Physics 207: Lecture 3, Pg 1 Physics 207, Lecture 3, Sept. 10 Assignment: 1.For Monday: Read Chapter 4 2.Homework Set 2 (due Wednesday 9/17) l Goals (finish

Physics 207: Lecture 3, Pg 1

Physics 207, Physics 207, Lecture 3, Sept. 10Lecture 3, Sept. 10

Assignment: Assignment:

1.1. For Monday: Read Chapter 4For Monday: Read Chapter 4

2.2. Homework Set 2 Homework Set 2 (due Wednesday 9/17)(due Wednesday 9/17)

Goals (finish Chap. 2 & 3) Goals (finish Chap. 2 & 3) Understand the relationships between position, Understand the relationships between position,

velocity & acceleration in systems with 1-dimensional velocity & acceleration in systems with 1-dimensional motion and non-zero acceleration (usually constant)motion and non-zero acceleration (usually constant)

Solve problems with zero and constant acceleration Solve problems with zero and constant acceleration (including free-fall and motion on an incline)(including free-fall and motion on an incline)

Use Cartesian and polar coordinate systemsUse Cartesian and polar coordinate systems Perform vector algebraPerform vector algebra


Position, velocity & acceleration for motion Position, velocity & acceleration for motion along a linealong a line

If the position x is known as a function of time, then we can find both the instantaneous velocity vx and instantaneous acceleration ax as a function of time!

vx

t

x

t

ax

t

2

2va

dtxd

dtd x

x

dtdx

x v

] offunction a is [ )( txtxx


Position, displacement, velocity & accelerationPosition, displacement, velocity & acceleration All are vectors and so vector algebra is a must !

These cannot be used interchangeably (different units!)

(e.g., position vectors cannot be added directly to velocity vectors)

But we can determined directions “Change in the position” vector gives the direction of

the velocity vector “Change in the velocity” vector gives the direction of

the acceleration vector

Given x(t) vx(t) ax (t) Given ax (t) vx (t) x(t)

v

a


And given a And given a constant accelerationconstant acceleration we we can integrate to get explicit can integrate to get explicit vvxx and and aaxx

x

ax

vx

t

t

t

txxx avv0

221

0 a v0

ttxx xx

consta x

2

2va

dtxd

dtd x

x

dt

dxx v

] offunction a is [ )( txtxx

0

x0


A “biology” experiment Hypothesis: Older people have slower reaction

times Distance accentuates the impact of time differences Equipment: Ruler and four volunteers

Older student Younger student Record keeper Statistician Expt. require multiple trials to reduce statistical

errors.

20 )( Exploiting txx


Rearranging terms gives two other relationshipsRearranging terms gives two other relationships

For constant acceleration:

From which we can show (caveat: a constant acceleration)

)v(v21

v

)x(x2av v

xx(avg)x

0x2

x2x

0

0

txxx avv0

221

0 a v0

ttxx xx

consta x

ax

t


AccelerationAcceleration Changes in a particle’s motion often involve acceleration

The magnitude of the velocity vector may change The direction of the velocity vector may change

(true even if the magnitude remains constant) Both may change simultaneously

vv11vv00 vv33 vv55vv22 vv44

aa aa aa aa aa aa

vat

t0

v(t)=v0 + a t

a t = area under curve = v

t


Acceleration has its limitsAcceleration has its limits

“High speed motion picture camera frame: John Stapp is caught in the teeth of a massive deceleration. One might have expected that a test pilot or an astronaut candidate would be riding the sled; instead there was Stapp, a mild mannered physician and diligent physicist with a notable sense of humor. Source: US Air Force photo


Free FallFree Fall

When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity.

This acceleration on the Earth’s surface, caused by gravity, is typically written as “little” g

Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant.


When throwing a ball straight up, which of the When throwing a ball straight up, which of the following is true about its velocity following is true about its velocity vv and its and its acceleration acceleration aa at the highest point in its path? at the highest point in its path?

A.A. BothBoth v = 0v = 0 andand a = 0a = 0

B.B. v v 0 0, but , but a = 0a = 0

C.C. v = 0v = 0, but , but a a 0 0

D.D. None of the aboveNone of the above

y

Exercise 1Exercise 1Motion in One DimensionMotion in One Dimension


When throwing a ball straight up, which of the following is When throwing a ball straight up, which of the following is true about its velocity true about its velocity vv and its acceleration and its acceleration aa at the highest at the highest point in its path?point in its path?

A.A. BothBoth v = 0v = 0 andand a = 0a = 0

B.B. v v 0 0, but , but a = 0a = 0

C.C. v = 0v = 0, but , but a a 0 0

D.D. None of the aboveNone of the above

y

Exercise 1Exercise 1Motion in One DimensionMotion in One Dimension


In driving from Madison to Chicago, initially my speed is at a In driving from Madison to Chicago, initially my speed is at a constant 65 mph. After some time, I see an accident ahead of me on constant 65 mph. After some time, I see an accident ahead of me on I-90 and must stop quickly so I decelerate increasingly “fast” until I I-90 and must stop quickly so I decelerate increasingly “fast” until I stop. The magnitude of mystop. The magnitude of my acceleration acceleration vsvs time time is given by,is given by,

AA

AA

AA

a

t

Exercise 2 Exercise 2 More complex Position More complex Position vs.vs. Time Graphs Time Graphs

• Question: My velocity vs time graph looks like which of the following ?

v

t

v

v


In driving from Madison to Chicago, initially my speed is at a In driving from Madison to Chicago, initially my speed is at a constant 65 mph. After some time, I see an accident ahead of me on constant 65 mph. After some time, I see an accident ahead of me on I-90 and must stop quickly so I decelerate increasingly fast until I I-90 and must stop quickly so I decelerate increasingly fast until I stop. The magnitude of mystop. The magnitude of my acceleration acceleration vsvs time time is given by,is given by,

AA

AA

AA

a

t

Exercise 2 Exercise 2 More complex Position More complex Position vs.vs. Time Graphs Time Graphs

• Question: My velocity vs time graph looks like which of the following ?

v

t

v

v


Gravity facts:Gravity facts: g does not depend on the nature of the

material ! Galileo (1564-1642) figured this out

without fancy clocks & rulers!

Feather & penny behave just the same

in vacuum

Nominally, g = 9.81 m/s2 At the equator g = 9.78 m/s2

At the North poleg = 9.83 m/s2


Gravity Map of the Earth Gravity Map of the Earth (relief exaggerated)(relief exaggerated)

A person off the coast of India would weigh 1% less than at most other places on earth.


Gravity map of the USGravity map of the US

Red: Areas of stronger local g Blue: Areas of weaker local g Due to density variations of the Earth’s crust and mantle


Exercise 3 Exercise 3 1D Freefall1D Freefall

A.A. vvAA < < vvB B

B.B. vvAA = = vvBB

C.C. vvAA > > vvBB

Alice and Bill are standing at the top of a cliff of heightAlice and Bill are standing at the top of a cliff of height HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, Alice straight, Alice straight downdown and Bill straightand Bill straight upup. The speed of the balls when . The speed of the balls when they hit the ground arethey hit the ground are vvAA andand vvBB respectivelyrespectively..

vv00

vv00

BillBillAliceAlice

HH

vvAA vvBB


Exercise 3 Exercise 3 1D Freefall1D Freefall

A.A. vvAA < < vvB B

B.B. vvAA = = vvBB

C.C. vvAA > > vvBB

Alice and Bill are standing at the top of a cliff of heightAlice and Bill are standing at the top of a cliff of height HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, Alice straight, Alice straight downdown and Bill straightand Bill straight upup. The speed of the balls when . The speed of the balls when they hit the ground arethey hit the ground are vvAA andand vvBB respectivelyrespectively..

vv00

vv00

BillBillAliceAlice

HH

vvAA vvBB


Problem Solution Method:Problem Solution Method:Five Steps:

1) Focus the Problem- draw a picture – what are we asking for?

2) Describe the physics- what physics ideas are applicable

- what are the relevant variables known and unknown

3) Plan the solution- what are the relevant physics equations

4) Execute the plan- solve in terms of variables

- solve in terms of numbers

5) Evaluate the answer- are the dimensions and units correct?

- do the numbers make sense?


Example of a 1D motion problemExample of a 1D motion problem A cart is initially traveling East at a constant speed of

20 m/s. When it is halfway (in distance) to its destination its speed suddenly increases and thereafter remains constant. All told the cart spends a total of 10 s in transit with an average speed of 25 m/s.

What is the speed of the cart during the 2nd half of the trip? Dynamical relationships:

txxx avv0

221

0 a v0

ttxx xx

consta x)v(v

21

v

)x(x2av v

xx(avg)x

0x2

x2x

0

0

) timetotal(

)ntdisplaceme( velocityaveragev

tx

And


The pictureThe picture

Plus the average velocity Knowns:

x0 = 0 m

t0 = 0 s

v0 = 20 m/s

t2 = 10 s

vavg = 25 m/s

relationship between x1 and x2

Four unknowns x1 v1 t1 & x2 and must find v1 in terms of knowns

t2t0 t1

v1 ( > v0 )

a1=0 m/s2

v0

a0=0 m/s2

x0 x1 x2

02

02vttxx

tx

0

vv xx txx x v

00

0a x


Using Using

Four unknowns Four relationships

t2t0 t1

v1 ( > v0 )

a1=0 m/s2

v0

a0=0 m/s2

x0 x1 x2

)(

v

0221

1

02

02

xxx

ttxx

tx

txx x v0

0

)( v 01001 ttxx )( v 12112 ttxx


Using Using

Eliminate unknowns

first t1

next x1

t2t0 t1

v1 ( > v0 )

a1=0 m/s2

v0

a0=0 m/s2

x0 x1 x2

221

1 xx

0 0 00 tx

101 v tx )( v 12112 ttxx

2

2vt

x

)( v 0

1

v2112xtxx 1 & 2

3

2

4

)( v 0

221

v21221

2xtxx 3

1


Now Algebra and Relationship 4Now Algebra and Relationship 4

Algebra to simplifyt2

t0 t1

v1 ( > v0 )

a1=0 m/s2

v0

a0=0 m/s2

x0 x1 x2

2

2

t

xv

4

)(2 vv

)(2 vv

)( v

)( v

0

0

2

0

221

0

221

v1

v212

v21221

v21221

2

v

tv

x

x

tt

tx

txx

)( v vv2 vv

1 0

0


FiniFini

Plus the average velocity Given:

v0 = 20 m/s

t2 = 10 s

vavg = 25 m/s

t2t0 t1

v1 ( > v0 )

a1=0 m/s2

v0

a0=0 m/s2

x0 x1 x2

m/s 3.33

v

v

15m/s 005

m/s 25m/s 20 2m/s 20 m/s 25

1

vv2 vv

1 0

0


Tips:Tips: Read !

Before you start work on a problem, read the problem statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem.

Watch your units (dimensional analysis) !Always check the units of your answer, and carry the

units along with your numbers during the calculation.

Ask questions !


See you MondaySee you Monday

(Chapter 3 on Monday….)(Chapter 3 on Monday….)

Assignment:Assignment:

For Monday, Read Chapter 4For Monday, Read Chapter 4

Mastering Physics Problem Set 2Mastering Physics Problem Set 2

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Physics 207: Lecture 3, Pg 1 Physics 207, Lecture 3, Sept. 10 Assignment: 1.For Monday: Read Chapter 4 2.Homework Set 2 (due Wednesday 9/17) l Goals (finish