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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

113 phy fall lecture phy 113 lecture fall 2012 2013 velocity lamb problem time mass textbook displacement constant relationships

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Slide1

8/29/2013

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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PHY 113 C Fall 2012 -- Lecture 2

2

Some updates/announcementsSlide3

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PHY 113 C Fall 2012 -- Lecture 2

3

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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PHY 113 C Fall 2012 -- Lecture 2

4

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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5

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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6

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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7

Motion in one-dimension Slide8

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8

Graphical representation of position (displacement)

x(t)

t (s)Slide9

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9

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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10

Graphical representation of position (displacement):

x(t)

 time rate of change of displacement = velocity:

v(t)

t (s)Slide11

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PHY 113 C Fall 2012 -- Lecture 2

11

Instantaneous velocity

t (s)

v

A

v

B

v

C

v

E

v

F

v

DSlide12

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12

Demonstration of tangent line limitSlide13

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13

Average velocity versus instantaneous velocitySlide14

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14

Average velocity

i

clicker

exercise:

This results is:

Exact

ApproximateSlide15

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PHY 113 C Fall 2012 -- Lecture 2

15

Webassign

ExampleSlide16

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16

Instantaneous velocity using calculus

x

vSlide17

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17

Anti-derivative relationship

t

xSlide18

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18

VelocitySlide19

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19

AccelerationSlide20

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20

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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21

Anti-derivative relationshipSlide22

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22

Examples

v(t)

x(t)Slide23

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23

Examples

v(t)

x(t)Slide24

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24

SummarySlide25

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25

x(t)

v(t)

Another exampleSlide26

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26

From

webassign

:Slide27

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27

Another example

(m/s)

(s)Slide28

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28

Another

example -- continued

(m/s)

(s)Slide29

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29

Special relationships between

t,x,v,a

for constant

a:Slide30

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30

Special relationships between

t,x,v,a

for constant

a:Slide31

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PHY 113 C Fall 2012 -- Lecture 2

31

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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32

Helpful mathematical

relationships

(see Appendix B of your text)