Chapter 1
Introduction and
Mathematical Concepts
1.1 The Nature of Physics
Physics predicts behavior of nature  applications
Newton’s Laws → Rocketry
Maxwell’s Equations → Telecommunications
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1.2 Units
units
meter (m): unit of length
kilogram (kg): unit of mass
second (s): unit of time
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1.3 The Role of Units in Problem Solving
CONVERSION OF UNITS
1 ft = 0.3048 m
1 mi = 1.609 km
1 hp = 746 W
1 liter = 10-3 m3
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1.3 The Role of Units in Problem Solving
Example 1 The World’s Highest Waterfall
The highest waterfall in the world is Angel Falls in Venezuela,
with a total drop of 979.0 m. Express this drop in feet.
Since 3.281 feet = 1 meter, it follows that
(3.281 feet)/(1 meter) = 1
 3.281 feet 
Length  979.0 meters 
  3212 feet
 1 meter 
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1.3 The Role of Units in Problem Solving
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1.3 The Role of Units in Problem Solving
Example 2 Interstate Speed Limit
Express the speed limit of 65 miles/hour in terms of meters/second.
Use 5280 feet = 1 mile and 3600 seconds = 1 hour and
3.281 feet = 1 meter.
feet
 miles 
 miles  5280 feet  1 hour 
Speed   65
11   65


 95
hour 
hour  mile  3600 s 
second


feet 
feet  1 meter 
meters


Speed   95
1   95

  29
second
 second 
 second  3.281 feet 
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1.3 The Role of Units in Problem Solving
DIMENSIONAL ANALYSIS
[L] = length
[M] = mass
[T] = time
Is the following equation dimensionally correct?
x  vt
1
2
2
L 2
L   T  LT
T 
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1.4 Trigonometry
ho
sin  
h
 ho 
  sin  
h
1
ha
1  ha 
cos  
  cos  
h
h
Pythagorean theorem:
h h h
2
2
o
ho


h

1
o
tan  
 


tan
ha
 ha 
2
a
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1.4 Trigonometry
 ho 
  tan  
 ha 
1
 2.25m 

  tan 
  9.13
 14.0m 
1
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1.5 Scalars and Vectors
A scalar  a single number:
temperature, speed, mass
A vector  magnitude and direction:
velocity, force, displacement
8 m/s
4 m/s
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1.6 Vector Addition and Subtraction
3m
5m
8m
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1.6 Vector Addition and Subtraction
R  2.00 m   6.00 m 
2
R
2
2
2.00 m  6.00 m
2
2
 6.32m
R
2.00 m
6.00 m
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1.6 Vector Addition and Subtraction
tan   2.00 6.00
  tan
1
2.00 6.00  18.4

6.32 m
2.00 m

6.00 m
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1.6 Vector Addition and Subtraction

B
 
AB

A

A

B
 
AB
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1.7 The Components of a Vector
Vector components
 

A  Ax  Ay
scalar components

A  Ax xˆ  Ay yˆ
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1.7 The Components of a Vector
Example
A displacement vector has a magnitude of 175 m and points at
an angle of 50.0 degrees relative to the x axis. Find the x and y
components of this vector.
sin   y r


y  r sin   175 m sin 50.0  134 m
cos   x r
x  r cos   175 mcos 50.0   112 m

r  112 mxˆ  134 myˆ
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1.8 Addition of Vectors by Means of Components
  
C AB

B  Bx xˆ  By yˆ

A  Ax xˆ  Ay yˆ
C y  Ay  By

C  Ax xˆ  Ay yˆ  Bx xˆ  B y yˆ
Cx  Ax  Bx
  Ax  Bx xˆ  Ay  B y yˆ
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