Transcript
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:
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:
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
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
20ms=?mph
20ms 1mile1609m 3600 s
1hr =44.75mph
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
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
Distance vs. Displacement:
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
Distance vs. Displacement:
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
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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
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y
x=60o
R=10m@60o
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y−axis
x−axisz−axis
i
j
k
Vector Notation
31
i
j Vector Notation
A=4m i
B=3m j
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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
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x
yr=x 2 y2
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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
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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
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i
j Vector Notation
4m
3m
R=ABR=4m i3m jR=5m@36.87o
∣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
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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∣@A=5m@53.13o
∣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 x iB y j=−4m i2m j
B
∣B∣=−4m 22m 2=4.47m
III
IVIII
53
Vector Notation
i
j
B
Bx=−4m
B y=2m
B=B x iB y j=−4m i2m j
∣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=B x iB y j=−4m i2m j
∣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∣@B=4.47m@153.43o
∣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
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Vector Notation
i
j
C
C x=−3m
C y=−6m
C=C x iC y j=−3m i−6m jC=∣C∣@C=6.71m@251.57oII
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
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Adding Vectors
5m
5m
−5m
−5m A
B
A=5m i0m jB=0m i6m j
C C=AB=5m i6m j
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Adding Vectors
5m
5m
−5m
−5m A
B
A=5m i0m jB=0m i6m j
C C=AB=5m i6m j
∣C∣=5m 26m 2=7.81m
C=tan−16m5m =50.19o
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Adding Vectors
5m
5m
−5m
−5m A
A=5m i0m jB=3m i5m j
B
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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
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Adding Vectors
5m
5m
−5m
−5m A
B
A=5m i0m jB=−5m i5m j
70
Adding Vectors
5m
5m
−5m
−5m A
B
A=5m i0m jB=−5m i5m j
71
Adding Vectors
5m
5m
−5m
−5m A
B
A=5m i0m jB=−5m i5m j
C
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Adding Vectors
5m
5m
−5m
−5m A
B
A=5m i0m jB=−5m i5m j
C
C=AB=0m i5m j
∣C∣=0m 25m 2=5.00mC=90.00o
73
Adding Vectors
5m
5m
−5m
−5m
AB
A=7m i3m jB=−5m i5m j
74
Adding Vectors
5m
5m
−5m
−5m
AB
A=7m i3m jB=−5m i5m j
C
75
Adding Vectors
5m
5m
−5m
−5m
AB
A=7m i3m jB=−5m i5m j
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
A=7m i3m jB=−5m i5m j
C
C=4m i−2m j
D=ABC=6m i6m j
79
Adding Vectors
5m
5m
−5m
−5m
AB
A=7m i3m jB=−5m i5m j
C
C=4m i−2m j
D=ABC=6m i6m jAB
80
Adding Vectors
5m
5m
−5m
−5m
A=7m i3m jB=−5m i5m j
C
C=4m i−2m j
D=ABC=6m i6m jAB
ABC
81
Adding Vectors
5m
5m
−5m
−5m
A=7m i3m jB=−5m i5m j
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
Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.
v y=5m / s
x=100m
y=50mv
86
x=100m
Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.
v y=5m / s
x=100m
y=50mv
v=v x iv y jv=10m / s i5m / s j
87
x=100m
Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.
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
Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.
v y=5m / s
x=100m
y=50mv
v=11.18m /s@26.57o
r= x 2 y 2
r=100m 250m 2=111.80m
89
x=100m
Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.
v y=5m / s
x=100m
y=50mv
v=11.18m / s@26.57o
r=111.80m
v= r t
=111.80m10 s
=11.18m / s
90
x=100m
Tugboat with broken rudder. Can only go straight. It takes ten seconds to cross lake. Now consider strong current.
v y=5m / s
x=100m
y=50mv
v=11.18m / s@26.57o
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
v=17.20m / s@54.46o
92
Adding Vectors
10m / s
10m / s
−10m / s
−10m / s
v=17.20m / s@54.46o
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
Vector (Cross) ProductVector (Cross) Product
100
Vector (Cross) ProductVector (Cross) Product
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
1)Travel 6 m in 3s.2)Travel 24 m in 3s.3)Stop for 2 s.4)Travel 10 m in 2s.
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)
1)Travel 6 m in 3s.2)Travel 24 m in 3s.3)Stop for 2 s.4)Travel 10 m in 2s.
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.
v=x f−x it f−t i
=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.
v=x f−x it f−t i
=30.0m−30.0m8.0s−6.0 s
=0.0m /s
4) Travel 10 m in 2s.
v=x f−x it f−t i
=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
Special Case: Acceleration is constant.
a=constantv=voat
x=xovo t12a t2
v2=vo22a x
Special Case: Acceleration is constant.
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
Special Case: Acceleration is constant.
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
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x−x ot
=vvo
22x−xo=vvo t
2x−xo=[voat ]vo t
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x−x ot
=vvo
22x−xo=vvo t
2x−xo=[voat ]vo t2 x−xo=vo tat
2vo t
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x−x ot
=vvo
22x−xo=vvo t
2x−xo=[voat ]vo t2 x−xo=vo tat
2vo t2x−xo=2vo tat
2
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x−x ot
=vvo
22x−xo=vvo t
2x−xo=[voat ]vo t2 x−xo=vo tat
2vo t2x−xo=2vo tat
2
x−xo=vo t12at 2
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x−x ot
=vvo
22x−xo=vvo t
2x−xo=[voat ]vo t2 x−xo=vo tat
2vo t2x−xo=2vo tat
2
x−xo=vo t12at 2
x=xovo t12at 2
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t2
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t2 v=voat
t=v−voa
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t2
x=xovo v−voa
12a v−voa
2
v=voat
t=v−voa
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t2
x−xo=vo v−voa
12a v−voa
2
v=voat
t=v−voa
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
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
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t 2
x−xo=vov−voa 12av−voa
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
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t 2
x−xo=vov−voa 12av−voa
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
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
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
Special Case: Acceleration is constant. x i=xo , vi=voat t i=0.0s
a=constantv=voat
x=xovo t12a t2
x−xo=vov−voa 12av−voa
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
Special Case: Acceleration is constant.
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
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 .
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
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 .
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
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 .
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
1.2
1.4
Acelleration vs. Time
Time(s)
a ( m
/s2
)
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 .
0 5 10 15 20 250
5
10
15
20
25
30
Velocity vs. Time
Time(s)
v (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 .
0 5 10 15 20 250
50
100
150
200
250
300
350
Position vs. Time
Time(s)
x (m
)
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 .
0 5 10 15 20 250
50
100
150
200
250
300
350
Position vs. Time
Time(s)
x (m
)
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 .
0 5 10 15 20 250
50
100
150
200
250
300
350
Position vs. Time
Time(s)
x (m
)
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