BLACKBOARD COURSE
(Developed by Dr. V.S. Boyko)
PHYS 1433
PHYSICS 1.2
SCI Core
4 cl hrs, 2 lab hrs, 4 cr
Basic concepts and principles of mechanics, heat and sound for liberal arts
students and technology students. Cover statics, kinematics, dynamics, work
and energy, circular and rotational (?) motion, fluids, temperature, heat
transfer and wave motion.
CONTENT
1. Introduction……………………………………………………
2. Kinematics in One Dimension……………………………….
3. Vectors; Kinematics in Two Dimensions……………………
4. Dynamics: Newton’s Law of Motion…………………………..
5. Static Equilibrium………………………..……………………
6. Circular Motion; Gravitation…………………………………..
7. Work and Energy………………………………………………
8. Linear Momentum……………………………………………...
9. Fluids……………………………………………………………
10 Temperature, Thermal Expansion, Ideal Gas Law, and Kinetic
Theory…………………………………………………………
11. Heat…………………………………………………………….
12.The Laws of Thermodynamics………………………………….
13.Vibrations and Waves……………………………………………
14.Sound…………………………………………………………….
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1. INTRODUCTION.
Physics is the most basic of the sciences. It deals with the behavior and
structure of matter. It is the most advanced science and base of all sciences.
Physics formed technology basis of our civilization. Physics describe all
levels of Nature from elementary particles to the Universe at whole. Why
Physics is so dramatically successful? The answer is the scientific method
developed in physics and then used by all other sciences. This method
consists of three parts: observations of natural events, which include the
design and carrying out of experiments; the creation of theories to explain
the observations; testing of theories to see if their predictions are confirmed
by the experiments. Despite the mathematical beauty of some of its most
complex abstract theories, including of those of elementary particles and
general relativity, physics is above all an experimental science. Any
beautiful and flawless from the mathematical point of view theory could be
accepted if it is supported by experiment. General statements about how
nature behaves are called the law. The building blocks of physics are the
physical quantities that used by physicists to express the laws of physics.
Among them are length, time, speed, acceleration, mass, force and so on.
Laws of physics usually have the form of a relationship or equation between
the physical quantities.
It was written above that physics is experimental study and results from
measurements. But, at the same time, no measurement is absolutely precise.
There is an uncertainty associated with every measurement. This
uncertainty stemmed from the limited accuracy of our instruments, from
uncontrollable changing of the conditions of measurements and so on. When
giving the result of a measurement, it is important to show estimated
uncertainty. If the estimated uncertainty for a physical quantity A is a then
the result of measurement can be written as A  a. This means that the actual
value of A is most likely lies between (A – a) and (A + a). The percent
uncertainty is (a/A)  100%. Other form of stating uncertainty often is used
by keeping in a quantity the correct number of significant figures.
Significant figures are the number of reliably known digits in a numerical
value of physical quantity. The significant figures of the experimentally
measured value include all numbers that can be read directly from the
instrument scale plus one estimated number. So the last digit is estimated.
When doing calculations with measured physical quantities, you need to
follow the rule that final result of a multiplication or division should
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have only as many digits as the number with the least number of
significant figures used in the calculation. The same we can say for the
addition or subtraction of the numerical values of physical quantities: the
final result is no more accurate that the least accurate number used. Using a
calculator, remember that not all produced digits are accurate.
However, to obtain the most accurate result, you can keep all calculator
digits in intermediate results and round off final result to the proper number
of significant figures. Remember that it is treated as a mistake when
students write final result of the problem with all digits displayed on the
student’s calculator. Give final result only in terms of significant figures.
Remember, this is physics. Precision of the result is determined by the
precision of the experimental data. We could not increase precision of result
by more precise calculations.
It is not clear sometimes how many significant figures some number has. For
example, if there are zeros at the end of the number, we need additional
information to know whether these zeros are significant digits or just zero
holders. The ambiguity can be avoided if we write numbers in “scientific
notation” or “powers of ten”. To write a number in scientific notation,
express it as a number containing exactly one nonzero digit to the left of the
decimal point (total number of digits should be exactly equal to the number
of significant digits of this physical quantity) multiplied by the appropriate
power of ten. This approach is extremely useful in physics because physics
describe nature at all levels from elementary particles to universe at whole.
Studying physics, we will encounter in very small or very large numbers.
For example, the mass of electron in regular notation can be written as me =
0.00000.... (30 zeros in all)…000911 kg. Mass of the Sun can be written as
M = 19900…(28 zeros in all)…00… kg. In scientific notations these
physical quantities will be written as following: me = 9.1110^(-3); M =
19910^(+30) kg.
The founder of scientific method Galileo Galilei stated, “the book of Nature
is written in math”. Therefore Mathematics is extremely important in
Physics. Our course is algebra-based course. We will need some basic
concepts of elementary geometry and trigonometry as well. All mathematics
that we need is described in the Mathematical Review (see the section
Appendices in the textbook).
Measurement
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The experiment is first of all the measurement of physical quantities.
Measurement means comparison of measured quantities with some
standards of these quantities. It is therefore critical that those who make
precise measurements be able to agree on standards in which to express the
results of those measurements, so that they can be communicated from one
laboratory to another and verified. We begin our study of physics by
introducing some of the basic units of physical quantities and the standards
that are accepted for their measurement.
If you will analyze the textbook you will see that in each chapter new
physical quantities are introduced and accordingly new units should appear.
How can we organize this multitude of units? But it is very interesting that
there are few base units through which all other units (derived units) can be
expressed. It looks like a mystery. We will try to explain this mystery on the
example of the first branch of physics that we will study - Mechanics.
Mechanics study the motion of material objects in space and time. The base
quantities of mechanics are length (the characteristic of space), mass (the
characteristic of matter), and time. All other units in mechanics result from
multiplication and /or division of fundamental units. Derived unit can be
determined from the formula that defined corresponding physical quantity.
For example, unit of speed can be determined from the corresponding
formula
EXAMPLE 1.1. Speed Definition
Speed = (distance covered)/(time needed)
(1.1)
We might operate with units by the same way as with numbers. Therefore to
find unit of speed we should divide a unit of length by a unit of time.
System of Units
Depending of what units were chosen as the base units, there are different
systems of units. We will use international system of units, which is
abbreviated SI. In the SI system, the base units: unit of length is 1 meter
= 1 m; unit of time is 1 second = 1 s; unit of mass is 1 kilogram = 1 kg.
Because meter is a base unit of this system, it SI system sometimes is called
a metric system. In the United States, British Engineering System is in use
for every day life and partially in technology. The base units in this system
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are as following: the unit of length is 1 foot = 1; the unit of time is 1 second
= 1 s; the unit of force 1 pound = 1 lb.
Usually we will use the SI (metric) system because it has some advantages.
First of all, this system of units is used in science through the all world.
Second, in the SI system, the larger and smaller units are defined in
multiples of 10 from the standard unit, and this makes calculation easy.
Adding a prefix to the name of the fundamental unit derives the names of the
additional units. For example, the prefix “kilo”, abbreviated k, always means
a unit larger by a factor of 1000:
1 kilometer = 1 km= 10^3 meters = 10^3 m
1 kilogram = 1 kg = 10^3 grams =1 10^3 g
Her are several examples of the use of multiples of 10 and their prefixes
with the units of length, mass, and time.
LENGTH:
1 nanometer = 1 nm = 10^(-9) m (a few times the size of the largest atom)
1 micrometer = 1 m = 10^(-6) m (size of some bacteria and living cells)
1 millimeter = 1 mm = 10^(-3) m (diameter of the point of a ballpoint pen)
1 centimeter = 1 cm = 10^(-2) m (diameter of your little finger)
Compare these relationships with relationships between units of the length in
the British Engineering System: 1 mi = 5280 ft, 1 ft = 12 in.
TIME:
1 ns = 10^(-9) s (time for light to travel 0.3 m)
1 s = 10^(-6) s (time for a computer to perform few addition operations)
1 ms = 10^(-3) s (time for sound to travel 0.33 m in the air)
MASS:
1 g = 10^(-9) kg (mass of a very small dust particles
1 mg = 10^(-3) g = 10^(-6) kg (mass of a grain of salt)
1 g = 10^(-3) kg (mass of a paper clip)
The prefixes can be applied to any other metric units. For example,
1 megawatt = 1 MW = 10^6 watts = 10^6 W.
STANDARDS OF UNITS
For any unit we use, we need to define a standard, which defines exactly this
unit value. It is important that standards be reproducible and accessible.
With the progress of the methods of measurement, more precise standards
are introduced. The good example is the history of unit of length in SI
system -- meter that evolved over the years. When the metric system was
established in 1791 by the French Academy of Sciences, the meter was
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defined as one ten-millionth of the distance from the North Pole to the
equator. A platinum rod to represent this length was made. The last
definition of the meter was established in 1983: “The meter is the length of
path traveled by light in vacuum during a time interval o 1/(299,792,458) of
a second. This provides a much more precise standards of length than
before. But you can see that it includes the new standard of time. For many
years the second was defined as 1/(86,400) of a mean solar day. The present
standard, adopted in 1967, is much more precise. It is based on an atomic
clock, which uses the energy difference between two lowest energy states of
the cesium atom. When bombarded by microwaves of precisely the proper
frequency, cesium atoms undergo a transition from one of these states to the
other. One second is defined as the time required for 9,192,631,770 cycles of
this radiation.
Unit Conversion
Remember, that physical quantity must always include number and
unit. When you will solve problems, always write as for any physical
quantity the numerical value and unit. Do not forget to write unit. Physical
quantity written without unit is treated as a mistake disregarding whether its
numerical value is wrong or write. Solving problems, start with analysis of
data. You can get write answer for a problem only if you use consistent
system of units. It means that all physical quantities in the problem must be
expressed in the units of the same system of units. If it is not so you should
make conversion of units. How the conversion of units can be made?
Units are multiplied and divided just like ordinary algebraic symbols. This
gives us an easy way to convert a quantity from one system of units to
another. The key idea is that we can express the same physical quantity in
two different units and form equality. For example, When we say that 1 m =
3.28 ft, we do not mean that number 1 is equal to the number 3.28; we mean
that 1m represents the same length as 3.28 ft. For this reason, the ratio
(1m)/(3.28 ft) equals 1, as does its reciprocal (3.28 ft)/(1m). Since
multiplying or dividing by 1 does not affect the value of any quantity, we
can multiply any physical quantity by either of these factors without
changing this quantity’s physical meaning. These ratios (1m)/(3.28 ft) =1 or
(3.28 ft)/(1m) = 1 are called conversion factors. To illustrate the technique of
converting from a base unit of one system of units to a base unit of another,
we will solve the following example.
EXAMPLE 1.2: The Empire State Building versus the Eiffel Tower.
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(a) The Empire State Building in New York City is 1250 ft high without its
television tower. Express the height of the Empire State Building in meters.
(b) The Eiffel Tower in Paris is approximately 300 m tall. Express the height
of the Eiffel Tower in feet.
You should take the relationship between units in question from textbook.
(A Table containing many relationships between units can be found inside
the front cover of this book). In our case, it is 1m = 3.28 ft. Then prepare
from this relationship two conversion factors :(1m)/(3.28 ft) =1 and (3.28
ft)/(1m) = 1. Choose the right conversion factor, this is one in which
desirable unit is in the numerator. Multiply the given height by conversion
factor. Because units may be treated as any algebraic quantity they can be
cancelled. If you do unit conversions correctly, unwanted units (feet in the
question (a) of our example, and meters in the question (b) of this example)
would be cancelled out.
(a) (1250 ft)  (1m)/(3.28 ft) = 381 m.
(b) (300 m) (3.28 ft)/(1 m) = 984 ft.
If instead you had multiplied by wrong conversion factor you would get
senseless answer.
(1250 ft) (3.28 ft)/(1 m) = 4100 (ft)^2/(1 m)
When you need to convert derived unit from one system to another, prepare
so many conversion factors how many base units in derived unit you should
convert. Multiply given quantity by all these conversion factors. Undesirable
units should be cancelled out. To illustrate the technique of converting from
a derived unit of one system of units to a derived unit of another, we will
solve the following example.
EXAMPLE. 1 3. Speed limit. According to “Driver’s Manual”, you must
obey the posted speed limit, or if no limit is posted, drive no faster than 55
mi/h (55 miles per hour). Express this speed limit (a) in km/h and (b) in m/s.
(a) In this case we need to convert only one base unit. Therefore we
should use only one conversion factor to transfer mi to km. From
textbook 1mi = 1.609 km. From this relationships we can get two
conversion factors: (1 mi)/(1.609 km) = 1 and (1.609) and (1.609
km)/(1 mi) =1. To get rid of undesirable unit we should use second
conversion factor
55 (mi)/(h)  (1.609 km)/(mi) = 88. 5 (km)/(h)
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Desirable unit (in this case mi) disappear and we get answer in (km)/(h).
(b) Now we need to convert two units, so we should prepare conversion
factors for each of them. The right factors are (1000 m)/(I km) = 1 and
(3600 s)/(1 h) = 1. The answer will be
88.5 (km/h) (1000m)/(1 km)(1 h)/(3600 s) = 24.6 m/s
This is important statement, and it will be repeated again. When a
problem requires calculations using numbers with units, always write the
numbers with the correct units and carry the units through the calculation.
This provides a very useful check for calculations. If at some stage in a
calculation you find, that an equation or an expression has inconsistent
units, you know you have made an error somewhere. This technique is
called Dimensional Analysis. The dimensions of a quantity are type of
units or base quantities that make it up. For example, the dimensions of a
volume is always length cubes, abbreviated {L^3], Dimension of speed
are always a length [L] divided by time [T], so [v] = [L]/[T]. In any
equation, expressing physical relationships, after the mathematical
procedures of addition, subtraction, multiplication, division, and
cancellation have been performed, the units on one side of equation must
equal the units on the other side, the same statement can be said for each
term in equation.
These are some basic concepts of the language of physics, now we will
begin to study real physics. We will start with the branch of physics called
Mechanics. Mechanics describe a motion.
2. DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION.
Mechanics, the oldest branch of physics, is the study of the motion of
material objects in space and the related concepts of force, work and energy.
Mechanics is consists of two main parts: kinematics and dynamics.
Kinematics describes how objects are moving including geometry of
motion: trajectory (imaginary line along which the object is moving);
position of an object on this trajectory; how this position is changing with
time and so on. Dynamics try to answer question why objects are moving
including the causes of motion.
The motion of a material object could be very complicated. Because of this
we will start with the simplest case: the translational motion of a particle
along the straight line. Translational motion means motion without rotation.
In the translational motion all points of an object are moving by the same
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way and we can consider this motion as the motion of a particle –object
without size.
The motion can be measured as a change of position with respect to some
particular frame of reference. In physics, the coordinate systems are used as
a frame of reference. Coordinate system used more often in physics is the
Cartesian coordinate system firstly introduced by French philosopher
mathematician Descartes when he developed analytical geometry
(representation of geometrical objects by some analytical expressions) by
merging algebra and geometry. For three-dimensional space (3D case)
Cartesian system of coordinate represents by three rectangular axes X, Y, Z
emerging from one point that is called origin of coordinate. It would be
very difficult for us to start our study with consideration of motion in general
3D dimensional case. Motion in a plane (2D case) it is also difficult to
consider now. Because of this we will start with 1D case – the motion along
straight line. It is not just abstraction that we choose only because of the
simplicity. The motion along straight part of a highway is a good application
of this consideration.
Thus we have the origin of coordinates 0. We have coordinate axis
emerged from this origin. One direction we designate as +X, opposite as –X.
We will choose appropriate scale and plot distance from the origin of
coordinates on the axis. Positive in the direction +X, negative in the
direction –X. We introduce the physical quantity: the position of an object.
It is designated by x. The unit in which we measure the position in SI
system is meter. Its x coordinates give the position of an object at any
instant of time.
The change in position is called displacement. If an object at an instant of
time t1 has the position x1 and at instant of time t2 its position changes to x2
the displacement of this object can be written as follows
x = x2 – x1
(2.1)
Symbol  designates the change in a physical quantity that is written just
after this symbol, namely final value of a physical quantity minus initial
value of this quantity. This change in position occurs during interval of time
t = t2 – t1
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(2.2)
Distinguish between distance covered by an object and its displacement. For
example, object moved 100 m ahead, then it came back exactly to the same
place. Distance covered by this object is 100 m, but its displacement is 0.
Units for displacement measurement are units of length. In SI system the
unit of displacement is meter. Displacement is the vector quantity (we will
study vector quantities in details later in our course). This means that it is
characterized not only by numerical value (magnitude) but also by direction
in space. In 1D case only two directions are possible. When an object is
displaced in the positive direction of the X-axis, so x2 > x1, and
x > 0, displacement is positive. When an object is displaced in the negative
direction of the X-axis, then x2 < x1, and x < 0, displacement is negative.
One of the most important features of an object in motion is how fast it is
moving. We introduced the physical quantity speed by relationship (1).
Speed is a useful concept, because it indicates how fast an object is moving.
However, speed does not reveal anything about the direction of the motion.
To describe both how fast a motion is and what is its direction we need
vector quantity. It is called the velocity. Velocity signifies the magnitude
(numerical value) of how fast an object is moving and also the direction of a
motion. The average velocity is defined in terms of displacement (not in the
terms as the distance traveled as in the concept of the speed):
Average velocity = displacement/(time elapsed)
(2.3)
SI unit of average velocity is meter/second (m/s).
As definition of the average velocity we can use the following expression:
v = x / t
(2.4)
Where v is the average velocity (bar over a symbol of any physical quantity
signifies the average value).
SI unit for average velocity is meter/second (m/s)
We assume that our clocks always running forward (t2 -- t1 > 0), then the
average velocity is positive when object is moving in the positive direction
of X-axis and negative otherwise.
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The concept of the average velocity is not productive for each situation. For
example, suppose you are living at the distance 30 mi from college. You
reach college by car through 1 hr. Suppose once upon a time you were late,
try to be in time and would pass other car. Police officer stop s your car and
gives you the ticket. If you would like tell him that your average speed v =
30 mi/h is smaller than speed limit and there were no violation of traffic
rules, you would fail. Therefore having the concept of an average velocity
could not describe all features of the motion. We need to introduce the
concept of an instantaneous velocity, which is the velocity at any instant of
time. Strict definition of the instantaneous velocity is as the following; the
instantaneous velocity at any instant of time is the average velocity during
an infinitesimally short interval of time. Symbolically this definition can be
written as following
v = lim t0 (x / t)
(2.5)
The notation lim t0 (x / t) means that the ratio (x / t) is defined by the
limiting process in which smaller and smaller values of t are used, so small
that they tend to approach zero. But we would not divide x by zero,
because x also tends to zero, but their ratio does not become zero. It
approaches the instantaneous velocity. By analogy, we can introduce the
concept of instantaneous speed –- speed at an instant of time. Because the
distance covered and the magnitude of the displacement becomes the same
when they tend to be infinitesimally small, instantaneous speed always
equals the magnitude of the instantaneous velocity. Therefore you can derive
the magnitude of instantaneous velocity from the information shown on the
speedometer of a car.
SI unit of instantaneous velocity (and of instantaneous speed) is
meter/second (m/s).
Students often ask, how we will use the formula (2.4) when solving
problems. Actually we will not use this representation of instantaneous
velocity solving problems. It is useful to understand the concept of the
instantaneous velocity. For brevity, we will use the word velocity to mean
instantaneous velocity in this course.
The velocity of a real moving object may change in a number of ways. For
example, the car moving along a straight street should be brought to stop
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before the red traffic light and accelerated after changing traffic light from
red to green. To describe this changing of velocity in time we introduce the
concept of average acceleration. By analogy to average velocity it can be
written as following:
a = v / t
(2.6)
By analogy to instantaneous velocity we introduce the concept of an
instantaneous acceleration – acceleration at the current instant of time:
a = lim t0 (v / t)
(2.7)
SI unit for average acceleration and for instantaneous acceleration is
meter/(second squared) or (m/s^2).
When the velocity of an object increases, acceleration is positive and
directed in the same direction as velocity. When velocity decreases,
acceleration is negative and directed in the direction that is opposite to the
direction of the velocity. In this case acceleration sometimes is called
deceleration.
There are a lot of situations in which acceleration changes. But we will not
introduce new physical quantity that describes the rate of change of
acceleration. We will choose more productive way. We will consider one
specific case – motion with constant acceleration along the straight line. This
is specific case but there are important situations in which acceleration is
constant. The examples: the motion of a breaking car; the motion of falling
objects near the surface of the Earth and so on. However, keep in mind that
this is special situation, and the results that we will derive now are
applicable only to the case when a = const. Examples of cases with non
constant acceleration: a swinging pendulum bob; raindrop falling against air
resistance and so on.
To achieve our goal, we will use general formulas (2.4) and (2.6) but specify
them to the considered situation. If a = const, then a = a. To simplify
consideration, we will suppose that object start at instant of time t 1 = 0. Its
position at this instant of time is called an initial position and designated as
x0. The position of an object at time in question (current time t) is designated
just as x. The velocity of an object at an instant of time t1 = 0 is designated as
v0 and it is called an initial velocity. The velocity of an object at time in
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question (current time t) is designated just as v. using these designations we
can rewrite the equations (2.4) and (2.6) as the following:
v = x / t = (x – x0) / t
a = a = v / t = (v – v0)/t
(2.8)
(2.9)
We will use now these equations to derive a set of equation that relate the
position x, velocity v and acceleration a with time, velocity v and
acceleration a with position for the case when a = const. From (2.6) we can
derive the equation that relates velocity, initial velocity, and acceleration
with time
v = v0 + a t
(2.10)
This is the first equation from the set of desired equations. In the case of
motion with the constant acceleration, velocity of an object is directly
proportional to the time.
The displacement at the time t can be obtained from equation (2.8)
x = x0 + v t
(2.11)
When a = const, velocity accordingly to (2.10) is directly proportional to the
time (increases a constant rate). Therefore the average velocity will be
midway between initial and final velocities and (2.11) can be written as the
following.
v = (v + v0) / 2
(2.12)
Now we can substitute v in (2.11) by (2.12)
x= x0 + [(v + v0) / 2] t
(2.13)
Substituting v in (2.13) by expression (2.10) we finally get expression that
relates position and time:
x= x0 + v0 t + (1/2) a t^2
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(2.14)
This equation allows us to find displacement of an object at any instant of
time if x0, v0, and a are known. In some problems, time is not known.
Nevertheless, we can find displacement using expression (2.13). We will
exclude time from this expression. The expression for time derived from
(2.10) is as following:
t = (v – v0) / a
(2.15)
Substituting t in (2.13) by expression (2.15), we will get
x = x0 + (v^2 – v0^2)/ (2 a)
(2.14)
Solving this expression for v^2, we will get equation
v^2 = v0^2 + 2 a (x – x0)
(2.15)
If we know initial velocity v0, initial position x0, and acceleration we can
find the velocity v at any position x.
Thus, we derive the beautiful set of kinematical equations completely
describing motion of an object along the straight line with the constant
acceleration. We will collect all these equations together:
Motion with constant acceleration: a = const
v = v0 + a t
(2.16a)
x= x0 + v0 t + (1/2) a t^2
(2.16b)
v^2 = v0^2 + 2 a (x – x0)
(2.16c)
v = (v + v0) / 2
(2.16d)
_____________________________________________________________
Equations (2.16a) and (2.16b) are useful for analysis of kinematics as an
initial value problem: the acceleration and the initial conditions (x0, v0) we
can find velocity v and position x at any instant of time. Sometimes in
problems, time is not given but there is information about position of an
object. In these cases we can find from equation (2.16c) velocity of an object
at any position. Equation (2.16d) can be useful in some cases. For example,
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if acceleration is not given, we can find position of an object combining
equations (2.11) and (2.16d).
GRAPHICAL ANALYSIS OF THE MOTION.
We used successfully the coordinate system to describe the motion in the
space. But during the motion all its characteristics also are changing in time.
But there is no direct possibility in this approach to analyze in parallel these
changes. Therefore along with the space information about a motion (the
coordinate analysis), another approach – graphical analysis of time
dependence of position, velocity, and acceleration is in a wide use. In
constructing a graph the physical quantity versus time, you should not be
afraid of graphs in physics. Remember, that these graph representations are
constructed by the sane way as you did in mathematical course that we
studied before. In these courses, you represented functional dependences
between variables. Absolutely the same you should do in physics.
Differences are only in the designations. For example, direct proportionality
between variables y and x analytically expressed as y = b + ax, where a and
b are constants, can be represented graphically as straight line (Fig. ). (This
is why this type of dependence sometimes is called linear dependence). If we
analyze Equation (2.16a), we can see that this equation represented the same
type of dependence between velocity v and time t. Analyzing Equation
(2.16b), we can deduce that the dependence of position x on time is
quadratic. This parabolic type of dependence can be represented by the
curve like parabola displaced from the origin of coordinate by x0 along the
positive direction of X-axis. The acceleration is supposed to be a constant at
any instant of time. Because of this the graph of a versus time is just a
straight line parallel to the t-axis. These graphs are shown on the Fig. 1, Fig.
2, and Fig. 3 correspondingly.
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Fig. 1. Graph of position as a function of time (x vs. t) for the motion with
constant acceleration.
Fig. 2. Graph of velocity as function of time (v vs. t) for the motion with
constant acceleration.
16
Fig. 3. Graph of acceleration as function of time (a vs. t) for the motion with
constant acceleration.
Take into account that the graphs are in the wide use in the all branches of
physics. They are a very convenient way to represent the overall trend of the
dependences between physical quantities.
Now we consider the some particular but important case of the motion of an
object with constant acceleration along straight line. Let us consider case
when acceleration is constant but equal to zero, namely a = const = 0. To get
the set of equation describing this case, we will put zero value of
acceleration into set of equations (2.16). As a result we will see that in this
case v is constant, and the set of equations describing the motion of an
object with constant velocity along the straight line can be written as
following:
____________________________________________________________
If acceleration is constant and zero (a = const = 0), then an object is moving
with constant velocity.
v = v0 = const
(2.17a)
x= x0 + v t
(2.17b)
17
_____________________________________________________________
We can complete these analytical relationships with the graphs of time
dependences of position, velocity, and acceleration. From equations (2.17),
we can deduce that in the case of motion with constant velocity we have
direct proportionality between position and time. Graph of velocity will be
just straight line parallel to t-axis. Graph of acceleration as function of time
will be straight line lying exactly on the t-axis. All these graphs are
represented on the Fig. 4, Fig. 5, and Fig. 6.
Fig. 4. Graph of position as a function of time (x vs. t) for the motion with
constant velocity.
18
Fig. 5. Graph of velocity as a function of time (v vs. t) for the motion with
constant velocity
Fig. 6. Graph of acceleration as a function of time (a vs. t) for the motion
with constant velocity.
19
Now we are ready to solve problems related to the motion with constant
acceleration, constant velocity, or a combination of these types of motion.
Before this we shortly outline some key steps of strategy used in solving
problems.
SOLVING PROBLEMS.
1. Read and reread the whole problem. The good method to check your
understanding of problem is as following. Try to formulate the problem by
yourself with closed textbook.
2. Draw a simple sketch of the situation outlined in the problem.
3. Usually sketch includes a coordinate system. You can choose the origin of
the coordinate system at any convenient location. Try to choose origin of
coordinate so that x0 = 0. In this case the equations of motion (2.16) and
(2.17) become somewhat simplified. You can choose either direction of
coordinate axis to be positive. Usually, it is more convenient to choose
positive direction of X-axis in the direction of motion. Remember that your
choice of the positive axis direction automatically determines the positive
directions for v and a. If x is positive to the right of the origin, then v and a
will be positive toward the right The choice of the origin and direction of the
coordinate axis must remain the same throughout the solution of the problem
in question.
4. The graphical analysis of motion sometimes is needed. It is especially
useful if an object changes the type of motion through some interval of time.
Or there are objects in problem that simultaneously are moving with
different type of motion.
5.Write down what quantities are known and then what you should find.
6. Try to derive all information from conditions both explicit and implicit.
Examples of implicit information: if in conditions is written that object starts
from rest, it means that v0 = 0; if in conditions is written that objects is
brought to stop, it means that v = 0; if there is no any mention in conditions
about the position of an object before the clock in the problem begins to run,
then x0 = 0.
7. Analyze units of the physical quantities given in conditions. If there is no
consistent system of units in the problem, make conversion of units
(preferable to the SI system). Transfer from subunits to the units if they even
belong to the same system of units (for example, if some distances in the
problem expressed in meters, other -- in kilometers, make all distances
expressed in meters).
20
8. Try to understand which principles of physics should be applied to solve
the problem (by other words, what formulas can be used to solve this
problem). Remember each formula that we use in our course has its own
range of validity. For example, Eq. (2.17) can be used only when object is
moving with constant velocity, Eq. (2.16) only when object is moving with
constant acceleration.
9. Try to choose an equation from the system of relevant equations that
contains only one from desirable unknowns. Solve the equation algebraically
for desirable unknown. Sometimes several sequential calculations should be
done.
10. Substitute known values and compute the values of the unknowns. Keep
one or two extra digits, but round off the final answers to the correct
numbers of significant figures.
11. In calculations, write numerical values together with units and keep track
of units. The units on each side of equality must be the same. If it is not so,
mistake has been made. Unfortunately, this analysis tells you only if you are
wrong, not if you are right. Because of this, important is the next step.
12. Analyze carefully your result. Is it reasonable or no? Apply your
intellect, experience and common sense. Remember, this is physics, not
mathematics. Mathematics sometimes can bring you unexpected solutions.
For example, the Eq. (2.16b) is quadratic equation with respect to t.
Therefore it has two solutions t1 and t2, but only one of these solutions is
physical solution. Analyze both of them with respect to conditions of the
problem and choose only one -- physical solution (solution that corresponds
conditions of the problem in question).
Let us solve some examples of the typical problems, using the
recommendations as a template.
EXAMPLE 2.1. Pontiac versus Toyota (starting from rest). Pontiac G6
GT starts from rest and after 7.9 s acquire velocity 60 mi / h. Toyota Camry
LE starts from rest and after 8.3 s acquire velocity 26,8 m / s. Suppose that
cars are moving with the constant acceleration. (a) Find the accelerations of
these cars. b) Compare these accelerations (find just ratios of corresponding
accelerations).
Situations considered in the problem are of the same type. It is written that
objects are moving with the constant acceleration Therefore we have right to
use set of formulas (2.16) and graphs shown in figures: Fig. 4, Fig. 5, Fig. 6.
Take into account implicit information about initial velocities. They equal
zero.
21
Now we can write what is given and what we want to find.
vp0 = 0
vp = 60 mi/h
tp = 7.9 s
vt0 =0
vt = 26.8 m/s
t = 8.3s
___________
(a) ap ?, at ?
(b) ap/at ?
Analysis of the data shows that there is no consistent system of units in the
problem. We need to make conversion. We can do it by multiplying vp by
two conversion factors.
vp = (60 mi/h) (1609 m/ 1 mi)( 1 h/ 3600 s) = 26.8 m/s
(a) Cars are moving with the constant acceleration. Therefore we will try to
choose an equation from the system of equations (2.16) that contains only
known quantities and desirable unknown. This is Eq. (2.16a). Solving this
equation algebraically for desirable unknown, we will get:
a = (v – v0)/t
Substituting in this formula data for Pontiac G6 GT and then data for Toyota
Camry LE we will get answer for question (a):
ap = (26.8 m/s – 0 m/s) / (7.9 s) = 3.39 m/ s^2 (x = 105m?)
at = (26.8 m/s – 0 m/s)/(8.3 s) = 3.23 m/s^2
(b)
ap / at = (3.39 m/s^2) / (3.23 m/s^2) = 1.050
EXAMPLE 2.2. Pontiac versus Toyota (slowing down). When Pontiac G6
GT is moving with the velocity 60 mi/h, it could be brought to stop through
a braking distance 141 ft. If Toyota Camry LE is moving with the velocity
26,8 m / s, it could be brought to stop through a braking distance 41.2 m.
Suppose that cars are moving with the constant acceleration. (a) Find the
accelerations of these cars. (b) Compare these accelerations (find just ratios
of corresponding accelerations).
22
Situations considered in this problem are of the same type. Objects are
moving with the constant acceleration Therefore we have right to use set of
formulas (2.16). Take into account implicit information about final
velocities. They equal zero. The principal difference with respect to the
Example 2.1 is that velocities now are decreasing during the motion. From
this fact we can deduce that both accelerations and slopes of the straight
lines in the graph are negative.
Write what is given and what we want to find.
vp0 = 60 mi/h = 26.8 m/s
vp = 0
xp = 141 ft
tp = 7.9 s
xp0 = 0
vt0 = 26.8 m/s
vt = 0
xto = 0
xt = 41.2 m
___________
(a) ap ? at ?
(b) ap/at ?
There is no consistent system of units in the problem, and there is necessity
to make conversion. We can do it by multiplying xp by conversion factor.
xp = (141 ft)(1 m / 3.28 ft) = 43.0 m
(a) Cars are slowing down with the constant negative acceleration. Therefore
we will try to choose an equation from the system of equations (2.16) that
contains only known quantities and desirable unknown. This is Eq. (2.16c).
Solving this equation algebraically for desirable unknown, we will get:
a = (v^2 – v0^2)/ [2(x---x0)]
Substituting in this formula data for Pontiac G6 GT and then data for Toyota
Camry LE we will get answer for question (a):
ap = -- 8.35 m / s^2
at = -- 8.72 m / s^2
23
(b)
ap / at = (--8.35 m/s^2) / (--8.72 m/s^2) = 0.958
Situations are more complicated when an object in the problems is moving
by different types of motion in turn.
EXAMPLE 2.3. A person driving her car Toyota Camry LE at constant
speed vr = 55 mi / h approaches intersection just as the traffic light turns red.
Reaction time to get her foot on the brake is tr = 1.0 s. During reaction time
the speed is constant, so the acceleration is zero: ar = 0. Then the driver uses
brakes. The car slows down with constant acceleration ab = -- 8.72 m / s^2
and comes to a stop. (a) Sketch the v – t graph of motion of the car. (b) What
is the distance covered by the car during the reaction time? (c) What is the
total stopping distance? “The Defensive Driving Course Guide”
recommends to have 346 ft total stopping xst’ = 346 ft distance for speed 55
mi / h. (d) Is this recommendation fulfilled in our example?
vr = 55.0 mi/h = const
tr = 1.00 s
ar = 0
ab = -- 8.72 m/s^2
_______________
(a) v – t graph?
(b) xr ?
(c) x st?
(d) xst > xst’ or xst < xst’?
Analysis of data shows that we need to perform conversion. After this we
will get: vr = 24.6 m/s, xst’ = 105 m.
(a) Preparing v – t graph of motion, we should take into account that part
of time car is moving with constant velocity. This period of time the
graph should be similar to graph from the Fig. 5. Next interval of time
the car is moving with constant acceleration. Therefore this part of the
graph should be similar to the graph in Fig. 2 but with negative slope
(acceleration is negative – car is slowing down)). As a result we will
get the graph shown in Fig. 2.7.
24
Fig.2.7. Example 2.3. Graph of velocity as function of time (v vs t).
We face here the situation in which the car is moving different parts of the
total distance according to the different laws of motion. Because of this we
could not to use the same formulas to describe motion at a whole. In those
cases, it is recommended to consider parts of motion separately. For each
part of motion we will use only formulas relevant for this specific type of
motion. During reaction time, car is moving with constant velocity.
Therefore the distance covered by car can be determined by means of Eq.
(2.17b). Note that the initial position for the motion during reaction time can
be chosen as zero.
xr = vr tr = (24.6 m/s) (1.00 s) = 24.6 m
(b) The output from the first part of motion can be used as an input to the
next part. Distance xr traveled during reaction time can be treated as
initial position xo for the next part –- motion with constant negative
acceleration. The velocity of the car during reaction time can be
treated as initial velocity for the second part of motion. Because this
motion is the motion with constant with constant acceleration, we can
use Eq. (2.16c) (more precisely, Eq. (2.14) that can be easily derived
from Eq. (2.16c)) in which we will take v0 = vr and x0 = xr.
x = x0 + (v^2 – v0^2)/ (2 a)  xst = xr + (0 -- vr^2)/ (2 a)
25
= 24.6 m + (--24.6 m/s)^2 / [(2)(-- 8.72m/s^2)]
xst = 59.2 m
(c)
xst < xst’
We can face another sequence of events in a problem that is represented in
the following example.
EXAMPLE 2.4. A Pontiac G6 GT accelerates from rest to v = 26.8 m/s,
moving 105 m away from the starting point. The acceleration is supposed to
be constant during this initial stage of motion. Then the car continues to
move 100 s with constant velocity v = 26.8 m/s. (a) Sketch the v – t graph of
motion of the car. Determine: (b) the acceleration during the initial stage of
motion; (c) the total distance traveled by the car.
v1 = 26.8 m/s
x01 = 0
v2 = 26.8 m/s = v1
x1 = 105 m
x02 = x1 = 105 m
tc = 100 s
______________________
(a) Graph v – t ?
(b) a ?
(c) xtot ?
(a) The motion of the car consists of two parts. The first one is the motion
with constant acceleration. Second one is the motion with constant
velocity. Preparing v – t graph of motion, we should take into account
that part of time car is moving with constant acceleration. This period
of time the graph should be similar to graph from the Fig. 2. Next
interval of time the car is moving with constant velocity. Therefore
this part of the graph should be similar to the graph in Fig. 5. As a
result we will get the graph shown in Fig. 2.8.
26
Fig.2.8. Example 2.4. Graph of velocity as function of time (v vs t).
We face here the situation in which the car is moving different parts of the
total distance according different laws of motion. Because of this we could
not to use the same formulas to describe motion at whole. In those cases, it is
recommended to consider parts of motion separately. For each part of
motion we will use only formulas relevant for this specific type of motion.
(b) Therefore the acceleration of the car during initial stage of motion can be
determined by means of Eq. (2.16c).
a = (v^2 – v0^2)/ [2(x --xo)] = [(26.8 m/s)^2 –0]/[2 (105 m –0)]
a = 3.42 m/s^2
(c) The output from the first part of motion can be used as an input to the
next part – motion of the car with constant velocity.
xtot = x1 + v tc = 105 m + (26.8 m/s) (100 s) = 2785 m
In problems, we can face sometimes situation, when two objects are moving
simultaneously but by different types of motion.
27
EXAMPLE 2.5. A speeding motorist traveling 33.5 m/s passes a stationary
police officer. The officer immediately begin pursuit at a constant
acceleration 3.00 m/s^2. (a) Sketch the x – t graph of motion of the car.
(b) How much time will it take for the police officer to catch the motorist?
(c) What is the distance each vehicle traveled at that point?
x0s = 0
x0p = 0
vs = 33.5 m/s = const
as = 0
vpo = 0
ap = 3.00 m/s = const
(a) x – t graph of motion?
(b) td?
(c) xd?
(a) The sketch of the situation in space attached to the coordinate axis and
the graph of position x versus time t will be very useful to solve this
problem.
x
Speeder
Police officer
t
t1
Fig.2.11. Example 2.5. Graph of position as function of time (x vs t).
(b) It is easy to see from this graph, that at t = td the distance covered by
speeder xds equals to the distance covered by the police officer xdp:
xds = xdp
Speeder is moving with constant velocity and covered distance xds is
described by the Eq. (2-17b). Police officer is moving with the constant
acceleration and covered by him distance xdp is described by the Eq.
(2.16b). Therefore we can find td using this circumstance.
x0s + vs td = x0p + v0p td + ½(ap) (td^2)
28
33.5 m/s td = ½(3.00m/s^2) (td^2)
td = 2 vs/a = 2 (33.5 m/s)/(3.00 m/s^2) = 22.3 s
(c) The covered distance can be easily found now using Eq. (17b) or Eq.
(16b). For example, using Eq. (17b) we will get
xd = vs td = (33.5 m/s) (22.3 s) = 747 m
FREE FALL MOTION.
Good example of the motion with constant acceleration is fall motion -motion of an object allowed to fall freely nears the Earth surface. This
simple type of motion attracted the attention of scientist from the ancient
times. Based on every day experience Aristotle supposed that velocity of the
object is proportional to the weight of the object. But Galileo disagreed with
this statement. Experiment shows that object dropped from larger height
drive a stake into the ground further than the same stone dropped from
smaller height. Galileo concluded that velocity is changing during free fall,
so it is accelerated motion. According to a legend, Galileo experimented by
dropping cannonballs of the same shape and size made from wood and iron.
These cannonballs released simultaneously on the top of the Pisa Leaning
Tower reached the ground at the same time disregarding their weight.
Galileo developed mathematical description of the motion of objects with
the constant acceleration. According this description, distance covered by
the object d should be proportional to the time squired (see Eq. (2.16b)).
Performing experiments of the motion of the object along incline plane (it
was treated by him as a particular example of falling) Galileo confirmed his
mathematical consideration. Finally he makes a conclusion that at the given
location on the Earth surface and at in the absence of air resistance, all
objects fall with the same constant acceleration. This type of motion is
called Free Fall, although it includes rising as well as falling motion.
This acceleration is called the acceleration due to gravity. It is designated as
g. In our problems, we usually will take g = 9.80 m/s^2. Actually at different
locations on the Earth surface g is different. For example, in New York City
g = 9.803 m/s^2, at North Pole g = 9.830 m/s^2. On the surface of the Moon,
g is six times smaller. Strictly speaking, free fall should be studied in
vacuum conditions. American astronaut David Scott made good
demonstration of this phenomenon on the surface of the Moon. He released
simultaneously a feather and a geology hammer. There is vacuum on the
29
surface of the moon. The feather and the hammer simultaneously reached
the ground. Millions of people on the Earth had opportunity to see this free
fall experiment through TV.
At first glance, the fact that all object disregarding their weight are falling
with the same acceleration looks like mystery. We will try to explain this
mystery later in our course. Now we will write the set of equation describing
free fall motion. Actually this is the same system of equations (2.16) but
specified for free fall motion. This motion is also the motion with the
constant acceleration along straight line, but now this line is not horizontal
but vertical and can be used as y-coordinate axis. The acceleration now is
specified: this is acceleration due to gravity. Magnitude of this acceleration
is g (in the most of problems related to the motion near the Earth surface g =
9.80 m/s^2). It is always positive. If we choose direction up as the positive
direction of y-axis, then acceleration in free fall motion will be a = -- g, and
the system of equations describing the free fall motion can be written as
following.
_____________________________________________________________
Free Fall Motion: a = -- g
v = v0 -- g t
(2.18a)
y= y0 + v0 t -- (1/2) g t^2
(2.18b)
v^2 = v0^2 -- 2 g (y – y0)
(2.18c)
Solving problems related to the free fall motion, we should remember that
there is some additional information in those problems, which could be very
helpful for their solution.
EXAMPLE 2.6. An arrow is fired from the ground level straight upward
with an initial speed of 10 m/s. (a) How long is the arrow in the air? (b) To
what maximum elevation does the arrow rise? (c) What time is needed for
the arrow to reach the maximum elevation h? (d) With what velocity will it
hit the ground?
v0 = 10 m/s
g = 9.80 m/s^2
yo = 0
y=0
30
(a) ttot ?
(b) ymax = h?
(c) th ?
We will use the implicit information that relates the vertical position of the
object and time, or the vertical position of the object and velocity.
(a) When t -= ttot, the object reaches the ground, so its vertical position is y =
0 Then we should choose equation that relates position and time. This is the
Eq. (18b). Inserting data in this equation gives us relationship
0 = 0 + v0 t – ½ gt^2 = 0 +(10 m/s) t – ½ (9.80 m/s^2) t^2  (v0 –-1/2 g t) t
=0
This quadratic equation has two solutions. This is interesting point. Solving
physical problems, we sometimes encounter in situations in which math
bring us some additional nonphysical solutions. In those cases we should
analyze solutions with what we are asked to find in the conditions. For
example, in considered problem, the first solution is t1 = 0. It is obvious
result that at t = 0, the arrow is at the origin of coordinates at the ground
level y = 0. This is not what we are asked about in the conditions. Therefore
t1 is mathematically right but not physical solution. The second solution is
t2 = 2 v0/g = 2.04 s
This is reasonable result from the point of view of conditions. We can treat
it as a physical solution.
(b) When object is ascending, its velocity directed up. When object
descending, its velocity directed down. At the highest elevation
y = ymax = h, v =0. Therefore we can use Eq. (2.18c) that relates
position and the velocity. In our case we will get
0 = v0^2 -- 2 g yh  yh = v0^2/(2 g) = (10 m/s)^2/[2 (9.80 m/s^2)] = 5.10 m
(c) At the highest elevation y = yh, vh = 0, t = th. To find th we can use the
Eq (2.18a) that relates velocity and time, or Eq (2.18b) that relates the
position and time. If we use, for example, Eq (2.18a), we will get
0 = v0 – g t  v0 = g th  th = v0 / g = 1.02 s = ½ ttot
Pay attention that th = ½ ttot. It means that the motion is absolutely
symmetrical. How many times is needed for the object to ascend, so
much time is needed for the object to fall down.
31
2. KINEMATICS IN TWO DIMENSIONS; VECTORS.
We studied in details the motion of an object along straight line – onedimensional motion. Now we will try to consider more general case – the
motion in two dimensions. But to reach this goal we need additional
mathematical technique. We do not need this technique to work with
physical quantities that have no relation to the direction in space. These
physical quantities are called scalars. They can be characterized only by
magnitude (number with unit). Examples of these quantities are as
following: time, mass, density etc. We can use just arithmetic’s to operate
with them. Other situation is with the physical quantities that have a
direction in space. When object is moving along straight line there are only
two directions: in term of coordinate axis these are directions in the positive
or negative direction of X-axis. In the two-dimensional case, we have two
mutually perpendicular axes: horizontal X-axis and vertical Y-axis. Object
can move now in any direction laying in the XOY plane, so the physical
quantities that have direction in space – displacement, velocity, acceleration,
and others now can be directed in any direction. The physical quantities that
characterized by magnitudes (some number with unit) but also by the
direction in space are called vectors. Vector is a quantity that incorporates
magnitude and direction. To operate with them we could not use just
arithmetic. To add, subtract, multiply vectors we need to become familiar
with some basic concepts of the special branch of mathematics – vector
algebra.
ADDITION OF VECTORS BY GRAPHICAL METHODS.
Each vector we represent as an arrow directed in space. The length of this
arrow in some scale represents the magnitude of a vector. A vector quantity
we will designate by corresponding letter with small arrow above. For

example, the vector of velocity is represented by designation v , the
magnitude of this vector designated just v. The vector quantity can be
designated by the bold font also like v. The vector of acceleration is

represented by designation a or a, the magnitude of this vectoris designated
as a. For the vector of displacement we have correspondingly D or D, and
D. Due the spatial character of vectors it is reasonable to use graphical
methods to operate with them.
32
Graphical Methods.
First of all we identify equality of two vectors. Two vectors supposed to be
equal when their magnitudes are equal and these vectors are directed in the
same direction. This definition allows us to displace vector without changing
its magnitude and direction in any other position in space. This is the key
point in the graphical methods techniques. The first of them is called tail to
tip methods of adding vectors. In this method, the tail of the added vector
is brought to the tip of initial vector. Vector that results from this operation
(it is called resultant) directed from the tail of the initial vector to the tip of
an added vector. This operation is shown in Fig. 3.1a. Object was displaced
from the point 1 to the point 2 (displacement vector D1), and then from the
point 2 to the point 3 (displacement vector D2). Vector DR results from this
vector addition (vector that results from vector addition sometimes is called
resultant). By this method we can add more than two vectors (in this case,
this type of vector addition is called polygon method. Addition of three
vectors is shown in Fig. 3.1b as an example of application of the polygon
method. Other wide used method of vector addition is called parallelogram
method. In this method their tales brings two vectors together. The figure is
completed to the parallelogram. Resultant vector is a diagonal of this
parallelogram (see Fig. 3.1c). Comparison of Fig.3.1a and Fig.3.1c shows
that tail to tip method and parallelogram method bring absolutely identical
results.
Before considering the vector subtraction we introduce the concept of
negative vector. The negative vector has magnitude equals to the magnitude
of initial vector but directed in opposite direction. Now we can consider
vector subtraction as following: A – B = A + (-- B)
We just add to the vector A vector (-- B).
33
Fig. 3.1 Vector addition by the graphical methods: (a) Tail to Tip method;
(b) Polygon Method; (c) Parallelogram method.
34
The result of the multiplication of vector by positive scalar is the vector
directed in the same direction as initial vector but with magnitude equals to
the magnitude of initial vector multiplied by scalar. If scalar is negative, the
direction will be opposite to the direction of initial vector.
Summarizing the consideration of the graphical methods of vector
operations, we can notice that these methods are very clear and helpful to
understand spatial nature of vector operations.
At the same time they have serious disadvantageous. These methods are not
precise. Their precision is limited by accuracy of instruments used (meter
sticks and protractors). Most important, graphical methods are not applicable
to the general three-dimensional case. To perform vector operations, we will
use much more convenient and powerful method – Method of Components.
ADDING AND SUBTRACTING VECTORS BY COMPONENTS.
Key point in consideration is as following: we bring tail of vector at the
origin of coordinate axes (see Fig. 3.2).
y
A
Ay

x
Ax
Fig. 3.2. Resolution of a vector into components.
The length of the arrow in some scale represents the magnitude A of the
vector A. The direction of the vector is determined by an angle θA between
the vector and the positive direction of the X-axis. What are the
components? The projection of vector on the X-axis Ax is called xcomponent of the vector A. The projection of vector on the Y-axis Ay is
called y-component of the vector A. Actually we have a right triangle
situated at the origin of coordinates in which A plays role of diagonal. Side
adjacent to the angle θA is Ax. Side opposite to the angle θA equals to the
component Ay. The following trigonometric functions can be defined for this
triangle:
sin θA = Ay/A
35
(3.1a)
cos θA = Ax/A
(3.1b)
tan θA = Ay/Ax
(3.1c)
Using these relationships we can get following useful formulas for
operations that is called resolution vector into components. This operation
allows finding components of vector when its magnitude and direction are
given:
Ax = A cos θA
(3.2)
Ay = A sin θA
(3.3)
If we are given by components and need to find magnitude of a vector and
its direction (angle between the vector and the positive direction of the Xaxis) can use the theorem of Pythagoras’:
________
V = √ Ax² + Ay²
(3.4)
θA = tan ‾¹(Ay//Ax)
(3.5)
Component situated on the positive part of the coordinate axis is positive,
situated on the negative part of the coordinate axis is negative. Vector addition
by components can be performed as following. For example, we are given
magnitudes and directions of two vectors A and B. It means that we are given
their magnitudes A and B, and their angles with the positive direction of X-axis
θA and θB. Suppose we are asked to find vector C. Actually we need to solve
vector equation
C=A+B
(3.6)
We could not solve it just substituting vectors by their magnitudes. This
equation is only symbolical representation of operation. To really solve this
equation we need to resolve vectors A and B into components using formulas
(3.2) and (3.3). Than we should find components of resultant vector C
according following formulas
Cx = Ax + Bx
(3.7)
Cy = Ay + By
(3.8)
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Then we can find the magnitude of vector C and its direction (angle with
respect the positive direction of X-axis) using formulas (3.4) and (3.5):
________
C = √ Cx² + Cy²
(3.9)
θA = tan ‾¹(Cy/Cx)
(3.10)
If we need to subtract one vector from another, for example, to find vector D
D=A–B
(3.11)
In this case, we can find components of vector D as following
Dx = Ax – Bx
(3.12)
D y = Ay – B y
(3.13)
Then the magnitude and direction of vector D then can be found by the same
way as for the vector C (see formulas (3.9) and (3.10)).
EXAMPLE 3.1.
Each of the displacement vectors A and B (Fig. 3.3) has a magnitude of 5.00 m.
Their directions relative to the positive direction of the X- axis are 30.0º for the
vector A and 45.0º for the vector B. Find: (a) x- and y-components of the
vectors A and B; (b) x- and y- components of the vector D = B – A; (c) the
magnitude and direction of the vector D.
37
Fig. 3.3. Example 3.1.
The same approach we can use to solve more complicated vector equation. For
example, equation
E=F+G–H+I
(3.14)
We should to resolve each of the given vectors into components using formulas
(3.7) and (3.8), and then to find components of the vector E as following
Ex = Fx + Gx – Hx + Ix
(3.15)
E y = F y + Gy – H y + I y
(3.16)
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When the components of the vector E become known, its magnitude and
direction can be found using in the formulas (3.9) and (3.10) components Ex
and Ey.
PROJECTILE MOTION.
We will use the technique of the vector addition to describe one type of 2D
motion that is called Projectile Motion (Fig. 3.4).
y
v0

h
O
P x
X
Fig. 3.4. Projectile motion (general case).
Projectile can be any object launched into space. We are not interested to
know how and who launched the projectile into space. We only need to
know the initial velocity v0 of the projectile (the magnitude of the initial
velocity v0, and its direction -- the angle θ0 of the velocity with respect of the
positive direction of the X-axis). To simplify the consideration, we will
neglect the air resistance and assume that only the gravity influences the
motion of the projectile. Nevertheless, the situation remains very
complicated because the direction of the velocity is changing during all time
of projectile motion. The solution of this complicated problem was found by
Galileo Galilei. He performed the experiment in which to objects
simultaneously start to move from the same height. One of them was falling
vertically down experiencing free fall motion. Another object was
simultaneously launched in the projectile motion. Both of them
simultaneously reached the ground. Galilei deduced that motions of the
object in vertical direction and in horizontal direction are absolutely
independent. In accordance with our today’s knowledge, we can say, that we
can independently consider the x- and y-components of the projectile
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motion. The gravity influences the motion of the projectile along vertical (ycomponent). We studied before this type of motion. This is the motion with
constant acceleration (acceleration due to gravity) or free-fall motion. We
studied before this type of motion, but now this approach is applied only to
the description of the y-component of the position, velocity and acceleration
of the projectile.
ay = -- g = const
(3.17)
vy = vy0 -- g t
(3.18)
y= y0 + vy0 t -- (1/2) g t^2
(3.19)
vy ² = vy0² -- 2 g (y – y0)
(3.20)
Where vy0 is the y-component of the initial velocity
vy0 = v0 sin θ0
(3.21)
If the air resistance can be ignored, no forces influence the motion of the
projectile along the horizontal. Therefore the projection of the projectile
motion on the horizontal (x-component of the projectile motion) is the
motion with the constant velocity. We also studied this type of motion, but
now it is applicable only to the x-component of projectile motion:
ax = = const = 0
(3.22)
v = vx0 = const
(3.23)
x= x0 + vx0 t
(3.24)
Where vx0 is the x-component of the initial velocity
vx0 = v0 cos θ0
(3.21)
Solving problems, we should remember that these are only components of
the motion. Real motion is occurred along the trajectory – some curve in
space – so in the any instant of time, the time t in equations is the same for
both of components. There is some times additional implicit information in
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conditions of problems related to the projectile motion: when object is
situated at ground level, it means that y = 0; when object reaches its highest
elevation, its y-component of velocity is zero; x-component of velocity
always is the same. How to apply this information in problems we can see
for the following example.
EXAMPLE 3.2. Projectile launched horizontally.
A rescue plane is flying horizontally at a speed of 120 m/s and drops
package at an elevation of 2000 m (Fig. 3.5). (a) How much time is required
for the package to reach the ground? (b) How far does it travel horizontally
(in other words, what is the largest distance – range covered by the projectile
before striking the ground)? (c) What is the x-component and y-component
of the velocity of the package just before it reaches the ground? What is the
magnitude of the velocity of the package just before it reaches the ground?
O
v0
x
h
y
Fig. 3.5. Example 3.2. Projectile launched horizontally.
x0 = 0
y0 = 2000 m
v0 = 120 m/s
θ0 = 0º
(a) tr ?
(b) xr ?
(c) vstr ?
41
(a) When package reaches the ground t = tr, y = 0. The equation that
relates these two physical quantities is (3.19). Solving this equation
for unknown tr, we will get tr, = 20.2 s.
(b) At the same time, when t = tr, x = xr. The equation that relates these
two physical quantities is (3.24). Solving this equation for unknown t r,
we will get xr = 2424 m.
(c) At the same time, when t = tr, v = vstr. To find magnitude of the vector
vstr, we need to find its components. The x-component of this vector is
all the time the same vxstr = 120 m/s. To find vystr at t = tr, we will use
Eq. (3.18) that relates these two physical quantities. As a result we
will get vystr = -- 198 m/s. Now the magnitude of the vector vstr can be
found using formula. Finally, we will get vstr = 231 m.
EXAMPLE 3.3. Projectile launched at nonzero angle to the horizon.
The projectile is launched with initial velocity v0 at some nonzero angle θ0
with the respect to the horizon. (Fig. 3.6). Find: (a) time needed to reach the
range; (b) largest distance along X-axis (range) covered by the projectile
before striking the ground; (c) time th needed for the projectile to reach
highest elevation yh; (d) highest elevation.
x0 = 0
y0 = 0
v0 = v0
θ0 = θ0
----------------
(a) tr?
(b) xr ?
(c) th?
(d) yh?
42
y
v0

O
x
R
Fig. 3.6. Example 3.3. Projectile launched with
some nonzero angle.
First two questions can be considered by the same way as it was done in the
Example 3.2.
(a) When projectiles reaches the range, t = tr, y = 0. The equation that
relates these two physical quantities is (3.19). Solving this equation
for unknown tr, we will get
tr = (2 v0 sin θ0)/g
(3.22)
(b) At the same time, when t = tr, x = xr. The equation that relates these
two physical quantities is (3.24). Solving this equation for unknown
xr, we will get
xr = ( v0² sin 2θ0)/g
(3.23)
The maximal value of the expression (3.23) will be when θ0 = 45º.
Therefore the largest range will be occurred when projectile will be
launched at the angle of 45º with the respect of the horizon.
(c) At the highest elevation (t = th, y = yh) the y-component of the velocity
equals zero. Using Eq. (3.18) related t and vy for this specific instant
of time we will get
th = (v0 sin θ0)/g
(3.24)
Comparing Eqs. (3.22) and (3.24), we come to conclusion that tr = 2 th. It
means that the motion of the projectile is absolutely symmetrical: how
many time is needed to reach the highest elevation, so many time then
needed to reach the ground.
(d) (d) When t = th, y = yh. Using the equation that relates these two
physical quantities (3.19), we can find
43
yh = ( v0² sin² θ0)/(2 g)
(3.23)
Discussing this beautiful Galileo’s solution of the projectile problem, we
need to remind that it is developed for the situations in which we can neglect
the air resistance. Actually it means that results are applicable only to the
case of comparatively small velocities of projectiles. Otherwise we should
apply ballistics – the scientific study of the fast projectiles (shells and
rockets), their ejection and flight through the air.
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