physics 201: lecture 2, pg 1 lecture 2 chapter chapter 2.1-2.4 define position, displacement &...

Physics 201: Lecture 2, Pg 1

Lecture 2ChapterChapter 2.1-2.4

Define Position, Displacement & Distance

Distinguish Time and Time Interval

Define Velocity (Average and Instantaneous), Speed

Define Acceleration

Understand algebraically, through vectors, and graphically the relationships between position, velocity and acceleration

Comment on notation


Informal Reading Quiz

Displacement, position, velocity & acceleration are the main quantities that we will discuss today.

Which of these 4 quantities have the same units

A. Velocity & position

B. Velocity & acceleration

C. Acceleration & displacement

D. Position & displacement

E. Position & acceleration


Range of LengthsDistance Length (m)Radius of Visible Universe 1 x 1026

To Andromeda Galaxy 2 x 1022

To nearest star 4 x 1016

Earth to Sun 1.5 x 1011

Radius of Earth 6.4 x 106

Willis Tower 4.5 x 102

Football Field 1 x 102

Tall person 2 x 100

Thickness of paper 1 x 10-4

Wavelength of blue light 4 x 10-7

Diameter of hydrogen atom 1 x 10-10

Diameter of proton 1 x 10-15


Range of Times

Interval Time (s)

Age of Universe 5 x 1017

Age of Grand Canyon 3 x 1014

Avg age of college student 6.3 x 108

One year 3.2 x 107

One hour 3.6 x 103

Light travel from Earth to Moon 1.3 x 100

One cycle of guitar A string 2 x 10-3

One cycle of FM radio wave 6 x 10-8

One cycle of visible light 1 x 10-15

Time for light to cross a proton 1 x 10-24

World’s most accurate timepiece: Cesium fountain Atomic Clock

Lose or gain one second in some 138 million years


One-Dimension Motion (Kinematics) Position, Displacement, Distance

Position: Reflects where you are. KEY POINT 1: Magnitude, Direction, Units KEY POINT 2: Requires a reference point (Origin)

Origins are arbitrary



Position: Reflects where you are. KEY POINT 1: Magnitude, Direction, Units KEY POINT 2: Requires a reference point (Origin)

Origins are arbitrary

Example: Where is Boston ?

Choose origin at New YorkNew York

Boston is 212 miles northeast of New York

OR

Boston is 150 miles east and 150 miles north of New York

Boston

New York (Origin)



Getting from New York to Boston requires a PATH Path defines what places we pass though

Displacement: Change in position Requires a time interval Any point on the path must be

associated with a specific time

( t1, t2, t3, ….) Path 1 and Path 2 have the

same change in position so theythe same displacement.

However the distance travelled isdifferent.

Boston

New York

Path 1

Path 2


Motion in One-Dimension (Kinematics) Position

Position along a line; references xi and ti :

At time = 0 seconds Pat is 10 meters to the right of the lamp Origin lamp Positive direction to the right of the lamp Position vector ( xi , ti) or (10 m, 0.0 s) Particle representation

10 meters

PatO

-x +x10 meters


Displacement

One second later Pat is 15 meters to the right of the lamp

At t = 1.0 s the position vector is ( xf , tf ) or (15 m, 1.0 s) Displacement is just change in position

x ≡ xf – xi

There is also a change in time

t ≡ tf – ti

10 meters 15 meters

Patxi

O

xfΔx


Displacement Putting it all together

x = xf - xi = 5 meters to the right !t = tf - ti = 1 second

Relating x to t yields average velocity

O Patxi

10 meters 15 meters

xfΔx


Average Velocity

Changes in position vs Changes in time

)increment time(

)ntdisplaceme( velocityaverageav, t

xv gx

• Average velocity = displacement per time increment , includes BOTH magnitude and direction

• Pat’s average velocity was 5 m / s to the right

)sec 1()right the tom 5(

avg, tx

vx


Average Speed

Average speed, vavgavg, reflects a magnitude “How fast” without the direction.

References the total distance travelled

t

d

tv

) timetotal(

path along taken distancespeed averageavg

• Pat’s average speed was 5 m / s

NOTE: Serway’s notation varies from other texts(There really is no standard)


Pat on tour (graphical representation) Pat is walking from and to the lamp (at the origin).

(x1 , t1) = (10 m, 0.0 sec)

(x2 , t2) = (15 m, 1.0 sec)

(x3 , t4) = (30 m, 2.0 sec)

(x4 , t4) = (10 m, 3.0 sec)

(x5 , t5) = ( 0 m, 4.0 sec)

Compare displacement distance avg. vel. avg. speed

t = 1 s x1,2 = x2 – x1 = 5 m d = 5 m vx,avg= 5 m/s vx,avg= 5 m/s



Here d = |x1,2 | + |x2,3 |+ |x3,4 | = 5 m + 15 m + 20 m = 40 mSpeed and velocity measure different things!

t (seconds)x

(met

ers

)10

30

20

1 2 430


Calculating path distance in general

dtdxdf

i

t

tdtdx || ||

|| ixd


Exercise 2 Average Velocity

x (meters)

t (seconds)

2

6

-2

4

What is the magnitude of the average velocity over the first 4 seconds ?

(A) -1 m/s (D) not enough information to decide.

(C) 1 m/s(B) 4 m/s

1 2 430


Average Velocity Exercise 3 What is the average velocity in the last second (t = 3 to 4) ?

A. 2 m/s

B. 4 m/s

C. 1 m/s

D. 0 m/s

x (meters)

t (seconds)

2

6

-2

4

1 2 43


Average Speed Exercise 4

What is the average speed over the first 4 seconds ?0 m to -2 m to 0 m to 4 m 8 meters total

A. 2 m/s

B. 4 m/s

C. 1 m/s

D. 0 m/s

x (meters)

t (seconds)

2

6

-2

4

1 2 43

turning point


Instantaneous velocity

• Limiting case as the change in time 0

dtdxv t

x

tx

)time()ntdisplaceme(

0lim

x

t0

• As t 0 velocity is the tangent to the curve (& path)

• Dashed green line is vx

• Yellow lines areaverage velocities instantaneous velocity at

t = 0 s


Instantaneous speed

• Just the magnitude of the instantaneous velocity

sv dtdx

tx

tx

|||lim||| )time(

)ntdisplaceme(

0


What is the instantaneous velocity at the fourth second?

(A) 4 m/s (D) not enough information to decide.

(C) 1 m/s(B) 0 m/s

x (meters)

t (seconds)

2

6

-2

4

1 2 43

Exercise 5Instantaneous Velocity


Special case: Instantaneous velocity is constant

• Slope is constant over a time t.

tx

dtdxvx

constant

x

t0

t

x (xi , ti)

(xf , tf)


Special case: Instantaneous velocity is constant

• Slope is constant over a time t.

x

t0

t

x (xi , ti)

(xf , tf)

ixf

xif

xtvx

tvxx

• Given t, xi and vx we can deduce xf

and this reflects the area under the velocity curve

txx

txv if

x


Now multiple vx ; Pat’s velocity plot

t (seconds)x

(met

ers

)10

30

20

1 2 430

v x (m

/s) 10

20

1 2 430

-10

-20t (seconds)

(x1 , t1) = (10 m, 0.0 s)

(x2 , t2) = (15 m, 1.0 s)

(x3 , t3) = (30 m, 2.0 s)

(x4 , t4) = (10 m, 3.0 s)

(x5 , t5) = ( 0 m, 4.0 s)


Home exercise 6 (and some things are easier than they appear)

A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to the finish at 30 km/hr. When the bird reaches the finish line, it turns around and flies back to the runner, and then turns around again, repeating the back-and-forth trips until the runner reaches the finish line.

How many kilometers does the bird travel?

A. 10 km

B. 15 km

C. 20 km

D. 30 km


Objects with slowly varying velocities

dt

dxvx

] offunction a is [ )( txtxx x

vx

t

t

Change of the change….

changes in velocity with time

give average acceleration

tv

a xx

avg,

t

vx


And finally instantaneous acceleration

2

2

adt

xddtdvx

x

t

ax

t

x

vx

t

t

tv

a xx

avg,

dtdv

a xx

tx tv

0lim


Example problem

A car moves to the right first for 2.0 sec at 1.0 m/s and then 4.0 seconds at 2.0 m/s.

What was the average velocity?

Two legs with constant velocity but ….

vx

t

221

avgx,

vvv


Example problem

A particle moves to the right first for 2.0 seconds at 1.0 m/s and then 4.0 seconds at 2.0 m/s.

What was the average velocity?

Two legs with constant velocity but ….

We must find the total displacement (x2 –x0) And x1 = x0 + v0 (t1-t0) x2 = x1 + v1 (t2-t1) Displacement is (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0) x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6.0 s or 1.7 m/s

vx

t2vv

v 21Avg


Position, velocity & acceleration

All are vectors! Cannot be used interchangeably (different units!)

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

But the directions can be determined “Change in the position” vector vs. time gives the

direction of the velocity vector “Change in the velocity” vector vs. time gives the

direction of the acceleration vector Given x(t) vx(t) ax (t) Given ax (t) vx (t) x(t)

v

a


Assignment

Reading for Tuesday’s class

» All of Chapter 2

physics 201: lecture 2, pg 1 lecture 2 chapter chapter 2.1-2.4 define position, displacement &...

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