phys 100: lecture 1 - university of illinois · physics 100 lecture 1, slide 7 embedded questions:...

Physics 100 Lecture 1, Slide Physics 100 Lecture 1, Slide 11

PHYS 100: Lecture 1

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (seconds)

Dis

pla

cem

ent

(m)

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (seconds)

Vel

oci

ty (

m/s

)

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (seconds)

Acc

eler

atio

n (

m/s

^2)

displacement

velocity acceleration

Kinematic Definitions


WELCOMECourse Website (source of most everything!)

http://online.physics.uiuc.edu/courses/phys100

• THE INSTRUCTION CYCLE:• Prelecture (web)• Preflight (web)• Lecture• Homework (web)• Discussion• Quiz (web)

SOMETHING NEW THIS SEMESTER: 2 CREDIT HOURS• All students currently registered for two hours credit

MIDTERM MAR 15

STUDENTS WHO DO WELL ON EXAMmay choose to leave course with one hour credit

OTHERSRemain in course for two hours credit


Q: What are the benefits of participating ?Q: What are the benefits of participating ?A: You learn moreA: You learn more

53%Blew through pre-lectures

62%Viewed pre-lectures

Final Exam averageStudents who…

Prelecture Data from PHYS 100 Fall 2008

PHYS 100 FINAL EXAM Fall08

0

1

2

3

4

5

6

7

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95

Score

Viewers

Non-Viewers


Clickers

• Research shows you learn more when you talk!

• 5 points for each lecture

• Maximum of 25 points (one hour option) (5/7 lectures) 55 points (two hour option) (11/13 lectures)

– (No EX’s but you can miss/forget your clicker for two lectures)

IMPORTANT:IMPORTANT:The class next door also is using i>clickersThe class next door also is using i>clickersWe will need to run at a different frequencyWe will need to run at a different frequency

1. Power On

2. Hold ON/OFF until POWER light flashes

3. Press BB


Music

Who is the Artist?Who is the Artist?

A)A) Eric ClaptonEric ClaptonB)B) Bill Bill FrisellFrisellC)C) Jimmy PageJimmy PageD)D) Jeff BeckJeff BeckE)E) Buddy GuyBuddy Guy

Why?Why?

A great way to start the semester!A great way to start the semester!

My vote for most memorable performance at My vote for most memorable performance at New Orleans New Orleans JazzfestJazzfest by a big name!by a big name!

Somewhere over the RainbowSomewhere over the RainbowPeople Get ReadyPeople Get ReadyAmazingAmazing……..

BB


Main Points• We use the Displacement vector to define the location of an object at a given time.

• The subsequent motion of the object is defined in terms of its Velocity and Acceleration.

dva

dt=

rrdx

vdt

=r

r

• Given the Displacement vector as a function of time, we can differentiate once to get the Velocity and twice to get the Acceleration.

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (seconds)

Dis

pla

cem

ent

(m)

displacement

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (seconds)

Vel

oci

ty (

m/s

)

velocity

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (seconds)

Acc

eler

atio

n (

m/s

^2)

acceleration


Embedded Questions: DisplacementEmbedded Questions: Displacement

(A)x1

x2

x1x2

(B)

x1

x2(C)

x1x2

(D)

The vector shown represents the change in displacement for an object between time t1 and t2.

Which of the possible pairs of displacement vectors (x(t1)= x1 and x(t2) = x2) could produce the above change in displacement vector?

∆x

121221 )()(),( xxtxtxttx rrrrr −=−=∆

-x1

∆x

-x1 ∆x

x2A)

x2B)

BB


Embedded Questions: DisplacementEmbedded Questions: Displacement

(A)x1

x2

x1x2

(B)

x1

x2(C)

x1x2

(D)

One more chance, if needed:

?

The vector shown represents the change in displacement for an object between time t1 and t2.

Which of the possible pairs of displacement vectors (x(t1)= x1 and x(t2) = x2) could produce the above change in displacement vector?

∆x

-x1∆x

x2D)

BB


Preflight 1Preflight 1

Instantaneous velocity (1) < Average velocity (0 Ø2) (A)

Instantaneous velocity (1) = Average velocity (0 Ø2) (B)

Instantaneous velocity (1) > Average velocity (0 Ø2) (C)

0

10

20

30

40

50

A B C

You said:

• The slope of the average velocity from t=0 to t=2 is much larger than the slope of the instantaneous velocity at t=1. Therefore, the instantaneous velocity at t=1 is less than the average velocity from t=0 to t=2.

• When t = 1, the slope, which would give us the velocity, of the displacement graph is very close to the average velocity because the slope seems greater than it is at t = 1 before t=1 and greater after, so it seems reasonable to see the t=1 as the point that has the same instantaneous velocity as the average velocity of the graph

• When you use the average velocity formula u can see that you need to divide the change of x by two which would be less than the instantaneous velocity at t=1.

BB



Instantaneous velocity (1) < Average velocity (0 Ø2) (A)

Instantaneous velocity (1) = Average velocity (0 Ø2) (B)

Instantaneous velocity (1) > Average velocity (0 Ø2) (C)

Instantaneous Velocitydefined at any single point

Average Velocitydefined between two points

)()( tdtxd

tvr

r =

Geometric: slope of x vs t curve

12

12

21

2121 ),(

),(),(

ttxx

tttttx

ttvavg −−=

∆∆=

rrrr

∆x∆t

Geometric: slope of line drawn between

the two endpoints


Preflight 3Preflight 3TRUE (A)

FALSE(B)An object experiences a positive acceleration, thus its speed must always increase.

0

20

40

60

80

A B

You said:

• Acceleration is the derivative of velocity (speed). Knowing that the speed is increasing, the velocity is positive causing the derivative to be positive as well.

• If the acceleration is positive, but the velocity is negative, the object could have a positive acceleration and be slowing down.

BB


Preflight 3Preflight 3TRUE (A)

FALSE(B)An object experiences a positive acceleration, thus its speed must always increase.

Is this statement true?

• A positive acceleration means a positive change in velocity.

This statent is TRUE!BUT speed ∫ velocity

speed = magnitude (velocity)

dtvd

ar

r =

v

t0

t1 t2

v1

v2

• acceleration positive• velocity is increasing ( v2 > v1 )• speed is decreasing ( |v2| < |v1|)

Example


• When is particle speeding up? speedup

speedup

• When is acceleration negative?

a < 0 a < 0

FollowFollow--UpUp

(A)

At what times is the acceleration negative, but the particle is speeding up?

only 10 << t only 21 << t(B)

20 << t(C) 32 << t(D) 43 << t(E)

BB



Which graph shows an object in 1-D motion moves in the positive direction and is slowing down?

0

20

40

60

80

A B C D E

You said:

• the object is slowing down, thus the slope of the tangent to the velocity curve must approach zero.In addition, it is moving in the positive direction, so it can only be a)

• The x values (in graph b) are positive so its moving in the positive direction and the x values are decreasing overtime so its slowing down.

BB


Preflight 5Preflight 5An object in 1-D motion moves in the positive direction and is slowing down.

• D and E can be eliminated• straight line fl constant velocity

X XFor graph C:(A) positive direction & speeding up(B) positive direction & slowing down(C) negative direction & speeding up(D) negative direction & slowing down

For graph B:(A) positive direction & speeding up(B) positive direction & slowing down(C) negative direction & speeding up(D) negative direction & slowing down

Slope is positivev points in positive directionx increases in time

“moves in + direction”

Magnitude of slope increases “speeds up”

Slope is negative “moves in negative direction”

Magnitude of slope increases “speeds up”

XX



Acceleration of particle is negative and it is speeding up (A)

Acceleration of particle is positive and it is speeding up (B)

Acceleration of particle is negative and it is slowing down (C)

Acceleration of particle is positive and it is slowing down (D)

0

20

40

60

80

A B C D

You said:

• The velocity magnitude is negative so its always negative however the velocity is increasing overtime so its speeding up.

•The object is speeding up therefore it must have a positive acceleration.

•Even though the velocity is negative, it is becoming less negative over time which means the acceleration is positive. It is always slowing down because the velocity does not change as rapidly in the end as it does in the beginning.

BB


Preflight 7 ModificationPreflight 7 Modification

Acceleration of particle is negative and it is speeding up (A)

Acceleration of particle is positive and it is speeding up (B)

Acceleration of particle is negative and it is slowing down (C)

Acceleration of particle is positive and it is slowing down (D)

Consider these possible student explanations

• Because the slope of this velocity is always positive, the object is always experiencing positive acceleration. However, because the slope is gradually getting less and less steep, this also implies that the object is slowing down as well.

• The slope is positive at all points t, so the accelartion of the particle is always positive. The magnitude decreases, so the particle slows down.

Suppose we change the plot by moving the t-axis downWILL THE ANSWER CHANGE?

t

Acceleration decreasing? YES

Slowing down? NOTNECESSARILY

BB

Slowing down or speeding up is determined by the change in the MAGNITUDE of the velocity


Rope Around the WorldRope Around the WorldThe Earth has an equitorial radius of 3963 miles. There are 5280 feet in one mile. Imagine a rope wraped around the equator of a perfectly smooth Earth. Suppose we now add 15 feet to the rope and shape this longer rope into a smoooth circle with its center at the center of the Earth.

How far will the rope now stand away from the surface of the Earth? For now just use your intuition.. Don’t calculate anything, we will do that next.

< 1mm (A) (B) ¼ inch 15 ft (E)2 ft (D)2 inches (C)

BB



How do we start this calculation? Which of the following options would you do first?

Use your calculator to convert radius of Earth to feet. (A)

(B) Use your calculator to determine the initial length of rope in miles

Write down a mathematical expression that relates the radius of the Earth to the original length of rope

(D)

Use your calculator to convert 15 feet into miles (C)

WHY?Always THINK first

Numbers can be substituted in at the end (IF NEEDED)

BB


Rope Around the WorldRope Around the WorldThe Earth has an equatorial radius of 3963 miles. There are 5280 feet in one mile. Imagine a rope wraped around the equator of a perfectly smooth Earth. Suppose we now add 15 feet to the rope and shape this longer rope into a smoooth circle with its center at the center of the Earth.

Which of the following equations relates RE, the radius of the Earth, to L, the initial length of rope?

L is circumference of circle of radius RE:

(A) none of these(D)2L

RE = (B)πL

RE = (C)π2L

RE =

ERL π2=

RE

L

BB



RE

L

Which of the following equations relates ∆L (the extra 15 feet) to ∆R (the amount the new rope stands above the surface of the Earth)?

(A) none of these(D)LLR

R E ∆=∆ (B) (C)LRL

RE

∆=∆π2L

R∆=∆

RE

L+∆L

R

∆R = R - RE

LLRRE ∆+=∆+ )(2π

LR ∆=∆π2

ftftL

R 4.22

152

==∆=∆ππ

Look Ma, No Calculator !!

π2L

RE =BB


Rope Around the WorldRope Around the World

< 1mm (A)

The Earth has an equitorial radius of 3963 miles. There are 5280 feet in one mile. Imagine a rope wraped around the equator of a perfectly smooth Earth. Suppose we now add 15 feet to the rope and shape this longer rope into a smoooth circle with its center at the center of the Earth.

How far will the rope now stand away from the surface of the Earth?

(B) ¼ inch 15 ft (E)2 ft (D)2 inches (C)

NOTE: Example of relevance of calculus:

RL π2=π21=

dLdR

LR ∆=∆π21

ftft

4.22

15 ==π


NotesNotes

• Web Homework DUE TUESDAY 8am

• Office Hour on Monday 11am in 271 Loomis

• Online Quiz DUE THURSDAY 8pm

• Prelecture 2 and Preflight 2 DUE FRIDAY 8am

phys 100: lecture 1 - university of illinois · physics 100 lecture 1, slide 7 embedded questions:...

Documents