fundamental physics i

Fundamental Physics I

FALL 2009-2010

Dr. Joseph Trout

Dimensions:Length: meter●1790 - One ten-millionth of the distance from the equator to either pole.●1889 - Platinum- iridium rod●1960 - 1,650,763.73 wavelength of orange light produced by krypton-86●1983 - Distance light travels in 1/ 299,792,458 of a second.

Time: second●1/86,400 of a mean solar day●9,192,631,770 oscillations of a cesium atom.

Mass: kilogram●platinum – iridium cylinder

Prefixes:

yotta Y 1024

zetta Z 1021

exa E 1018

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

hecto h 102

deka da 101

deci d 10-1

centi c 10-2

milli m 10-3

micro µ 10-6

nano n 10-9

pico p 10-12

femto f 10-15

atto a 10-18

zepto z 10-21

yocto y 10-24

Prefixes:

Mass

Length

Time

1kilogram=1kg=1000 g=1 X 103g

1centimeter=1cm=0.01m=1 X 10−2m1kilometer=1 km=1000m=1 X 103m

1microsecond=1 s=0.000001 s=1 X 10−6 s1nanosecond=1ns=0.000000001 s=1 X 10−9 s

1mile=1.609m1km=0.621mi1hr=3600 s1 year=3.156 X 107 s1kg=0.0685 slug1 lb=4.448 N

1mile=1.609km

1mile1.609km

=1

1.609km1mile

=1

Conversion Factors:

1mile=1.609 km

1mile1.609 km

=1

5 km=?miles

5 km1mile1.609 km =3.11mi

1mile=1.609 km

5 km=?miles

5 km1mile1.609 km =3.11mi

1hr=3600 s1mi=1609m

20ms=?mph

20ms 1mile1609m

1hr=3600 s1mi=1609m

20ms=?mph

20ms 1mile1609m 3600 s

1hr

1hr=3600 s1mi=1609m

20ms=?mph

20ms 1mile1609m 3600 s

1hr =44.75mph

1hr=3600 s1mi=1609m

60mph=?m /s

60mihr 1609m

1mi 1hr3600 s =26.82m /s

1mi=1609m

3mi3=?m3

3mi31609m1mi

3

=1.25 X 1010m3

Scalar – Magnitude only.

Example: mass, distance, speed Example: m, x, v

Vector – Magnitude and Direction.

Example: displacement,velocity, acceleration, forceExample: x ,v ,a , F

Distance – scalar – magnitude of the total distance traveled.

Displacement – vector – distance between final position and initial position AND the direction.

Displacement in One Dimension:

Direction will either be positive or negative.

Distance vs. Displacement:

X= 0.0 m X= 2.0 m X=4.0 mX=-2.0 mX= -4.0 m

distance= x=4.0m


X= 0.0 m X= 2.0 m X=4.0 mX=-2.0 mX= -4.0 m

distance= x=4.0mdisplacement=x=x f−x i=4.0m−0.0m=4.0m


X= 0.0 m X= 2.0 m X=4.0 mX=-2.0 mX= -4.0 m

distance=4.0m8.0m=12m

displacement x=x f−x i=−4m−0m=−4m

Start Finish

Marathon distance = 26 miles

Marathon displacement = -0.25 miles

+x

Displacement= x=x f−x i

28

Scalar – Magnitude ONLY

Vector – Magnitude and Direction

distance=5milesmass=6 kg

Example :Velocityv=40m / s Northv=40m / s in positive x direction.v=40m / s@30o

29

y

x=60o

R=10m@60o

30

y−axis

x−axisz−axis

i

j

k

Vector Notation

31

i

j Vector Notation

A=4m i

B=3m j

32

i

j Vector Notation

A=4m i

B=3m j

R=AB

33

i

j Vector Notation

A=4m i

B=3m j

R=ABR=4m i3m j

34

Pythagoras was a Greek philosopher who made important developments in mathematics, astronomy, and the theory of music. The theorem now known as Pythagoras's theorem was known to the Babylonians 1000 years earlier but he may have been the first to prove it.

35

x

yr 2=x 2 y2

36

x

yr=x 2 y2

37

x=4m

y=3m

r=4m 23m 2=16m29m2=25m2=5m

r

38x=6m

y=7m

r=6m 27m 2=36m249m2=85m2=9.22m

r

39

x

yr

sin= yr

cos=xr

tan= yr

r=x2 y2

40

adj

opphyp

sin=opphyp

cos=adjhyp

tan=oppadj

41

i

j Vector Notation

4m

3m

R=ABR=4m i3m j

∣R∣=4m 23m 2=25m2=5m

tan=oppadj

=tan−13m4m =36.87o

42

i

j Vector Notation

4m

3m

R=ABR=4m i3m [email protected]

∣R∣=4m 23m 2=25m2=5m

tan=oppadj

=tan−13m4m =36.87o

43

i

j Vector Notation

R x

R y

∣R∣=10m

R=40o

R=10m@40o

44

i

j Vector Notation

R x

R y

∣R∣=10m

R=10m@40o

R=40o

cos=adjhyp

cosR=R x∣R∣

R x=∣R∣cosR=10mcos 40o =7.66m

45

i

j Vector Notation

R x=7.66m

R y

∣R∣=10m

R=10m@40o

R=40o

sin=opphyp

sinR=R y∣R∣

R y=∣R∣sinR=10m sin 40o =6.43m

46

i

j Vector Notation

R x=7.66m

R y=6.43m

∣R∣=10m

R=10m@40oR=R x iR y j=7.66m i6.43m j

R=40o

47

i

j Vector Notation

R x=7.66m

R y=6.43m

∣R∣=10m

R=10m@40oR=R x iR y j=7.66m i6.43m j

R=40o

Check:∣R∣=R x2R y2∣R∣=7.66m 2 6.43m 2=10m

tan =R yR x

=tan−16.43m7.66m =40o

48

Vector Notation

i

jA

A x=3m

A y=4m A=Ax iA y j=3m i4m j

III

IVIII

49

Vector Notation

i

jA

A x=3m

A y=4m A=Ax iA y j=3m i4m j

∣A∣=3m 24m 2=5m

A=tan−14m3m =53.13o

III

IVIII

50

Vector Notation

i

jA

A x=3m

A y=4m A=Ax iA y j=3m i4m jA=∣A∣@[email protected]

∣A∣=3m 24m 2=5m

A=tan−14m3m =53.13o

A

III

IVIII

51

Vector Notation

i

j

B

Bx=4m

B y=2m

B=B x iB y j=−4m i2m j

B

III

IVIII

52

Vector Notation

i

j

B

Bx=−4m

B y=2m


B

∣B∣=−4m 22m 2=4.47m

III

IVIII

53

Vector Notation

i

j

B

Bx=−4m

B y=2m


∣B∣=−4m 22m 2=4.47m

B=tan−12m−4m =−26.57o

B

?????

B '

III

IVIII

54

Vector Notation

i

j

B

Bx=−4m

B y=2m


∣B∣=−4m 22m 2=4.47m

B=tan−12m−4m =−26.57o

B

?????

B '

B=180oB 'B=180o−26.57o=153.43o

III

IVIII

55

Vector Notation

i

j

B

Bx=−4m

B y=2m

B=B x iB y j=−4m i2m jB=∣B∣@[email protected]

∣B∣=−4m 22m 2=4.47m

B=tan−12m−4m =−26.57o

B

?????

B '

B=180oB 'B=180o−26.57o=153.43o

III

IVIII

56

Vector Notation

i

j

C

C x=−3m

C y=−6m

C=C x iC y j=−3m i−6m j

C

III

IVIII

57

Vector Notation

i

j

C

C x=−3m

C y=−6m

C=C x iC y j=−3m i−6m j

C

CII

IVIII ∣C∣=−3m 2−6m 2=6.71m

C=tan−1−6m−3m =71.57o ?????

58

Vector Notation

i

j

C

C x=−3m

C y=−6m

C=C x iC y j=−3m i−6m jII

IVIII ∣C∣=−3m 2−6m 2=6.71m

C=tan−1−6m−3m =71.57o ?????

C

I

C '

C=180oC 'C=180o71.57o=251.57o

59

Vector Notation

i

j

C

C x=−3m

C y=−6m

C=C x iC y j=−3m i−6m jC=∣C∣@[email protected]

IVIII ∣C∣=−3m 2−6m 2=6.71m

C=tan−1−6m−3m =71.57o ?????

C

I

C '

C=180oC 'C=180o71.57o=251.57o

60

Vector Notation

i

j

D

D x=3m

D y=−3m

D=Dx iD y j=3m i−3m jD=∣D∣@D=4.24m@−45oII

IVIII ∣D∣=3m 2−3m 2=4.24m

D=tan−1−3m3m =−45o ?????

D

I

61

Vector Notation

i

j

D

D x=3m

D y=−3m

D=Dx iD y j=3m i−3m jD=∣D∣@D=4.24m@−45oII

IVIII ∣D∣=3m 2−3m 2=4.24m

D=tan−1−3m3m =−45o ?????

D

I

D '

62

Vector Notation

i

j

D

D x=3m

D y=−3m

D=Dx iD y j=3m i−3m jD=∣D∣@D=4.24m@−45o=4.24m@315oII

IVIII ∣D∣=3m 2−3m 2=4.24m

D=tan−1−3m3m =−45o ?????

D

I

D 'D=360oC 'D=360o−45o=315o

63

Adding Vectors

64

Adding Vectors

5m

5m

−5m

−5m A

B

A=5m i0m jB=0m i6m j

65

Adding Vectors

5m

5m

−5m

−5m A

B


C C=AB=5m i6m j

66

Adding Vectors

5m

5m

−5m

−5m A

B


C C=AB=5m i6m j

∣C∣=5m 26m 2=7.81m

C=tan−16m5m =50.19o

67

Adding Vectors

5m

5m

−5m

−5m A


B

68

Adding Vectors

5m

5m

−5m

−5m A

B

A=5m i0m j

C=AB=8m i5m j

∣C∣=8m 25m 2=9.43m

C=tan−15m8m =32.00o

B=3m i5m j

C

69

Adding Vectors

5m

5m

−5m

−5m A

B

A=5m i0m jB=−5m i5m j

70

Adding Vectors

5m

5m

−5m

−5m A

B


71

Adding Vectors

5m

5m

−5m

−5m A

B


C

72

Adding Vectors

5m

5m

−5m

−5m A

B


C

C=AB=0m i5m j

∣C∣=0m 25m 2=5.00mC=90.00o

73

Adding Vectors

5m

5m

−5m

−5m

AB


74

Adding Vectors

5m

5m

−5m

−5m

AB


C

75

Adding Vectors

5m

5m

−5m

−5m

AB


C

C=AB=2m i8m j

∣C∣=2m 28m 2=8.25m

C=tan−18m2m =75.96o

C

76

Adding Vectors

5m

5m

−5m

−5m

A

A=7m i3m jB=−5m i5m jC=4m i−2m j

B

C

77

Adding Vectors

5m

5m

−5m

−5m

A

B

A=7m i3m jB=−5m i5m jCC=4m i−2m j

D=ABC=6m i6m jD

78

Adding Vectors

5m

5m

−5m

−5m

AB


C

C=4m i−2m j

D=ABC=6m i6m j

79

Adding Vectors

5m

5m

−5m

−5m

AB


C

C=4m i−2m j

D=ABC=6m i6m jAB

80

Adding Vectors

5m

5m

−5m

−5m


C

C=4m i−2m j

D=ABC=6m i6m jAB

ABC

81

Adding Vectors

5m

5m

−5m

−5m


C

C=4m i−2m j

D=ABC=6m i6m jABC=D

82

x=100m Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross calm lake.

v x= x t

=100m10 s

=10m / s

83

x=100m Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross calm lake.

v x= x t

=100m10 s

=10m / s

v x=10m / s

84

x=100m

Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.

v x=10m / s

v y=5m / s

x=100m

y=v y t=5m / s 10 s =50m

85

x=100m


v y=5m / s

x=100m

y=50mv

86

x=100m


v y=5m / s

x=100m

y=50mv

v=v x iv y jv=10m / s i5m / s j

87

x=100m


v y=5m / s

x=100m

y=50mv

v=v x iv y jv=10m /s i5m / s j

∣v∣=10m /s 25m /s 2=11.18m /s

=tan−15m /s10m /s =26.57o

88

x=100m


v y=5m / s

x=100m

y=50mv

v=11.18m /[email protected]

r= x 2 y 2

r=100m 250m 2=111.80m

89

x=100m


v y=5m / s

x=100m

y=50mv

v=11.18m / [email protected]

r=111.80m

v= r t

=111.80m10 s

=11.18m / s

90

x=100m


v y=5m / s

x=100m

y=50mv


r=111.80m

v= r t

=111.80m10 s

=11.18m / s

v x=10m / s

v y=5m / s=26.57o

∣v∣=11.18m / s

91

Adding Vectors

10m / s

10m / s

−10m / s

−10m / s


92

Adding Vectors

10m / s

10m / s

−10m / s

−10m / s


v x=∣v∣cosv x=17.20m / s cos54.46o

v x=10m / s

v y=∣v∣sinv y=17.20m / s sin 54.46o

v y=14m / s

93

Projectile Motion

ymaxyo

Rangev f

vo

94

Projectile Motion

=30o

∣vo∣=10m / s

95

Projectile Motion

∣vo∣=10m / s

=30o

v x

v y

vox=10m / s cos 30o=8.66m / svo y=10m /s sin 30o=5.00m / s

96

Projectile Motion

∣vo∣=10m / s

=30o

v x

v y

vox=10m / s cos 30o=8.66m / svo y=10m /s sin 30o=5.00m / s

∣vo∣=vox2 vo y2 =9.9997m / s≈10m / s

=tan−1vo yvox =30o

97

Scalar (Dot) ProductScalar (Dot) Product

98

Vector (Cross) ProductVector (Cross) Product

99


100


Displacement=x=x f−x i

average speed=vave= xt

average velocity=vave=x t

=x f−x it f−t i

vave= x t

If you travel 300 miles in 6 hours:

vave

= 300 mi / 6 hours = 50 mph

1)Travel 6 m in 3s.2)Travel 24 m in 3s.3)Stop for 2 s.4)Travel 10 m in 2s.

vave=6.0m24.0m0m10.0m 3.0 s3.0 s2.0 s2.0 s

=4.0ms


0 2 4 6 8 10 120

4

8

12

16

20

24

28

32

36

40

Position vs. Time

Time (s)

Pos

ition

(m)


vave=distancetime

=6m24m0m10m3s3s2s2 s

=4ms

v=40m−0m10 s−0 s =4ms

Define Instantaneous Velocity:

v inst=v=limx0

x t

v=d xdt

Define Instantaneous Velocity:v inst=v=lim

x0x t

v=d xdt

x=6mst5m

v=dxdt

=ddt 6ms t5m=6m

s

Define Instantaneous Velocity:v inst=v=lim

x0x t

v=d xdt

x=−3ms2 t

22mst7m

v=dxdt

=ddt −3m

s2 t22mst7m

v=−3ms2 t2m

s

Define Instantaneous Velocity:

0 2 4 6 8 10 120

4

8

12

16

20

24

28

32

36

40

Position vs. Time

Time (s)

Posi

tion

(m)v inst=v=lim

x0 x t

1)Travel 6 m in 3s.

v=x f−x it f−t i

=6.0m−0.0m3.0s−0.0 s

=2.0m / s

2) Travel 24 m in 3s.


=30.0m−6.0m6.0s−3.0 s

=8.0m/ s

0 2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

Position vs. Time

Time (s)

Pos

ition

(m)

3) Stop for 2s.


=30.0m−30.0m8.0s−6.0 s

=0.0m /s

4) Travel 10 m in 2s.


=40.0m−3.0m10.0s−8.0 s

=5.0m / s

0 2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

Position vs. Time

Time (s)

Pos

ition

(m)

Equation of a Line:

0 1 2 3 4 5 60

5

10

15

20

25

Y vs X

X ( m )

Y ( m

)

y=m xb

m=slope= y x

=y f− yix f−xi

b= y intercept

Equation of a Line:

0 1 2 3 4 5 60

5

10

15

20

25

Y vs X

X ( m )

Y ( m

)

y=m xb

m= y x

=y f− yix f−x i

=15m−5m5m−0m

=2

b=5m

Equation of a Line:

0 1 2 3 4 5 60

5

10

15

20

25

Y vs X

X ( m )

Y ( m

)

y=m xb

m= y x

=y f− yix f−x i

=15m−5m5m−0m

=2

b=5m

y=2 x5

Equation of a Line:

m= y x

=y f− yix f−x i

=−16m−2m6m−0m

=−3

b=2m

y=−3 x2

0 1 2 3 4 5 6 7-20-18-16-14-12-10

-8-6-4-20246

Y vs X

X ( m )

Y (

m )

Equation of a Line:

y=−3 x2

0 1 2 3 4 5 6 7-20-18-16-14-12-10

-8-6-4-20246

Y vs X

X ( m )

Y (

m )

What is the value of y at x = 4 m ?

Equation of a Line:

y=−3 x2

0 1 2 3 4 5 6 7-20-18-16-14-12-10

-8-6-4-20246

Y vs X

X ( m )

Y (

m )

What is the value of y at x = 4 m ?

y=−342=−10

0 1 2 3 4 5 60

4

8

12

16

20

24

28

32

36

40

Position vs. Time

Time ( s )

Posi

tion

(m)

Velocity = change in position over the change in time

Velocity = slope of position vs. time plot

v=36m−4m5 s−0 s

=6ms

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 4605

10152025303540455055606570

Position vs. Time

Time (s)

Posi

tion

(m)

A

B

C

D

E

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 4605

10152025303540455055606570

Position vs. Time

Time (s)

Posi

tion

(m)

A

B

C

D

E

vc=5m−35m24 s−14 s

=−3ms

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 4605

10152025303540455055606570

Position vs. Time

Time (s)

Posi

tion

(m)

A

B

C

D

E

vd=65m−5m40 s−28 s

=5ms

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 4605

10152025303540455055606570

Position vs. Time

Time (s)

Pos

ition

( m

)

A

BC

D

E

0.0 m 10.0 m 20.0 m 30.0 m 40.0 m 50.0 m

A BC

DE

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time ( s )

Posi

tion

(m)

A

BC

DE

FG

H

I

J

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time (s)

Posi

tion

(m)

A

BC

E

F H

I

J

D

G

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time ( s )

Posi

tion

( m )

A

BC

E

F H

I

J

D

G

Stopped at B, D, G, I

Velocity is zero.

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time ( s )

Posi

tion

(m)

A

BC

E

F H

I

J

D

G

Moving forward at A, C, H

Velocity is positive.

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time ( s )

Posi

tion

(m)

A

BC

E

F H

I

J

D

G

Moving backward at E, F, J

Velocity is negative.

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time ( s )

Posi

tion

(m)

A

BC

E

F H

I

J

D

G

vave=−30m−5m26 s−0 s

=−1.3ms

Entire Trip

0 2 4 6 8 10 12 14 16 18 20 22 24 26-40

-30

-20

-10

0

10

20

30

40

50

60

70

Position vs. Time

Time (s)

Posi

tion

(m)

A

BC

E

F H

I

J

D

G

vE=45m−50m11 s−9 s

=−2.5ms

Acceleration: Time rate change of velocity.

aave=v t

=v f−v it f−t i

ainst=a= t0v t

= t0v f−vit f−t i

[a ]=ms2

Special Case: Velocity equals a constant.

v=constant

v= x t

=x f−x it f−t i

=constant

Special Case: Velocity equals a constant.

v=constant

v= x t

=x f−x it f−t i

=constant

Let: x i=xoat t i=0.0s

v=x−xot

x=xovt

A runner starts at x = +20 m and runs at a constant velocity of +5 m/s. Where will the runner be at 100 s?

x=v txox=5m

s100 s20m=520m

A runner starts at x = +20 m and runs at a constant velocity of +2 m/s. Where will the runner be at 100 s?

x=v txox=2m

s100 s20m=220m

0 20 40 60 80 100 120 1400

40

80

120

160

200

240

280

Position vs. Time

Time (s)

Pos

ition

(m)

Special Case: Acceleration is constant.

a=constant

a=v t

=v f−vit f−t i

v f=via t f−t i


a=constantv=voat

x=xovo t12a t2

v2=vo22a x


a=constant

a=v t

=v f−vit f−t i

v f=via t f−t i

x i=xo , vi=voat t i=0.0s

v=voat


x i=xo , vi=voat t i=0.0sa=constantv=voat

If acceleration is a constant and we know the initial velocity, then we can predict the velocity at any time (t).

We would also like to predict the position if we know the initial position.

Special Case: Acceleration is constant.x i=xo , v i=voat t i=0.0s

a=constantv=voat

vaverage=x−xot

Special Case: Acceleration is constant.x i=xo , v i=voat t i=0.0s

a=constantv=voat

vaverage=x−xot

=vvo

2

Special Case: Acceleration is constant. x i=xo , v i=voat t i=0.0s

a=constantv=voat

x−xot

=vvo

2

Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s

a=constantv=voat

x−xot

=vvo

22x−xo=vvot


a=constantv=voat

x−x ot

=vvo

22x−xo=vvo t

2x−xo=[voat ]vo t


a=constantv=voat

x−x ot

=vvo

22x−xo=vvo t

2x−xo=[voat ]vo t2 x−xo=vo tat

2vo t


a=constantv=voat

x−x ot

=vvo

22x−xo=vvo t


2vo t2x−xo=2vo tat

2


a=constantv=voat

x−x ot

=vvo

22x−xo=vvo t



2

x−xo=vo t12at 2


a=constantv=voat

x−x ot

=vvo

22x−xo=vvo t



2

x−xo=vo t12at 2

x=xovo t12at 2


a=constantv=voat

x=xovo t12a t2


a=constantv=voat

x=xovo t12a t2 v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t2

x=xovo v−voa

12a v−voa

2

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t2

x−xo=vo v−voa

12a v−voa

2

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t 2

x−xo=vov−voa 12av−voa

2

x−xo=vo v−vo

2

a1

2av−vo2a2

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t 2


2

x−xo=vo v−vo

2

a1

2av−vo2a2

x−xo=vo v−vo

2

a1

2v−vo

2

a

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t 2


2

x−xo=vo v−vo

2

a1

2av−vo2a2

x−xo=vo v−vo

2

a1

2v−vo

2

a

x−xo=vo v−vo

2

a1

2 v2−2v vovo

2

a

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t 2

x−xo=vov−voa 12a v−voa

2

x−xo=vo v−vo

2

a 12av−vo2a2

x−xo=vo v−vo

2

a 12v−vo

2

a

x−xo=vo v−vo

2

a 12 v

2−2v vovo2

a 2a x=2vo v−2vo

2v2−2v vovo2

2a x=v2−vo2

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t2


2

x−xo=vo v−vo

2

a1

2av−vo2a2

x−x o=vo v−vo

2

a1

2v−vo

2

a

x−xo=vo v−vo

2

a1

2 v2−2v vovo

2

a 2a x=2vo v−2vo

2v2−2v vovo2

2a x=v2−vo2

v2=vo22a x

v=voat

t=v−voa


a=constantv=voat

x=xovo t12a t2

v2=vo22a x

Special Case: Velocity is constant.

Review:

a=0v=constantx=x ovo t

Special Case: Acceleration is constant in One Dimension in the +x and or -x direction.

Moving forward or backward.

ax=constantv x=v xoa x t

x=xov xo t12a x t

2

v x2=v xo

2 2a x x

Special Case: Acceleration is constant in One Dimension in the +y and or -y direction.

Moving up or down.

a y=constantv y=v yoa y t

y= yov yo t12a y t

2

v y2=v yo

2 2a y y

Special Case: One Dimensional Motion near the surface of the earth.

a y=−g=−9.8 ms2=constant

g=9.8 ms2

Special Case: One Dimensional Motion near the surface of the earth.

a y=−g=−9.8 ms2=constant

g=9.8 ms2

v y=v yo−g t

y= yov yo t−12g t2

v y2=v yo

2 −2g x

FREEFALL

A runner with a mass of 60 kg is in the starting blocks to run a 300 meter race along a straight track. The starter's gun goes off at the time t = 0.0 s. The runner starts from rest and accelerates with a constant acceleration of 1.20 m /s2 . 1. What is the time required for the runner to run the race?2. How fast is the runner running when he crosses the finish line?3. What is his velocity at the half way mark?4. Where is he at 2.0 seconds?

A runner with a mass of 60 kg is in the starting blocks to run a 300 meter race along a straight track. The starter's gun goes off at the time t = 0.0 s. The runner starts from rest and accelerates with a constant acceleration of 1.2 m /s2 . 1. What is the time required for the runner to run the race?

x=xovo t12a t 2

x=0012a t2

x=12a t2

t=2 xa=2300m

1.2 ms2

=22.36 s

A runner with a mass of 60 kg is in the starting blocks to run a 300 meter race along a straight track. The starter's gun goes off at the time t = 0.0 s. The runner starts from rest and accelerates with a constant acceleration of 1 m /s2 .

2. How fast is the runner running when he crosses the finish line?

v2=vo22a x

v2=2a x

v=2a x=21.20 ms2 300m=26.83 m

s

A runner with a mass of 60 kg is in the starting blocks to run a 300 meter race along a straight track. The starter's gun goes off at the time t = 0.0 s. The runner starts from rest and accelerates with a constant acceleration of 1.2 m /s2 .

2. How fast is the runner running when he crosses the finish line?

A second method: We know that it took 22.36 s to complete the race?v=voat

v=01.20ms

22.36 s

v=26.83ms


3. What is his velocity at the half way mark?

v2=vo22a x

v2=2a x

v=2a x=2 1.20 ms2 150m=18.97 m

s


4. Where is he at 2.0 seconds?

x=x ovo t12a t2

x=12a t2

x=121.2ms2 2.0 s 2=2.4m


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Acelleration vs. Time

Time(s)

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/s2

)


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Velocity vs. Time

Time(s)

v (m

/s)


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Position vs. Time

Time(s)

x (m

)

fundamental physics i

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