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,

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you need to follow the rule that final result of a multiplication or division should 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 = 9.1110^(-3); M = 19910^(+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 m1 kilogram = 1 kg = 10^3 grams =1 10^3 gHer 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 particles1 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/sThis 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 (2.2)

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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.

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

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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 t0 (x / t) (2.5)

The notation lim t0 (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 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

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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 t0 (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 t1 = 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 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:

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v = x / t = (x – x0) / t (2.8)

a = a = v / t = (v – v0)/t (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 (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

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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, if acceleration is not given, we can find position of an object combining equations (2.11) and (2.16d).

GRAPHICAL ANALYSIS OF THE MOTION.

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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.

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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)

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_____________________________________________________________

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.

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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.

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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).

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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.

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Now we can write what is given and what we want to find.vp0 = 0vp = 60 mi/htp = 7.9 svt0 =0vt = 26.8 m/st = 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).

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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/svp = 0xp = 141 fttp = 7.9 sxp0 = 0vt0 = 26.8 m/svt = 0xto = 0xt = 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

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(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 = consttr = 1.00 sar = 0ab = -- 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.

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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)

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= 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/sx01 = 0v2 = 26.8 m/s = v1

x1 = 105 mx02 = x1 = 105 mtc = 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.

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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.

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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 = 0x0p = 0vs = 33.5 m/s = constas = 0vpo = 0ap = 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.

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

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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) 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.

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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 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

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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/sg = 9.80 m/s^2yo = 0y = 0

(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

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(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.

2. KINEMATICS IN TWO DIMENSIONS; VECTORS.

We studied in details the motion of an object along straight line – one-dimensional 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

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example, the vector of velocity is represented by designation , 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 or a, the magnitude of this vector is designated as a. For the vector of displacement we have correspondingly or D, and D. Due the spatial character of vectors it is reasonable to use graphical methods to operate with them.

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).

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Fig. 3.1 Vector addition by the graphical methods: (a) Tail to Tip method; (b) Polygon Method; (c) Parallelogram method.

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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).

Ay

A

Ax

y

x

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 x-component 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

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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 (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 X-axis) 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-axisθA 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

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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)

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)

Dy = Ay – By (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.

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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)

Ey = Fy + Gy – Hy + Iy (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).

x

h

yv0

O XP

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

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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 motion. The gravity influences the motion of the projectile along vertical (y-component). 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)

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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 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?

x

h

y

v0O

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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 ?

(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 tr,

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

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v0 = v0

θ0 = θ0

----------------

(a) tr?

(b) xr ?

(c) th?

(d) yh?

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)

x

yv0

O R

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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

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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