PHY 113 C Fall 2012 Lecture 2 1 PHY 113 C General Physics I 11 AM 1215 PM TR Olin 101 Plan for Lecture 2 Chapter 2 Motion in one dimension Time and Position Time and Velocity Time and Acceleration ID: 379806
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Slide1
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PHY 113 C Fall 2012 -- Lecture 2
1
PHY 113 C General Physics I
11 AM – 12:15 PM TR Olin 101
Plan for Lecture 2:
Chapter 2 – Motion in one dimension
Time and Position
Time and Velocity
Time and Acceleration
Special relationships for constant accelerationSlide2
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Some updates/announcementsSlide3
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iclicker
exercises:
Webassign
Experiences so far
Have not tried it
Cannot login
Can loginHave logged in and have completed one or more example problems.Textbook Experiences
Have no textbook (yet)
Have complete physical textbook
Have electronic version of textbook
Textbook is on order
OtherSlide4
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Comment on
Webassign
#1 problem:
7. A
pet lamb grows rapidly, with its mass proportional to the cube of its length. When the lamb's length changes by 16%, its mass increases by 16.8 kg. Find the lamb's mass at the end of this process.
Problem solving steps
Visualize
problem – labeling variables
Determine which basic physical principle(s) apply
Write down the appropriate equations using the variables defined in step 1
.
Check
whether you have the correct amount of information to solve the problem (same number of
knowns
and unknowns).
Solve the equations.
Check whether your answer makes sense (units, order of magnitude, etc
.).Slide5
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7. A
pet lamb grows rapidly, with its mass proportional to the cube of its length. When the lamb's length changes by 16%, its mass increases by 16.8 kg. Find the lamb's mass at the end of this process.
LSlide6
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i>clicker exercise:
What did you learn from this problem?
It is a bad idea to have a pet lamb
This was a very hard question
I hope this problem will not be on an exam
My physics instructor is VERY mean
All of the above.Slide7
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Motion in one-dimension Slide8
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Graphical representation of position (displacement)
x(t)
t (s)Slide9
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Comment:
Your text mentions the notion of a “scalar quantity” in contrast to a “vector quantity” which will be introduced in Chapter 3. In most contexts, a scalar quantity – like one-dimensional distance or displacement can be positive or negative.
Another comment:
Your text describes the time rate of change of displacement as “velocity” which, in one-dimension is a signed scalar quantity. In general “speed” is the magnitude of velocity – a positive scalar quantity.Slide10
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Graphical representation of position (displacement):
x(t)
time rate of change of displacement = velocity:
v(t)
t (s)Slide11
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Instantaneous velocity
t (s)
v
A
v
B
v
C
v
E
v
F
v
DSlide12
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Demonstration of tangent line limitSlide13
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Average velocity versus instantaneous velocitySlide14
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Average velocity
i
clicker
exercise:
This results is:
Exact
ApproximateSlide15
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Webassign
ExampleSlide16
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Instantaneous velocity using calculus
x
vSlide17
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Anti-derivative relationship
t
xSlide18
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VelocitySlide19
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AccelerationSlide20
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Rate of acceleration
iclicker
exercise
How many derivatives of position are useful for describing motion:
1
(dx/
dt
)
2
(
d
2
x/dt
2
)
3
(
dx
3
/dt
3
)
4
(
dx
4
/dt
4
)
Slide21
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Anti-derivative relationshipSlide22
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Examples
v(t)
x(t)Slide23
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Examples
v(t)
x(t)Slide24
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SummarySlide25
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x(t)
v(t)
Another exampleSlide26
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From
webassign
:Slide27
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Another example
(m/s)
(s)Slide28
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Another
example -- continued
(m/s)
(s)Slide29
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Special relationships between
t,x,v,a
for constant
a:Slide30
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Special relationships between
t,x,v,a
for constant
a:Slide31
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Special case: constant velocity due to earth’s gravity
In this case, the “one” dimension is the vertical direction with “up” chosen as positive:
a = -g = -9.8 m/s
2
y(t) = y
0
+ v0t – ½ gt2
y
0
= 0 v
0
= 20 m/s
At what time t
m
will the ball hit the ground
y
m
= -50m ?
Solve:
y
m
= y
0
+ v
0
t
m
– ½ gt
m
2
quadratic equation
physical solution: t
m
= 5.83 sSlide32
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Helpful mathematical
relationships
(see Appendix B of your text)