PHYS 1443 – Section 003 Lecture #2

Monday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu

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PHYS 1443 – Section 003Lecture #2

Wednesday, Sept. 9, 2002Dr. Jaehoon Yu

1. One dimensional motion w/ constant acceleration2. Kinetimatic equations of motion3. Free fall4. Coordinate systems 5. Vector; its properties, operations and components6. Two dimensional motion

• Displacement, Velocity, and Speed• Projectile Motion• Uniform Circular Motion

Today’s homework is homework #3, due 1am, next Monday!!


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One Dimensional Motion• Let’s start with the simplest case: acceleration is constant (a=a0)• Using definitions of average acceleration and velocity, we can draw

equations of motion (description of motion, velocity and position as a function of time)

t

vv

tt

vva

xi

i

xix

xf

f

xf

If tf=t and ti=0 tavv xxixf

For constant acceleration, simple numeric average

t

xx

tt

xxv

i

i

ix

f

f

f

If tf=t and ti=0

2

2

1tatvxtvxx xxiixif

Resulting Equation of Motion becomes

tavtavvv

v xxixxixfxi

x

2

1

2

2

2

tvxx xif


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Kinetic Equations of Motion on a Straight Line Under Constant Acceleration tavtv xxixf

2

2

1tatvxx xxiif

tvvtvxx xixfxif 2

1

2

1

ifxf xxavv xxi 222

Velocity as a function of time

Displacement as a function of velocity and time

Displacement as a function of time, velocity, and acceleration

Velocity as a function of Displacement and acceleration

You may use different forms of Kinetic equations, depending on the information given to you for specific physical problems!!


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Example 2.8

• Let’s call the time interval for police to catch the car; T• Set up an equation:Police catches the violator when his final

position is the same as the violator’s.22 00.3

2

1

2

1TaTx Police

f )1(0.451 TTvx Carf

)1(0.4500.32

1 ; 2 TTxx Car

fPolicef

a

acbbx

cbxax

2

4

are 0for Solutions

2

2

A car traveling at constant speed of 45.0m/s (~162km/hr or ~100miles/hr), police starts chasing the car at the constant acceleration of 3.00m/s2, one second after the car passes him. How long does it take for police to catch the violator?

Ex02-08.ip

00.300.3000.1 2 TT

possible)(not 00.1or (possible) 0.31 TT


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Free Fall• Free fall is a motion under the influence of gravitational pull

(gravity) only; Which direction is a freely falling object moving?• Gravitational acceleration is inversely proportional to the

distance between the object and the center of the earth• The gravitational acceleration is g=9.80m/s2 on the surface of

the earth, most of the time.• The direction of gravitational acceleration is ALWAYS toward

the center of the earth, which we normally call (-y); where up and down direction are indicated as the variable “y”

• Thus the correct denotation of gravitational acceleration on the surface of the earth is g=-9.80m/s2


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Example 2.12Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof

of a 50.0m high building, 1. Find the time the stone reaches at maximum height (v=0)2. Find the maximum height3. Find the time the stone reaches its original height4. Find the velocity of the stone when it reaches its original height5. Find the velocity and position of the stone at t=5.00s

g=-9.80m/s2

st

ttavv yyif

04.280.9

0.20

00.080.90.20

1

)(4.704.200.50

)04.2()80.9(2

104.2200.50

2

1

2

2

m

tatvyy yyiif

2

st 08.4204.2 3 Other ways?

)/(0.2008.4)80.9(0.20 smtavv yyiyf 4

)(5.27)00.5()80.9(2

100.50.200.50

2

1

2

2

m

tatvyy yyiif

5-Position

)/(0.29

00.5)80.9(0.20

sm

tavv yyiyf

5-Velocity

Ex02-12.ip


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Coordinate Systems• Makes it easy to express locations or positions• Two commonly used systems, depending on convenience

– Cartesian (Rectangular) Coordinate System• Coordinates are expressed in (x,y)

– Polar Coordinate System • Coordinates are expressed in (r)

• Vectors become a lot easier to express and compute

O (0,0)

(x1,y1)=(r)r

sin

cos

ry

rx

How are Cartesian and Polar coordinates related?

1

1tan

21

21

x

y

yxr

y1

x1

+x

+y


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Example 3.1 Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m. Find the polar coordinates of this point.

y

x

(-3.50,-2.50)m

r

s

)(30.45.18

50.250.322

22

m

yxr

s 180

7

5

50.3

50.2tan

s

5.357

5tan 1

s

2165.35180180 s


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Vector and ScalarVector quantities have both magnitude (size) and direction

Scalar quantities have magnitude onlyCan be completely specified with a value and its unit

Force, gravitational pull, momentum

Normally denoted in BOLD BOLD letters, FF, or a letter with arrow on top FTheir sizes or magnitudes are denoted with normal letters, F, or absolute values: For F

Energy, heat, mass, weightNormally denoted in normal letters, E

Both have units!!!


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Properties of Vectors• Two vectors are the same if their and the are the

same, no matter where they are on a coordinate system.

x

y

AABB

EE

DD

CC

FF

Which ones are the same vectors?

A=B=E=DA=B=E=D

Why aren’t the others?

C:C: The same magnitude but opposite direction: C=-A:C=-A:A negative vector

F:F: The same direction but different magnitude

sizes directions


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Vector Operations• Addition:

– Triangular Method: One can add vectors by connecting the head of one vector to the tail of the other (head-to-tail)

– Parallelogram method: Connect the tails of the two vectors and extend– Addition is commutative: Changing order of operation does not affect the results

A+B=B+AA+B=B+A, A+B+C+D+E=E+C+A+B+DA+B+C+D+E=E+C+A+B+D

AA

BBAA

BB=AA

BBA+BA+B

• Subtraction: – The same as adding a negative vector:A A - B = A B = A + (-BB)

AA-B-B Since subtraction is the equivalent to adding

a negative vector, subtraction is also commutative!!!

• Multiplication by a scalar is increasing the magnitude A, BA, B=2A A

AA B=2AB=2A

AB 2

Fig03-07.ip

A+BA+BA+BA+B

A-BA-B

OR


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Example 3.2A car travels 20.0km due north followed by 35.0km in a direction 60.0o west of north. Find the magnitude and direction of resultant displacement.

N

E

r

22 sincos BBAr

20AA

BB

60cos

60sintan 1

BA

B

Find other

ways to solve this problem…

cos2sincos 2222 ABBA

cos222 ABBA

60cos0.350.2020.350.20 22

)(2.482325 km

60cos0.350.20

60sin0.35tan 1

N W wrt to9.385.37

3.30tan 1


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Components and Unit Vectors• Coordinate systems are useful in expressing vectors in their components

(Ax,Ay)AA

Ay

Ax

x

y

22

sin

cos

yx

y

x

AAA

AA

AA

AA

AAA

222

22

sincos

sincos

}Components(+,+)

(-,+)

(-,-) (+,-)• Unit vectors are dimensionless vectors whose magnitude are exactly 1

• Unit vectors are usually expressed in i,j,k or • Vectors can be expressed using components and unit vectors

kji , ,

jAiAjAiAA yx sincos So the above vector AA can be written as


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Examples 3.3 & 3.4Find the resultant vector which is the sum of AA=(2.0ii+2.0jj) and B B =(2.0ii-4.0jj)

mjiji

jijiBAC

0.20.40.40.20.20.2

0.40.20.20.2

Find the resultant displacement of three consecutive displacements: dd11=(15ii+30j j +12kk)cm, dd22=(23ii+14j j -5.0kk)cm, and dd11=(-13ii+15jj)cm

)(650.75925 222 cmD

)(5.4200.4160.20.4 22 mC

)(0.75925

0.512151430132315

cmkji

kjiD

jikjikji

dddD

15130.51423123015

321

270.4

0.2tantan 11

x

y

C

C


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Displacement, Velocity, and Acceleration in 2-dim

• Displacement: if rrr

• Average Velocity: if

if

tt

rr

t

rv

• Instantaneous Velocity: dt

rd

t

rv

t

0

lim

• Average Acceleration if

if

tt

vv

t

va

• Instantaneous Acceleration: 2

2

0lim

dt

rd

dt

rd

dt

d

dt

vd

t

va

t

How is each of these quantities defined in 1-D?

PHYS 1443 – Section 003 Lecture #2

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