
ANDatvv o tvvx o 21
tatvvx oo 21
221 2 attvx o 2
21 attvx o
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4 Kinematic Equations for Constant Acceleration•Kinematic Equations
• Equation #3• Combine Equation #1 and Equation #4.
We let the object be at the origin when the clock starts at zero seconds…
Eqn. 1
ANDatvv o tvvx o 21
a
vvt o Plug into Eqn. 4
a
vvvvx o
o
2
1
4 Kinematic Equations for Constant Acceleration•Kinematic Equations
• Equation #3 Continued• Combine Equation #1 and Equation #4.
We let the object be at the origin when the clock starts at zero seconds…
FOIL it!
THUS
a
vvvvx o
o
2
1
a
vvx o
2
22
222 ovvax xavv o 222
Physics 3310 Chapter 2 Notes
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4 Kinematic Equations for Constant Acceleration•Kinematic Equations
• 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?
• 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. If the distance of the fall is small compared to the radius of the Earth, then the accelerationremains essentially constant throughout the descent.
• This idealized motion is called free-fall and the acceleration of a freely falling body is called the acceleration due to gravity.
22 sft2.32or sm80.9 g
4 Kinematic Equations for Constant Vertical Acceleration•Kinematic Equations
• Equation #1
ygvv o 222gtvv o
• Equation #2
221 gttvy o
• Equation #3
• Equation #4
tvvy o 21
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2sm80.9g• Why negative??• If we consider UP as positive direction (+) then DOWN is the negative direction (-).
• Since gravity pulls everything downwards, acceleration due to gravity is negative.
Acceleration Due to Gravity•Chapter 2. Describing Motion:
• Example 8 Freely Falling Body
• A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement, y, of the stone?
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Vertical Acceleration Due to Gravity•Chapter 2. Describing Motion:
• Example 7 Acceleration and Velocity
sm0.0iv
NGf v
s 3.0t 2/8.9 smg my ???
m 1.44
s 00.3sm80.9s 00.3sm0 2221
221
gttvy o
Acceleration Due to Gravity•Chapter 2. Describing Motion:
• Example 9• 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?
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Vertical Acceleration Due to Gravity•Chapter 2. Describing Motion:
• Example 9 How High Does It Go???
sm0.5iv
smf /0.0v
Kt D 2/8.9 smg my ???
gyvv o 222 g
vvy o
2
22
m 28.1
sm80.92
sm00.5sm0
2 2
2222
g
vvy o
Vertical Acceleration Due to Gravity•Chapter 2. Describing Motion:
• Example 9 How High Does It Go???• 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?
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Vertical Acceleration Due to Gravity•Chapter 2. Describing Motion:
• Taking Advantage of Symmetry
• Does the ball in part a strike the ground beneath the cliff with a smaller, greater, or the same speed as the ball in part b?
Position vs. Time Graph•Graphical Analysis of Motion
• Constant Velocity
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Position vs. Time Graph•Graphical Analysis of Motion
• Constant Velocity
s 2
m 8 Slope
t
x
sm4Slope time-position v
sm4s 2
m 8 Slope
run Slope
rise
t
x
Graphical Analysis of Motion•Chapter 2. Describing Motion:
• Position vs. Time Graph
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Position vs. Time Graph•Graphical Analysis of Motion
• Constant Acceleration
Velocity vs. Time Graph•Graphical Analysis of Motion
• Constant Acceleration
Physics 3310 Chapter 2 Notes
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Velocity vs. Time Graph•Graphical Analysis of Motion
s 2
m/s 12 Slope
t
v
2sm6Slope time-velocity a
2sm6s 2
m/s 12 Slope
run Slope
rise
t
v
END of Chapter 2 Notes
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