Phys 1011 Module AAU-pages - General Physics PDF

General Physics Module Phys 1011 AAU 1 Preliminaries Learning Outcome After completing this Chapter, students are expected to: define a physical quantity, unit of measurement and uncertaint...

General Physics Module Phys 1011 AAU 1 Preliminaries Learning Outcome After completing this Chapter, students are expected to: define a physical quantity, unit of measurement and uncertainty in measurement. identify significant figures in measurements and calculations. define a unit vector and describe its purpose. distinguish between the vector components and the scalar components of a vector. apply dimensional analysis to determine the relation between different quantities. Solve problems related to the addition, components, magnitude and direction of vectors. Introduction This chapter introduces measurements of physical quantities and the uncertainties inherent to measurements. Accurate and precise measurements are important in the study of and research in physics. Since traditional ways of measurements do not give accurate and precise values, we use standard measurement techniques in science. For example, instead of saying that a string is 5 armlengths, we can specify its length using a standard measuring tape and say that the string is 2 meters long. In this chapter we use a standard known as the International System of Units (SI). How do we know that our measurements are accurate and precise? What we know for sure is that all measurements have some degree of error or uncertainty. No measurement is exact! A measurement is only an estimation of the true value. Some factors that cause measurement uncertainty and determining the amount of uncertainty will be discussed in this chapter. In physics experiments, measurements of different physical quantities are taken to verify or discover any relationship between them. Dimensional analysis is covered in this chapter to give students some ideas on how to check the dimensional consistency of equations and how new equation can be discovered. Also, geometric and algebraic methods of adding and resolving vectors will be discussed. 1.1 Physical Quantities and Units of Measurement Learning Outcome After completing this section, students are expected to: define a physical quantity and a unit of measurement. describe what measurement means in science. distinguish between quantities and units. solve problems related to units of measurement. 1.1.1 Quantities and Units A quantity is a definite or indefinite amount or size of something. Preliminaries 6 General Physics Module Phys 1011 AAU Examples Thing Amount Type Desire Strong Indefinite Mass of box Heavy Indefinite Mass of box 5 kg Definite Love Deep Indefinite Height of a person Tall Indefinite Height of a person 2m Definite Some of the quantities above are physical and some are non-physical. 1.1.1.1 Non-physical quantities Non-physical quantities (qualitative) such as love, hate, fear and hope are not a concern in physics studies and experiments and cannot be measured in the sense described below. However, there are research disciplines, such as psychology, that study and quantify such quantities as fear of exam or exam anxiety. 1.1.1.2 A physical quantity A physical quantity is a quantity that can be measured by defining its units of measurement or using a measuring instrument. A physical quantity is always expressed in terms of a numerical value (magnitude) and a unit. Physical quantity = (numerical value) unit Examples 1. speed of sound = 331 m/s, 2. mass of box = 5 kg 3. radius of hydrogen atom = 53 pm Exercise Give other examples of physical and non-physical quantities. 1.1.2 A unit of measurement A unit of measurement (defined and adopted by convention) is a standard by means of which the amount of a physical quantity is expressed. In the first example above the unit is “m/s” and the speed of sound is expressed as containing 331 such units. In the second example, the unit is “kg” and the mass of a box is expressed a containing 5 such units. Experiments in physics involve taking measurements of quantities and calculating some results. For measurements and calculations to be meaningful, units must be introduced. Physics without units is meaningless. Preliminaries 7 General Physics Module Phys 1011 AAU Exercises 1. Can your heart beat be used as a unit of time? Discuss. 2. Can your arm length or your step be used as units of length? Discuss. 1.1.3 Measurement A measurement is defined as the process of finding the size or amount of a physical quantity using the standard unit for that quantity (see below for standard units). Exercises 1. Is every act of finding the amount/size of a physical quantity a measurement? Consider the following: a. Counting students in a class, the amount of Birr you have, the number of stars in the sky; b. Measuring the length of a table using your arm; c. Estimating the distance between two towns; d. Comparing the length of a meter stick with your height. 1.1.4 Fundamental and Derived Units Physical quantities and their units are of two types: Fundamental (or Basic) and Derived. 1.1.4.1 Fundamental (basic) quantities and units In the SI system, there are seven basic physical quantities and units, listed in Table 1-1 (below). Table 1-1 The seven fundamental quantities and their SI units. Basic physical quantity Symbol for quantity Basic unit Symbol for unit Length 𝑙 metre 𝑚 Mass 𝑚 kilogram 𝑘𝑔 Time 𝑡 second 𝑠 Electric current 𝐼 ampere 𝐴 Temperature 𝑇 kelvin 𝐾 Amount of substance 𝑛 mole 𝑚𝑜𝑙 Luminous intensity 𝐼𝜈 candela 𝑐𝑑 Some unit symbols are in upper-case letter because they are named after scientists; for example, the unit of temperature (kelvin, K) reminds us Lord Kelvin who contributed a lot in thermodynamics. Note that the full names of the units are all in lower-case letters. Units of Time, Length, and Mass To give the student some insight into how standards are adopted we briefly discuss the three fundamental units of mechanics; the second, the meter and the kilogram. See your lab manual for more information about standard units of measurement. Preliminaries 8 General Physics Module Phys 1011 AAU The Second The SI unit for time, the second (abbreviated s) was first defined as 1/86,400 of a mean solar day. Since the solar day is getting longer due to the gradual slowing of the Earth’s rotation, a new standard of the second was adopted in terms of a non-varying physical phenomenon for greater accuracy. One such phenomenon is the steady vibrations of Cesium atoms, and these vibrations can be readily observed and counted. In 1967 the second was redefined as the time required for 9,192,631,770 of these vibrations (See https://physlibretexts.org). The Meter The SI unit for length, the meter (abbreviated m) was first defined as 1/10,000,000 of the distance from the equator to the North Pole. The meter was redefined more accurately to be the distance between two engraved lines on a platinum-iridium bar now kept near Paris. It was again redefined even more accurately in terms of the wavelength of light, so 1 m became 1,650,763.73 wavelengths of orange light emitted by krypton atoms. The present definition of the meter is the distance light travels in a vacuum in 1/299,792,458 of a second (See https://physlibretexts.org). The length of the meter will change if the speed of light is someday measured with greater accuracy. The Kilogram The SI unit for mass, the kilogram (abbreviated kg) was previously defined to be the mass of a platinum-iridium cylinder kept near Paris with exact replicas reserved at different parts of the globe. Since airborne contaminants slightly change the platinum-iridium mass over time, the scientific community adopted a more stable definition of the kilogram in May 2019. The kilogram is now defined in terms of the second, the meter, and Planck's constant, h (a quantum mechanical value that relates a photon's energy to its frequency). Exercises 1. Why do scientists keep redefining standards? 2. How are the standards for the other SI units defined (Table 1.1)? See the references given at the end of the course outline. 1.1.4.2 Derived quantities Derived quantities are combinations of two or more basic quantities. For example, volume is obtained by combining three lengths; speed is derived from length and time. Similarly, derived units are made by a combination of two or more of the fundamental units. The unit of volume is meter cubed (𝑚3 ); the unit of speed is meter per second (𝑚/𝑠). Table 1-2 (below). shows some examples of derived quantities and the corresponding derived units. Preliminaries 9 General Physics Module Phys 1011 AAU Table 1-2 Some derived quantities and their SI units Derived quantity Unit Symbol Area square meter 𝑚2 𝑚2 Volume cubic meter 𝑚3 𝑚3 Frequency Hertz 𝐻𝑧 𝑠 −1 Density kilogram per cubic metre 𝑘𝑔𝑚−3 𝑘𝑔𝑚−3 Force Newton 𝑁 𝑘𝑔𝑚𝑠 −2 Work, energy Joule 𝐽 (𝑁𝑚) 𝑘𝑔𝑚2 𝑠 −2 Power Watt 𝑊 (𝐽/𝑠) 𝑘𝑔𝑚2 𝑠 −3 Velocity (speed) metre per second 𝑚𝑠 −1 𝑚𝑠 −1 Note that in the rightmost column the derived units are expressed as powers of the fundamental units. These powers are called the dimensions of the physical quantity in the base units. The next sub-section discusses dimensional analysis. Exercise Express the following derived units in terms of the fundamental SI units given in Table 1-1 (above): The unit of acceleration The unit of moment of inertia The unit of linear momentum The unit of charge (coulomb, C) The unit of angular speed The unit of potential difference (vlot, V) The unit of torque The unit of magnetic field (tesla, T) 1.1.5 Dimension and dimensional analysis Every physical quantity can be expressed in terms of some powers of the fundamental SI quantities as shown in the rightmost column of Table 1-2 (above). These powers are called the dimensions of the physical quantity in question. The square brackets [ ] stand for “dimension of”. For example, [𝑚𝑎𝑠𝑠] or [𝑚] means “dimension of mass” and we write [𝑚𝑎𝑠𝑠] = [𝑚] = 𝑀. [𝑙𝑒𝑛𝑔𝑡ℎ] = [𝑙] = 𝐿 means “dimension of length” and we write [𝑙𝑒𝑛𝑔𝑡ℎ] = [𝑙] = 𝐿 [𝑡𝑖𝑚𝑒] = [𝑡] = 𝑇 means “dimension of time” and we write [𝑡𝑖𝑚𝑒] = [𝑡] = 𝑇 [𝑐𝑢𝑟𝑟𝑒𝑛𝑡] = 𝐼 means dimension of electric current and so on. In mechanics, a derived physical quantity 𝑥 can be expressed as [𝑥] = [𝑙]𝑎 [𝑚]𝑏 [𝑡]𝑐 = 𝐿𝑎 𝑀𝑏 𝑇 𝑐 Preliminaries 10 General Physics Module Phys 1011 AAU Examples 1. The dimensions of volume: [𝑉] = [𝑙𝑒𝑛𝑔𝑡ℎ] × [𝑤𝑖𝑑𝑡ℎ] × [ℎ𝑒𝑖𝑔ℎ𝑡] = 𝐿3 Volume has a dimension of 3 in length 2. The dimensions of density: [𝜌] = [𝑚]/[𝑉] = 𝑀𝐿−3 Density has a dimension of 1 in mass and a dimension of -3 in length 3. The dimensions of force: [𝐹] = [𝑚] × [𝑎] = 𝑀𝐿𝑇 −2 Force is said to have a dimension of 1 in mass, a dimension of 1 in length and a dimension of -2 in time 4. The dimensions of energy: [𝐸] = [𝐹] × [𝑠] = 𝑀𝐿2 𝑇 −2 Energy has dimensions of 1, 2 and -2 in mass, length and time, respectively Dimensional analysis is useful in deriving new formulas or checking existing formulas apart from dimensionless factors that may exist in the formulas. Examples 1. Verify that the following relation is correct apart from dimensionless factors: 𝑠 = 12𝑎𝑡 2 Solution [𝑠] = 𝐿 [𝑎𝑡 2 ] = [𝑎][𝑡 2 ] = 𝐿𝑇 −2 𝑇 2 = 𝐿 Therefore, [𝑠] = [𝑎𝑡 2 ] The given equation is dimensionally correct. 2. Show that the following equation is dimensionally consistent. 𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑎𝑠𝑠 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑠𝑞𝑢𝑎𝑟𝑒𝑑 Solution [𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒] = [𝑓𝑜𝑟𝑐𝑒][𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒] = 𝑀𝐿𝑇 −2 𝐿 = 𝑀𝐿2 𝑇 −2 𝐿 2 [𝑚𝑎𝑠𝑠 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑠𝑞𝑢𝑎𝑟𝑒𝑑] = [𝑚𝑎𝑠𝑠][𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦]2 = 𝑀 ( ) = 𝑀𝐿2 𝑇 −2 𝑇 The given equation is dimensionally correct. In general, the dimension of any physical quantity can be written as 𝐿𝑎 𝑀𝑏 𝑇 𝑐 𝐼 𝑑 𝛩𝑒 𝑁 𝑓 𝐽 𝑔 for some powers 𝑎,𝑏,c,𝑑,𝑒,𝑓, and 𝑔. We can write the dimensions of a length in this form with 𝑎 = 1 and the remaining six powers all set equal to zero: 𝐿1 = 𝐿1 𝑀0 𝑇 0 𝐼 0 𝛩0 𝑁 0 𝐽0. For a dimensionless quantity all seven powers are zero. Dimensionless quantities pure numbers. Table 1-3 (below) displays the symbols for all fundamental quantities. Preliminaries 11 General Physics Module Phys 1011 AAU Table 1-3 Fundamental Quantities and Their Dimensions Base Quantity Symbol for Dimension Length L Mass M Time T Current I Thermodynamic temperature Θ Amount of substance N Luminous intensity J Exercises Following the example above, check that the following equations are dimensionally correct. 𝑣2 1. 𝑣 2 = 𝑣02 + 2𝑎𝑠 2. 𝑎 = 𝑟 1 3. 𝑠 = 𝑣0 𝑡 + 𝑎𝑡 2 4. 𝑚𝑣 2 = 𝑘𝑥 2 2 5. 𝑣 = 𝑣0 + 𝑎𝑡 6. 𝑃 = 𝜌𝑔ℎ + 0.5𝜌𝑣 2 7. 𝑟𝐹 = 𝑚𝑙 2 8. 𝑓 = 𝜇𝑁 1.1.6 SI Prefixes and scientific notation The International System of Units (SI) is a decimal system in which units are divided or multiplied by 10 to give smaller or larger units. For example, it doesn’t make sense to give the length of a football field as 120000 millimetres or as 0.120 kilometer. A more appropriate unit is the metre. Telling people that the football field is 120 meters long gives them a better idea of the actual length of the field. It would be equally unsuitable to give the thickness of a human hair as 0.000 000 1 kilometer. In this case, the appropriate unit is the millimetre; Saying that the human hair is 1 millimetre thick would give people a far better idea of the actual thickness of the hair. Preliminaries 12 General Physics Module Phys 1011 AAU Table 1-4 lists the SI Prefixes in order with their names and symbols Preliminaries 13 General Physics Module Phys 1011 AAU Table 1-4 SI Prefixes Prefix Symbol Base Unit Multiplier In Words Exponential yotta Y 1,000,000,000,000,000,000,000,000 septillion 1024 zetta Z 1,000,000,000,000,000,000,000 sextillion 1021 exa E 1,000,000,000,000,000,000 quintillion 1018 peta P 1,000,000,000,000,000 quadrillion 1015 tera T 1,000,000,000,000 trillion 1012 giga G 1,000,000,000 billion 109 mega M 1,000,000 million 106 kilo k 1,000 thousand 103 hecto h 100 hundred 102 deca da 10 ten 101 (base unit) 1 one 100 deci d 0.1 tenth 10−1 centi c 0.01 hundredth 10−2 milli m 0.001 thousandth 10−3 micro μ 0.000001 millionth 10−6 nano n 0.000000001 billionth 10−9 pico p 0.000000000001 trillionth 10−12 femto f 0.000000000000001 quadrillionth 10−15 atto a 0.000000000000000001 quintillionth 10−18 zepto z 0.000000000000000000001 sextillionth 10−21 yocto y 0.000000000000000000000001 septillionth 10−24 Similarly, it is not convenient to write the electron mass as 0.0000000000000000000000000000091 kg or the diameter of the observable universe as 880000000000000000000000000 m. We rather use a scientific notation to express too big or too small numbers: 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑚𝑎𝑠𝑠 = 9.1 × 10−31 kg 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑏𝑙𝑒 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = 8.8 × 1026 m Preliminaries 14 General Physics Module Phys 1011 AAU Examples 1. Some illustrations of the use of Prefixes: a. Radius of hydrogen atom = 53 pm (picometers). b. Distance between Earth and Sun = 149.5 Gm (gigameters). c. Mass of electron = 0.000 91 yg (yoctograms). d. The mass of the earth = 5983 Yg (yottagrams). 2. Write the following physical quantities in scientific notation and using SI prefixes. 3270 g, 0.128 m, 65 000 000 W, 0.000056 s Solution Given Physical Quantity Scientific Notation Using SI Prefixes 3270 g 3.27 × 103 g 3.27 kg 0.128 m 1.28 × 10−1 m 128 mm 65 000 000 W 6.5 × 107 W 65 MW 0.000056 s 5.6 × 10−5 s 56 μs Exercises 1. Convert the following numbers into scientific notation: a. 27 000 000 b. 101 c. 007 12 d. 81 250 000 000 e. 821 f. 000 002 05 2. Write the following numbers using SI prefixes: a. 5.80 x 106 b. 2.52 x 10−3 c. 6.32 x 10−5 d. 6.10 x 10−11 e. 8.56 x 104 f. 6.25 x 10−24 g. 2.30 x 1010 h. 1.5 x 1019 3. Write the standard form of a. speed of light in a vacuum = 298 000 000km/s b. one light year = 10 000 000 000 000km Preliminaries 15 General Physics Module Phys 1011 AAU 1.2 Uncertainty in Measurement and Significant Figures Learning Outcome After completing this section, students are expected to: define measurement uncertainty, distinguish between error and uncertainty, explain the difference between accuracy and precision, give order of magnitude estimation of physical quantities identify significant digits in a measurement value, analyse errors in data and report results, 1.2.1 Uncertainty in measurement If a teacher overcounts the number of students in a class, she could easily correct the mistake by recounting the students two or more times. Miscounting is an example of error or a mistake that could easily be removed. On the other hand, an error made in physics measurements can never be removed no matter how the measurement is taken or how often it is repeated. Measurement errors in physics mean more than simple human mistakes; they are uncertainties inherent to the physical measurements. Although the terms “error” and “uncertainty” are used interchangeably, they are a bit different. 1.2.1.1 Error Error is defined as the difference between an observed value and a true value. 𝑬𝒓𝒓𝒐𝒓 = 𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 − 𝒕𝒓𝒖𝒆 𝒗𝒂𝒍𝒖𝒆 The “observed” value is either a result of direct measurement or a calculated value using other measured values in a formula. The “true” value exists but is unknown. Then how can one determine the error in measurements? The goal of measurement is to estimate the true value of a physical constant using experimental methods. True value Observed Error Quantity values Figure 1-1 A schematic representation of error Preliminaries 16 General Physics Module Phys 1011 AAU 1.2.1.2 Uncertainty Uncertainty is a quantification of the doubt about the measurement result. This quantification gives the range of values within which the true value is believed to lie with some level of confidence. Uncertainty is determined by statistical analysis of many values of measurement. 𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕 = (𝒃𝒆𝒔𝒕 𝒆𝒔𝒕𝒊𝒎𝒂𝒕𝒆 ± 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚) 𝒖𝒏𝒊𝒕 𝒐𝒇 𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕 TRUE value lies here Of TRUE value (usually the MEAN) uncertainty uncertainty best estimate values Measurement Figure 1-2 Uncertainty shows the area around the average value where the true value of the measurement is likely to be found Suppose the result of length measurement is (20.1 ± 0.1) cm. This means that the experimenter believes the true value to be closest to 20.1 cm but it could have been anywhere between 20.0 cm and 20.2 cm. 1.2.2 Sources and Types of Error Measurement errors can arise from three possible origins: the measuring device, the measurement procedure, and the measured quantity itself. Usually the largest of these errors will determine the uncertainty in the data. Errors can be divided into two types: Systematic and Random errors Systematic errors arise from procedures, instruments, bias or ignorance. Systematic errors bias every measurement in the same direction, causing your measurement to consistently be higher or lower than the accepted value. Example: An ammeter with zero error reads higher or lower values of current. Systematic errors can be estimated from understanding the techniques and instrumentation used in an observation. Figure 1-3 The zero error of an ammeter is an example of systematic error Preliminaries 17 General Physics Module Phys 1011 AAU Random errors are uncontrollable differences between measurements because of equipment, environment or other sources, no matter how well designed and calibrated the tools are. Random errors are unbiased small variations that have both positive and negative values. In general, making multiple measurements and averaging can reduce the effect of random errors. Exercise Identify the systematic and random errors in the list below: a. A metre ruler with worn ends b. A dial instrument with a needle that is not properly zeroed c. Human reaction time that is always either too late or too early d. Fluctuations in the readings of an instrument e. Parallax error (human error) as shown in the diagram below Figure 1-4 Examples of parallax error in (a) measuring length and (b) measuring volume. 1.2.3 Accuracy vs. Precision In physics, there are two distinct and independent aspects of measurement related to uncertainties: Accuracy refers to the closeness of a measured value to the ‘true’ (standard or known) value. It describes how well we eliminate systematic error. Example: if you measure the weight of a given substance as 3.2 kg, but the actual or known weight is 10 kg, then your measurement is not accurate. In this case, your measurement is not close to the known value. Precision refers to the closeness of repeated measurements to each other without referring to the ‘true’ value. It describes how well we suppress random errors. Example: if you weigh an object five times, and get 3.2 kg each time, then your measurement is very precise. The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurement increment, the more precise the tool. Preliminaries 18 General Physics Module Phys 1011 AAU Precision and accuracy are independent. A measurement can be precise but inaccurate, or accurate but imprecise as illustrated by the several independent trials of shooting at a bullseye target in Figure 1-5 (below). Figure 1-5 Illustration of the difference between accuracy and precision. Exercise Given the standard value of g = 9.80665 m/s 2 and the following sets of five measurement values, write “YES” or “NO” in response to the question of accuracy and precision and discuss your responses. Set of g values Accurate? Precise? 1 g = {9.800, 9.806, 9.807, 9.810, 9.811} 2 g = {10, 4, 15, 6, 32} 3 g = {9.80665, 10, 9.8, 9.8067, 9.81} 4 g = {19.80, 19.806, 19.807, 19.810, 19.799} Errors can also be classified as absolute and relative: 1.2.3.1 Absolute error Absolute error is the difference between the measured value and the accepted value. 𝑨𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒆𝒓𝒓𝒐𝒓 = |𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 – 𝒂𝒄𝒄𝒆𝒑𝒕𝒆𝒅 𝒗𝒂𝒍𝒖𝒆| 1.2.3.2 Relative error Relative error is a fractional error defined as 𝑨𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒆𝒓𝒓𝒐𝒓 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑬𝒓𝒓𝒐𝒓 = 𝑨𝒄𝒄𝒆𝒑𝒕𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 1.2.3.3 Percentage error Percentage error is relative error expressed as a percentage: 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒆𝒓𝒓𝒐𝒓 = 𝒓𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒆𝒓𝒓𝒐𝒓 × 𝟏𝟎𝟎% Preliminaries 19 General Physics Module Phys 1011 AAU Examples 1. Suppose the accepted value of gravity is 𝑔 = 9.80665 𝑚/𝑠 2. If the measured value is 𝑔 = 9.81 𝑚/𝑠 2 , what is the absolute error? Solution 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟 = |𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 – 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒| 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟 = |9.81 – 9.80665| 𝑚/𝑠 2 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟 = 0.00335 𝑚/𝑠 2 2. What is the relative (percentage) error in example 1 above? Solution 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟 0.00335 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐸𝑟𝑟𝑜𝑟 = = = 3.4 × 10−4 𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 9.80665 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 = 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 × 100% = 3.4 × 10−4 × 100% = 0.03% 3. A digital ammeter gives the value of a current as 456mA. The accuracy or absolute uncertainty of the meter is 1mA. How should the reading be expressed? as (456 ± 1) mA. Solution For this reading, 1mA is the absolute error and 456 mA is the estimated value. Therefore, the reading should be expressed as (456 ± 1) mA. Exercises 1. A measure of the length of two rods using a metre ruler gives: 𝐿𝑒𝑛𝑔𝑡ℎ 1 = 2.000𝑚 ± 1 × 10−3 𝑚 𝐿𝑒𝑛𝑔𝑡ℎ 2 = 0.100𝑚 ± 1 × 10−3 𝑚 a. Find the relative errors in the two measurements. b. Find the percentage errors in the two measurements. 2. A student measured the length of a laboratory table as 4.5 m accurate to 0.1 of a meter. Find the absolute, relative and percentage errors in this measurement. 3. The values below are results of a road test for a new car. Speed Speedometer correction (km/h) Indicated 60 80 100 111 Actual 59 78 96 104 Preliminaries 20 General Physics Module Phys 1011 AAU (a) Find the relative (or fractional) error at an actual speed of 59 km/h. (b) What is the percentage error at 59 km/h? (c) Find the relative (or fractional) error at an actual speed of 96 km/h. (d) What is the percentage error at 96 km/h? (e) Draw a graph of the relative error versus the speed of the car. (f) what is the relation of the relative errors to the speed of the car? 1.2.4 Quantifying Uncertainties We will now apply some basic statistics to quantify random errors. 1.2.4.1 The mean Suppose a quantity 𝑥 is measured 𝑁 times. A sample of the measured values is (𝑥1 , 𝑥2 ,... 𝑥𝑁 ). We want the mean, 𝜇, of the population from which such a data set was randomly drawn. We can approximate 𝜇 with the sample mean (average) of this particular set of N data points: 1 𝜇 ≅ 𝑥̅ = 𝑁 ∑𝑁 𝑖=1 𝑥𝑖 Note that 𝑥̅ is not the true mean of the population, because we only measured a small subset of the population. But it is our best guess and, statistically, it is an unbiased predictor of the true mean 𝜇. 1.2.4.2 The standard deviation The precision of the value of 𝑥 is determined by the sample standard deviation, 𝑠𝑥 ,defined as ∑𝑁 𝑖=1(𝑥𝑖 −𝑥̅ ) 2 𝑠𝑥 = √ 𝑁−1 The square of the sample standard deviation is called the sample variance, 𝑠𝑥2. The sample standard deviation is our best estimate of the true statistical standard deviation 𝜎𝑥 of the population from which the measurements were randomly drawn. 1.2.4.3 The standard error (uncertainty) If we do not care about the standard deviation of a single measurement 𝑥 but, rather, how well we can rely on the mean value, 𝑥̅ , then we should use the standard error or standard deviation of the mean 𝑠𝑥̅. This is found by dividing the sample standard deviation by √𝑁: 𝑠𝑥 𝑠𝑥̅ = √𝑁 1.2.4.4 Reporting Data Under normal circumstances, the best estimate of a measured value 𝑥 predicted from a set of measurements {𝑥𝑖 } is reported as 𝑥 = 𝑥̅ ± 𝑠𝑥̅ , where the standard error is now the statistical uncertainty 𝛿𝑥 = 𝑠𝑥̅. The uncertainties should be given to the same number of decimal places as the measured values. Example: 𝑥 = 𝑥̅ ± 𝑠𝑥̅ = (434.2 ± 1.6) nm. Preliminaries 21 General Physics Module Phys 1011 AAU Example: g revisited Suppose, in a physics lab session, students measured the acceleration due to gravity (g) 40 times. How well is the value of g determined by these measurements? Values of g measured in cm/s2 961 972 979 983 986 965 976 979 966 975 981 985 987 991 983 984 974 981 989 996 968 978 979 984 987 993 990 984 970 977 981 984 992 994 988 985 974 975 971 980 Solution ∑𝑁 𝑖=1 𝑔𝑖 39227 Mean of g: 𝑔̅ = = = 981 cm/s2 𝑁 40 Standard deviation First find the deviations of the values of g from their mean value of 981 cm/s2 -20 -9 -2 2 5 -16 -5 -2 -15 -6 0 4 6 10 2 3 -7 0 8 15 -13 -3 -2 3 6 12 9 3 -11 -4 0 3 11 13 7 4 -7 -6 -10 -1 Then find the sum of squared deviations and divide it by the number of values minus 1. Finally, take the square root to determine the standard deviation: ∑𝑁 (𝑔𝑖 − 𝑔̅ )2 2711 𝑠𝑔 = √ 𝑖=1 =√ = 8.33 c𝑚/𝑠 2 𝑁−1 39 The standard error (uncertainty in the measurement of g) 𝑠𝑔 8.33 𝑠𝑔̅ = = = 1.32 cm/s2 √𝑁 √40 Therefore, the students should report the value of g as 𝑔 = 𝑔̅ ± 𝑠𝑔̅ = (981 ± 1) 𝑐𝑚/𝑠 2. Exercise The temperature of air is measured at different times of a certain day and the following set of readings was recorded. Rec. No. 1 2 3 4 5 6 7 8 9 10 T (℃) 16 19 18 16 17 19 20 15 17 13 Find (a) the mean temperature of the day, (b) the standard deviation of the temperature data, (c) the standard error and (d) the final result in the form 𝑇 = 𝑇̅ ± 𝛿𝑇. Preliminaries 22 General Physics Module Phys 1011 AAU 1.2.5 Error Propagation Measurement uncertainties propagate through calculations that depend on several uncertain quantities. Suppose that you have two quantities 𝑥 and 𝑦, each with an uncertainty 𝛿𝑥 and 𝛿𝑦, respectively. What is the uncertainty of the quantity 𝑥 ± 𝑦 or 𝑥𝑦 (𝑜𝑟 𝑥 ⁄𝑦)? The rules for uncertainty propagation assume that the errors 𝛿𝑥 and 𝛿𝑦 are uncorrelated, i.e., they are completely random. 1. Multiplication by an exact number: If 𝑧 = 𝑐𝑥, then 𝛿𝑧 = 𝑐𝛿𝑥 2. Addition or subtraction by an exact number: If 𝑧 = 𝑐 + 𝑥, then 𝛿𝑧 = 𝛿𝑥 3. Addition or subtraction: If 𝑧 = 𝑥 ± 𝑦, then 𝛿𝑧 = √(𝛿𝑥)2 + (𝛿𝑦)2 𝛿𝑧 𝛿𝑥 2 𝛿𝑦 2 4. Multiplication or division: If 𝑧 = 𝑥𝑦 or 𝑧 = 𝑥/𝑦, then 𝑧 = √( 𝑥 ) + ( 𝑦 ) 𝛿𝑧 𝛿𝑥 5. Power: If 𝑧 = 𝑥 𝑐 , then 𝑧 =𝑐 𝑥 Examples 1. A measurement of the thickness of a pack of cards gives the value 𝑙1 = (32.3 ± 0.5) mm. Some cards are removed and the thickness is measured again, giving the value 𝑙2 = (20.4 ± 0.5) mm. what is the thickness of the removed cards together? Solution The estimated thickness of the removed cards together is 𝑙 = 32.3 − 20.4 = 11.9. Then by rule 3, 𝛿𝑙 = √(𝛿𝑙1 )2 + (𝛿𝑙2 )2 = √(0.5)2 + (0.5)2 = 0.7 mm 𝑙 = (11.9 ± 0.7 ) mm 2. A piece of paper is measured and found to be 5.63 ± 0.15mm wide and 64.2 ± 0.7mm long. What is the area of this piece of paper? Solution Data: length = 64.2 ± 0.77mm and width = 5.63 ± 0.15mm 𝐴𝑟𝑒𝑎 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ 𝐴 = (64.2 ± 0.7𝑚𝑚) × (5.63 ± 0.15𝑚𝑚) First work out the answer by just using the numbers, forgetting the errors 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝐴𝑟𝑒𝑎 (𝐴) = 𝑙 × 𝑤 = 64.2 × 5.63 = 361.446 Then, by rule 4: 𝛿𝐴 𝛿𝑙 2 𝛿𝑤 2 0.7 2 0.15 2 = √( ) + ( ) = √( ) +( ) = 0.0288 𝐴 𝑙 𝑤 64.2 5.63 𝛿𝐴 = 361.446 × 0.0288 = 10.4 Preliminaries 23 General Physics Module Phys 1011 AAU Report the area as 𝐴 = (361.4 ± 10.4) 𝑚𝑚2 The final answer should have as many decimal places as the data with least number of decimal places. 1.2.6 Significant Figures Significant figures (sig. figs) are those digits in a measurement that carry meaning and contribute to its precision. Significant figures express the precision of a measuring tool. Here are the rules for identifying significant figures in a measurement: 1. All non-zero figures are significant: 25.4 has three significant figures. 2. All zeros between non-zeros are significant: 30.08 has four significant figures. 3. Zeros to the right of a non-zero figure but to the left of the decimal point are not significant (unless specified with a bar): 109 000 has three significant figures. 4. Zeros to the right of a decimal point but to the left of a non-zero figure are not significant: 0.050, only the last zero is significant. 5. Zeros to the right of the decimal point and following a non-zero figure are significant: 304.50 have five significant figures. Example Determine the number of sig. figs. For the following numbers 21000 3250000 42210000 0.0012 469 1786 1.0 0.00843 508.6 0.18 0.234 0.6780 67 65.0 5.060 Solution Number sig. figs. Number sig. figs. Number sig. figs. 21000 Two (rule 3) 3250000 Three (rule 3) 42210000 Four (rule 3) 0.0012 Two (rule 4) 469 Three (rule 1) 1786 Four (rule 1) 1.0 Two (rule 5) 0.00843 Three (rule 4) 508.6 Four (rule 2) 0.18 Two (rule 4) 0.234 Three (rule 4) 0.6780 Four (rule 4) 67 Two (rule 1) 65.0 Three (rule 5) 5.060 Four (rule 5) Preliminaries 24 General Physics Module Phys 1011 AAU Exercises 1. Find the number of sig. figs for the following numbers 90000 70000000.0 84.10000 10082 0.0025 3008000 70000000 0.00008914 0.000339 2. Which are significant: Trailing zeros or leading zeros? When performing calculations, we must be careful about significant figures. When adding, subtracting, multiplying or dividing numbers, the answer should not be more precise than the number with the least number of significant figures. Examples 1. Find the difference 264.68 – 2.4711 Solution Put a question mark at all doubtful places and do the calculation: 264.68? ? −2.4711 = 262.21? ? = 262.21 In this calculation, the least number of sig. figs. is five so the final answer must have five sig. figs 2. Evaluate the product 2.345 × 3.56 = 8.3482 = 8.35. Solution Multiply the numbers using question marks at all doubtful places. The final answer has three sig. figs because the least number of sig. figs. in the operation is three Exercises 1. The following values are part of a set of experimental data: 618.5 cm and 1450.6mm. Write the sum of these values correct to the right number of significant figures. 2. The following values are part of a set of experimental data: 34.7cm and 19.65mm. How many significant figures would be present in the product and ratio of these two figures? Preliminaries 25 General Physics Module Phys 1011 AAU 1.2.7 Order of magnitude The order of magnitude of a number is the value of the number rounded to the nearest power of ten (no significant figures). It is used if you need to give only an indication of how large or small a number is, and only the power of ten is given. It also indicates that the accuracy of the measurement is limited. Examples 1. The velocity of light is 3.0 × 108 m/s. The order of magnitude of this velocity is 108. 2. The order of magnitude of 142 particles is 102. Since 142 in scientific notation is 1.42 × 102. 3. The order of magnitude for 10kV would be given as 104. 4. The mean free path of a nitrogen molecule at room temperature and one atmosphere is 59 nm. The order of magnitude is 10−8. 5. The number of molecules in a mole is of the order of 1023. 6. Planck’s constant is of the order of 10−34. Exercises 1. What is the order of magnitude of the gravitational constant? 2. What is the order of magnitude of the distance between the Earth and the Sun? 3. What is the order of magnitude of a charge of 100nC? 4. What is the order of magnitude of the electron charge? 5. What is the order of magnitude of the mass of a proton? 1.3 Vectors: Addition, Components, Magnitude and Direction Learning Outcome After completing this section, students are expected to: describe the difference between vector and scalar quantities. recognise quantities as either scalars or vectors. Use geometric methods to add vectors and find their magnitudes and directions in a plane. use algebraic method to find resultant of vectors and find components of vectors. define a unit vector. solve problems about vectors. Preliminaries 26 General Physics Module Phys 1011 AAU 1.3.1 Vectors Vectors are physical quantities such as force, velocity, acceleration and momentum that are expressed in terms of both magnitude (with a unit) and direction. Geometrically, they are represented by arrows in two or three dimensions because arrows have both characteristics of a vector: magnitude and direction. Note that quantities with a sense of direction such as angles and time are not vectors, because they do not obey the law of parallelogram. A physical quantity is a genuine vector if it adds to another vector according to the law of parallelogram. That is true vectors obey the law of parallelogram. Finite angles are physical quantities with a sense of direction (clockwise or counterclockwise), but they are not vectors because they violate the law of parallelogram. Physical quantities such as length, volume, mass, density, temperature and time can be expressed in terms of magnitude or size alone (together with a unit). These are called scalar quantities. Exercise Categorize each quantity as being either a vector or a scalar. Quantity Category Quantity Category Quantity Category 2 5m 256 bytes 9.8 m/s , up 30 m/sec, East 4000 Calories 10 μC 5 km, North 5 kg, down 45° South of East 20 degrees Celsius 5 N, down 45° clockwise 1.3.2 Vector notation There are many ways of writing the symbol of a vector. Vectors are denoted by bold-face letter or a letter with an arrow above it. For example, Bold face: 𝑨 Arrow above: 𝑣⃗ Harpoon above: 𝐹⃑ ̅̅̅̅ Overbar: 𝑃𝑄 1.3.3 Geometrical representation of vectors Vectors are geometrically represented by drawing arrows. The length of the arrow gives the magnitude of the vector and the arrowhead indicates direction of the vector. Figure 1-6(c) (below) shows that a ruler is used to measure the magnitude of the vector 𝐴𝐵 ⃗⃗⃗⃗⃗⃗ as 10.3 units, defining 1 cm as 1 unit. If, for example, the vector is a displacement vector, 10.3 cm on the ruler may represent 13.3 km on the ground; if it is a force vector, 10.3 cm may represent 10.3 N; if it is a velocity vector 10.3 cm may represent 10.3 m/s and so on. A protractor is used to specify the direction of the vector, which is shown to be 29.1o North of East. Preliminaries 27 General Physics Module Phys 1011 AAU Figure 1-6 A geometrical representation of a vector. (a) the displacement vector from the tent to ⃗⃗⃗. (c) The magnitude the park is represented by the arrow AB. (b) this displacement is denoted by 𝑫 and direction of a vector are determined using a ruler and a protractor. Figure 1-6(a)(above) clearly shows the difference between distance (scalar) and displacement (vector) between the tent and the camp. The distance between the tent and the camp depends on the route taken while the displacement from the tent to the camp has a fixed magnitude and a particular direction. 1.3.4 Equality of Two Vectors Two vectors are equal if they have the same magnitude and direction. In Figure 1-7 (below), vectors 𝐴⃗ and 𝐵 ⃗⃗ are equal whereas vectors 𝐶⃗ and 𝐷 ⃗⃗ are not equal eventhough they have the same magnitude. Note that two vectors need not be located at the same point in space to be equal. Moving a vector from one point in space to another doesn’t change its magnitude or its direction. ⃗⃗ , 𝑩 Figure 1-7 The vectors 𝑨 ⃗⃗⃗ and 𝑫 ⃗⃗ and vector 𝑫 ⃗⃗⃗ are equal vectors. Although vector 𝑪 ⃗⃗⃗ have the same magnitude they are not equal vectors as they have different directions. Exercises 1. Using a ruler and a protractor determine the magnitude and direction of the vector in Figure 1-6(b). 2. Using a ruler and a protractor check that three of the vectors in Figure 1-7 are equal vectors. Preliminaries 28 General Physics Module Phys 1011 AAU 1.3.5 Adding and Subtracting Vectors geometrically Two vectors can be added geometrically using the tail-to-head method (also called the triangle rule) or the parallelogram rule. 1.3.5.1 Tail-to-head method (triangle rule) To add vectors, place the tail of one vector at the head of the other vector. The resultant is obtained by joining the tail of the first vector to the head of the second vector. Figure 1-8 (b) and (c) show that vector addition is commutative. The tail-to-head method can be applied to three or more vectors (Figure 1-9). For any two vectors, ⃗⃗ + 𝑩 𝑨 ⃗⃗⃗ = 𝑩 ⃗⃗ , ⃗⃗⃗ + 𝑨 ⃗⃗ + 𝑩 𝑨 ⃗⃗⃗ ≠ 𝐴 + 𝐵, ⃗⃗ − 𝑩 𝑨 ⃗⃗⃗ ≠ 𝐴 − 𝐵 Figure 1-8 Head-to-tail method of vector addition. Geometry shows that ⃗𝑨⃗ + ⃗𝑩 ⃗⃗ + ⃗𝑨 ⃗⃗ = ⃗𝑩 ⃗⃗. Figure 1-9 Tail-to-head method for drawing the resultant of four vectors Preliminaries 29 General Physics Module Phys 1011 AAU 1.3.5.2 Parallelogram Rule Alternatively, place both vectors with their tails joined. Construct a parallelogram taking the two vectors as the two adjacent sides. The diagonal is the resultant vector (Figure 1-10). Figure 1-10 Parallelogram rule of vector addition Subtraction of vectors is the same as adding the negative of the second vector to the first as shown in Figure 1-11. Figure 1-11 Geometric subtraction of vectors, 𝐴⃗ − 𝐵 ⃗⃗ = 𝐴⃗ + (−𝐵 ⃗⃗). If the sum of two vectors is zero, one is said to be the negative of the other. That is, if 𝑨 + 𝑩 = 𝟎, then 𝑩 = −𝑨. Preliminaries 30 General Physics Module Phys 1011 AAU Example Three displacement vectors 𝐴⃗, 𝐵 ⃗⃗, and 𝐶⃗ are specified by their magnitudes in centimeters and by their direction angles with a horizontal line as shown. Choose a convenient scale and use a ruler and a protractor to find the following vector sums: (a) 𝑅⃗⃗ = 𝐴⃗ + 𝐵 ⃗⃗ = 𝐴⃗ − 𝐵 ⃗⃗, (b) 𝐷 ⃗⃗, and (c) 𝑆⃗ = 𝐴⃗ − 3𝐵⃗⃗ + 𝐶⃗. For parts (a) and (b) we use the parallelogram rule. For (c) we use the tail-to-head method. Solution For parts (a) and (b): R = 5.8 cm and 𝜃𝑅 ≈ 0°; D = 16.2 cm and 𝜃𝐷 = 49.3° For part (C) Preliminaries 31 General Physics Module Phys 1011 AAU Exercises 1. Using a ruler and protractor, find the sum and difference of the following vectors. Draw the vectors to scale, say 1 cm = 2 units of the given vector. a. ⃗A⃗ = 5 N, East and ⃗B⃗ = 8 N, North of East b. ⃗C⃗ = 4 m/s, East and ⃗D⃗ = 8 m/s, South c. ⃗⃗ = 4 m, North, F E ⃗⃗ = 8 m South ⃗⃗ = 4 m, East and G 2. Using the three displacement vectors 𝐵 ⃗⃗, 𝐶⃗ and 𝐹⃗ in the example above, choose a convenient scale, and use a ruler and a protractor to find vectors (a) 𝑆⃗ = 𝐶⃗ + 𝐹⃗ , (b) 𝐺⃗ = 𝐹⃗ − 𝐶⃗ and (c) 𝐽⃗ = 𝐴⃗ + 2𝐵 ⃗⃗ − 𝐹⃗. Use the tail-to-head method. 3. Discuss why the following equality does not hold: 𝐴⃗ + 𝐵 ⃗⃗ = 𝐴 + 𝐵, where A and B without arrows represent magnitudes only. 1.3.6 Components of a vector One way of finding the components of a vector uses the rectangular coordinate system as shown in Figure 1-12. When we know the scalar components 𝐴𝑥 and 𝐴𝑦 of a vector 𝐴⃗, we can find its magnitude 𝐴 and its direction angle 𝜃𝐴. The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis. The angle 𝜃𝐴 is measured in the counterclockwise direction from the +x-axis to the vector. y ⃗𝑨⃗ 𝑨𝒚 = 𝑨 𝐬𝐢𝐧 𝜽𝑨 𝜽𝑨 0 𝑨𝒙 = 𝑨 𝐜𝐨𝐬 𝜽𝑨 x Figure 1-12 Components of a vector. The vector 𝐴⃗ can be expressed as the sum of two vectors, 𝐴⃗𝑥 and 𝐴⃗𝑦 , which stand perpendicular to each other in the rectangular (aka Cartesian) coordinate system. That is, 𝐴⃗ = 𝐴⃗𝑥 + 𝐴⃗𝑦 𝐴⃗𝑥 and 𝐴⃗𝑦 are vector components along the x-axis and y-axis respectively. Applying simple trigonometry, we find the scalar components of the vector 𝐴⃗ in terms of its magnitude (𝐴) and direction angle (𝜃𝐴 ): 𝐴𝑥 = 𝐴 cos 𝜃𝐴 , 𝐴𝑦 = 𝐴 sin 𝜃𝐴 Given the rectangular components of a vector, we can also determine the magnitude and direction of any vector by the (inverse) equations: Preliminaries 32 General Physics Module Phys 1011 AAU 𝐴 𝐴 = √𝐴2𝑥 + 𝐴2𝑦 𝜃𝐴 = tan−1 ( 𝑦 ) 𝐴𝑥 Scalar components of a vector 𝐴⃗ may be positive or negative depending on the quadrant in which the vector lies. Vectors in the first quadrant (I) have both scalar components positive and vectors in the third quadrant (III) have both scalar components negative. The calculated angle θ in the first quadrant is identical to the direction angle 𝜃𝐴. The calculated angle θ in the IV quadrant is identical to the direction angle 𝜃𝐴. For vectors in quadrants II and III, the direction angle of a vector is given by 𝜃𝑨 = 𝜃 + 180° counterclockwise from the positive x-axis. In Figure 1-13 all possibilities are shown. Figure 1-13 Generally, the components of a vector can be positive and negative scalar components. In quadrants I and IV the direction angle of a vector is the same as calculator outputs; in quadrants II and III the direction angle is 180° plus the calculated values. Example Find the magnitude and direction of each of the four vectors given below. Vector x-component y-component A 3 units 4 units B -6 units 8 units C -9 units -12 units D 12 units -16 units Solution Magnitude and direction of vector A 𝐴 = √𝐴2𝑥 + 𝐴2𝑦 = √32 + 42 = 5 units 𝐴𝑦 4 𝜃 = tan−1 (𝐴 ) = tan−1 (3) = 53.13° [correct calculator answer] 𝑥 Vector A is in the first quadrant at an angle of 53.13° counterclockwise from +x-axis. Preliminaries 33 General Physics Module Phys 1011 AAU Magnitude and direction of vector B 𝐵 = √𝐵𝑥2 + 𝐵𝑦2 = √(−6)2 + 82 = 10 units 𝐵 8 𝜃 = tan−1 (𝐵𝑦 ) = tan−1 (−6) = −53.13° [incorrect calculator answer] 𝑥 The inverse tangent function of a calculator returns values between -90° and +90°, so the answer will be acceptable only for the first and fourth quadrants. To get the right angles for vectors in the second or third quadrant, add 180° to the calculator result. So, the correct angle for vector B is 𝜃 = −53.13 + 180 = 126.87° counterclockwise from +x-axis (second quadrant) Magnitude and direction of vector C 𝐶 = √𝐶𝑥2 + 𝐶𝑦2 = √(−9)2 + (−12)2 = 15 units 𝐶 −12 𝜃 = tan−1 (𝐶𝑦 ) = tan−1 ( −9 ) = 53.13° [incorrect calculator answer] 𝑥 The correct angle for vector C is 𝜃 = 53.13 + 180 = 233.13° counterclockwise from +x-axis (third quadrant) Magnitude and direction of vector D 𝐷 = √𝐷𝑥2 + 𝐷𝑦2 = √122 + (−16)2 = 20 units 𝐷𝑦 −16 𝜃 = tan−1 (𝐷 ) = tan−1 ( 12 ) = −53.13° [correct calculator answer] 𝑥 Vector D is in the fourth quadrant at an angle of 53.13° clockwise from +x-axis. Note that the calculator answer is correct only half of the time. A ball is shot at an angle of 35° with the horizontal. If the initial velocity of the ball has a magnitude of 50 m/s find its horizontal (x) and vertical (y) components. Solution Horizontal component: 𝑣𝑥 = 𝑣 cos 𝜃 = 50 × cos 35° ≅ 41 𝑚/𝑠 Vertical component: 𝑣𝑦 = 𝑣 sin 𝜃 = 50 × sin 35° ≅ 29 𝑚/𝑠 Exercises 1. The magnitude of vector 𝐴⃗ is 35.0 units and points in the direction 325° counter-clockwise from the positive x-axis. Calculate the x- and y-components of this vector. 2. A boy ran 3 blocks west, 5 blocks north. Find the magnitude and direction of his resultant displacement with reference to East. 3. The rectangular components of four vectors are given below. Find the magnitude and direction of each vector. Preliminaries 34 General Physics Module Phys 1011 AAU Vector x-component y-component A 3 units -4 units B 5 units 1 unit C -2 units -3 units D -12 units 6 units 1.3.7 Adding and Subtracting Vectors Algebraically Algebraic addition and subtraction of vectors makes use of the components of the vectors. The algebraic sum of two or more vectors is obtained by adding x and y components separately. So, if 𝑅⃗⃗ = 𝐴⃗ + 𝐵 ⃗⃗, then the components of the resultant vector are: 𝑅𝑥 = 𝐴𝑥 + 𝐵𝑥 𝑅𝑦 = 𝐴𝑥 + 𝐵𝑦 ⃗⃗ = 𝐴⃗ − 𝐵 Similarly, if 𝐷 ⃗⃗, then the components of difference vector are 𝐷𝑥 = 𝐴𝑥 − 𝐵𝑥 𝐷𝑦 = 𝐴𝑥 − 𝐵𝑦 Remember: Algebraic addition and subtraction of vectors are carried out component by component. Example On a certain day, a student goes to school by first walking 2.0 km 45.0° north of east from her home to her ant’s home to drop a message. Then she walks 0.8 km in a direction 60.0° south of east of north where her school is located. (a) Determine the components of the student’s displacements in the first and second parts of her walk. (b) Determine the components of her total displacement for the trip from home to school. Solution Let 𝐴⃗ and 𝐵 ⃗⃗ be the first and second displacements. Their components are: 𝐴𝑥 = 𝐴 cos 𝜃 = 2.0 cos 45° = 1.4 𝑘𝑚 𝐵𝑥 = 𝐵 cos 𝜃 = 0.8 cos(−60°) = 0.40 𝑘𝑚 𝐴𝑦 = 𝐴 sin 𝜃 = 2.0 sin 45° = 1.4 𝑘𝑚 𝐵𝑦 = 𝐵 sin 𝜃 = 0.8 sin(−60°) = −0.69 𝑘𝑚 Preliminaries 35 General Physics Module Phys 1011 AAU The components of the total displacement 𝑅⃗⃗ are: 𝑅𝑥 = 𝐴𝑥 + 𝐵𝑥 = 1.4 + 0.4 = 1.8 𝑘𝑚 𝑅𝑦 = 𝐴𝑦 + 𝐵𝑦 = 1.4 − 0.69 = 0.7 𝑘𝑚 The diagram shows the student’s displacements. Compare the calculated values of the components to the corresponding components on the diagram. Exercises 1. The magnitude and direction (with respect to the +x axis) of two force vectors are 𝐴(50 N, 30°) and B (25 N, -60°). Find a. The components of 𝐴⃗ and 𝐵 ⃗⃗, b. The components of the sum 𝑆⃗ = 𝐴⃗ + 𝐵 ⃗⃗ and difference 𝐷 ⃗⃗ − 𝐴⃗. ⃗⃗ = 𝐵 2. Find the magnitude and direction of the vectors whose xy-components are a. (7 m, 11 m)and b. (24 m/s, -15 m/s) 1.3.8 Unit Vectors A convenient way of analysing vectors is to first describe them in terms of unit vectors. 1.3.8.1 The unit vector defined Definition 1: A unit vector is a vector of magnitude one. So, if 𝑢̂ denotes a unit vector, then |𝑢̂| = 1. ⃗⃗ 𝑨 Definition 2: A unit vector in the direction of any vector 𝐴⃗ is given by 𝐴⃗ 𝑢̂𝐴 = 𝐴, where 𝐴 = |𝐴⃗| ෝ𝑨 𝒖 Definition 2 is consistent with Definition 1 because 𝐴⃗ 𝐴 |𝑢̂𝐴 | = | | = = 1 𝐴 𝐴 Preliminaries 36 General Physics Module Phys 1011 AAU Using Definition 2, we can write any vector 𝐴⃗ in terms of its magnitude and parallel unit vector: 1.3.8.2 Unit vectors of the rectangular xy-coordinate system In the rectangular 𝑥𝑦 coordinate system, unit vectors are defined in the directions of +𝑥 and +𝑦. The unit vector in the direction of +𝑥 is denoted by 𝑖̂ and the unit vector in the direction of +𝑦 is denoted by 𝑗̂. Unit vectors provide a convenient notation in vector algebra. In a rectangular (Cartesian) xy-coordinate system in a plane (Figure 1-14, below), the vector component 𝐴⃗𝑥 of a vector 𝐴⃗ can be described by the scalar component (𝐴𝑥 ) and the unit vector 𝑖̂: 𝐴⃗𝑥 = 𝐴𝑥 𝑖̂ Similarly, the vector component in the y direction is described as 𝐴⃗𝑦 = 𝐴𝑦 𝑗̂ Where (𝐴𝑥 , 𝐴𝑦 ) are the scalar components of the vector 𝐴⃗. Figure 1-14 Vector components and scalar components of a vector in the Cartesian coordinate system. The vector components 𝐴⃗𝑥 and 𝐴⃗𝑦 are multiples of the unit vectors, the multiplying factors being the scalar components 𝐴𝑥 and 𝐴𝑦. Suppose 𝐴⃗ and 𝐵 ⃗⃗ are any two vectors in the rectangular coordinate system given by 𝐴⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ and ⃗⃗ = 𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂ 𝐵 The sum or difference, 𝐴⃗ ± 𝐵 ⃗⃗, is carried out component-by-component as shown below: Preliminaries 37 General Physics Module Phys 1011 AAU 𝑆⃗ = 𝐴⃗ + 𝐵 ⃗⃗ = (𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂) + (𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂) = (𝐴𝑥 + 𝐵𝑥 )𝑖̂ + (𝐴𝑦 + 𝐵𝑦 )𝑗̂ ⃗⃗ = 𝐴⃗ − 𝐵 𝐷 ⃗⃗ = (𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂) − (𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂) = (𝐴𝑥 − 𝐵𝑥 )𝑖̂ + (𝐴𝑦 − 𝐵𝑦 )𝑗̂ Similarly, the scalar multiple of a vector 𝐴⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ can also be found by multiplying each component of 𝐴⃗ by the given scalar. So, if 𝐵 ⃗⃗ = 𝛼 𝐴⃗, then ⃗⃗ = 𝛼 𝐴⃗ = 𝛼(𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ ) = (𝛼𝐴𝑥 )𝑖̂ + (𝛼𝐴𝑦 )𝑗̂ = 𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂, 𝐵 where 𝐵𝑥 = 𝛼𝐴𝑥 and 𝐵𝑦 = 𝛼𝐴𝑦 Example Given three vectors (a) 𝐴⃗ = 3𝑖̂ + 4𝑗̂, (b) 𝐵 ⃗⃗ = −3𝑖̂ + 𝑗̂ and (c) 𝐶⃗ = 𝑖̂ − 2𝑗̂. Assume all the vectors start at the origin. 1. Show the three vectors in the xy-coordinate system 2. Find 𝐴⃗ + 𝐶⃗ and 𝐴⃗ − 𝐵 ⃗⃗ + 3𝐶⃗, ⃗⃗ such that 𝐷 3. Find he vector 𝐷 ⃗⃗ − 𝐶⃗ = 0 ⃗⃗ + 𝐵 Solution 1. The vectors in the xy-coordinate system y 4 A 2 B 0 x -2 C -4 -2 0 2 4 2. 𝐴⃗ + 𝐶⃗ = (3𝑖̂ + 4𝑗̂) + (𝑖̂ − 2𝑗̂) = 4𝑖̂ + 2𝑗̂ 𝐴⃗ − 𝐵 ⃗⃗ + 3𝐶⃗ = (3𝑖̂ + 4𝑗̂) − (−3𝑖̂ + 𝑗̂) + 3(𝑖̂ − 2𝑗̂) = 9𝑖̂ − 3𝑗̂ ⃗⃗ − 𝐶⃗ = 0 ⇒ 𝐷 ⃗⃗ + 𝐵 3. 𝐷 ⃗⃗ = 𝐶⃗ − 𝐵 ⃗⃗ = (𝑖̂ − 2𝑗̂) − (−3𝑖̂ + 𝑗̂) = 4𝑖̂ − 3𝑗̂ Exercises 1. Given 𝐴⃗ = 4𝑖̂ − 10𝑗̂ and 𝐵 ⃗⃗ = 7𝑖̂ + 5𝑗̂, find 𝑏 such that 𝐴⃗ + 𝑏𝐵 ⃗⃗ is a vector pointing along the x-axis (i.e. has no y component). ⃗⃗⃗ = 2𝑖̂ − 𝑗̂ and 𝑁 2. If 𝑀 ⃗⃗ = 4𝑖̂ + 3𝑗̂, find 𝑘 and 𝑙 such that 𝑘𝑀 ⃗⃗⃗ + 𝑙𝑁 ⃗⃗ = 2𝑖̂ − 6𝑗̂. Preliminaries 38 General Physics Module Phys 1011 AAU 3. If |11𝑖̂ − 𝑘𝑗̂| = |√5(3𝑖 + 𝑗)|, find the possible values for 𝑘. 4. Find the angle that each of the following vectors makes with the x-axis: a. 𝑖̂ + 𝑗̂ b. 2.4𝑖̂ − 2𝑗̂ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ c. (3.2𝑖̂ − 𝑗̂) + 4𝑖̂ + 3𝑗̂ d. 4.2𝑥⃗ + 𝑦⃗, where 𝑥⃗ = 4𝑖̂ + 3𝑗̂ and 𝑦⃗ = 6𝑖̂ − 8𝑗̂ 5. Find the unit vector in the direction of a. 𝐴⃗ = 𝑖̂ + 𝑗̂ ⃗⃗ = 2.4𝑖̂ − 2𝑗̂ b. 𝐵 c. 𝑥⃗ = 4𝑖̂ + 3𝑗̂ d. 𝑦⃗ = 6𝑖̂ − 8𝑗̂ 6. In the diagram below, 𝐴⃗ is a vector of magnitude 35 cm; 𝐵 ⃗⃗ is a vector of magnitude 13 cm. If tan 𝛼 = 4/3 and tan 𝛽 = 5/12, a. write A and B in terms of 𝑖̂ and 𝑗̂ b. Show that 𝐴⃗ + 𝐵 ⃗⃗ 𝑞makes an angle of 45° to the x-axis. Preliminaries 39 General Physics Module Phys 1011 AAU 2 Kinematics in one Dimensions Learning Outcome After completing this Chapter, students are expected to: Understand motion and position Define distance and displacement Define speed and velocity Identify the average velocity and instantaneous velocity Define the average and instantaneous acceleration Derive the equations of motion with constant acceleration Solve related problems Introduction The study of motion and of physical concepts such as force and mass is called dynamics. The part of dynamics that describes motion without regard to its causes is called kinematics. In this chapter, we study the basic physics of motion where the object (race car, tectonic plate, blood cell, or any other object) moves along a single axis. Such motion is called one-dimensional motion. Though everything around us and elsewhere on earth seems stationary, move with Earth’s rotation, Earth orbits around the Sun, the Sun orbits around the centre of the Milky Way galaxy, and that galaxy migrates relative to other galaxies. If the motion is along a straight line only the line may be vertical, horizontal, or slanted, but it must be straight. With this, questions like does the moving object speed up, slow down, stop, or reverse direction and if the motion does change, how is time involved in the change can be attempted. Generally, motion is a continuous change of position; whereas position is location of an object relative to some reference point, often the origin (or zero point) of an axis such as the x-axis. In one-dimensional motion, moving objects are restricted to motion along a straight line. To describe such motion a + or - sign are all that is needed to specify direction. The positive direction of the axis is in the direction of increasing numbers (coordinates), which is to the right, as in Fig. 2.1. The opposite is the negative direction. Kinematics in one Dimensions 44 General Physics Module Phys 1011 AAU Figure 2-1: Position on an axis that is marked in units of length. The aforementioned information is, therefore, essential to understand and explain the remaining topics such as distance, displacement, speed, velocity and acceleration. 2.1 Distance and Displacement It is important to recognize the difference between distance and displacement. 2.1.1 Distance Distance travelled by a particle is the length of the path a particle takes from its initial position to its final position. Distance is a scalar quantity that can be denoted by any English alphabet s, d…, usually and is always indicated by a positive number. 2.1.2 Displacement Displacement of a particle is the shortest distance between its initial and final positions. It is an example of a vector quantity. Many other physical quantities, including position, velocity, and acceleration, also are vectors. Displacement is usually denoted by s, in which other English alphabets can be also used. Moreover, one can define displacement as change in position in some time interval. As a particle moves from an initial position xi to a final position xf, along the +x - axis, the displacement becomes 𝑥𝑓 − 𝑥𝑖. We use the Greek letter delta (Δ) to denote the change in a quantity, therefore, 𝜟𝒙 = 𝒙𝒇 − 𝒙𝒊 (2-1) Note that ∆𝑥 is not the product of ∆ and x; it is a single symbol that means “the change in the quantity x.” Remark: Displacement is the change in position of a particle. It is positive if the change in position is in the direction of increasing x (the + x direction), and negative if it is in the -x direction. Distance is the total path length travelled by a particle and is a scalar physical quantity that can be denoted by any English alphabet s, d…, usually. Distance travelled by a particle cannot be zero where displacement can be. Kinematics in one Dimensions 45 General Physics Module Phys 1011 AAU Examples: 1. Consider the case of a particle moving from point A to point D, as shown. The shortest distance is the distance between A and D in the counter clock wise direction which is √(5 𝑘𝑚 − 2 𝑘𝑚)2 + (3 𝑘𝑚)2 = 3√2 𝑘𝑚, and hence is the displacement of the particle at 450 in the south east direction, whereas the distance traversed is that followed the longer path, 5 𝑘𝑚 + 3 𝑘𝑚 + 2 𝑘𝑚 = 10𝑘𝑚. 5 km B A 3km C D 2km Similarly, a car travelled 10 km from known initial position and then back to the same point covers 20 km of total distance whereas its displacement is zero. For only forward or backward straight-line motion distance can be magnitude of a displacement. 2. Find the distance and displacement referring to the following diagram, if an object travels 4m to the positive x-axis and then 3m to the y-axis. 3m θ 4m Solution: Distance, 𝑠 = 4𝑚 + 3𝑚 = 7𝑚 Magnitude of the Displacement, 𝑠 = √(4𝑚)2 + (3𝑚)2 = 5𝑚 and its direction will be Kinematics in one Dimensions 46 General Physics Module Phys 1011 AAU 3 θ = tan−1 = tan−1 0.75 = 370 4 Therefore, the displacement is 𝑠⃗ = 5𝑚, 370 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠. 2.2 Speed and Velocity 2.2.1 Speed and Average Speed Speed is the ratio of the distance travelled by any object, irrespective of its direction, to the time it takes to travel that distance. Speed involves both distance and time. Therefore, its unit in SI system is meter per second (m/s or ms-1). There are other non-SI units like cm/s, km/h, km/min, ft/s,...etc. Ordinarily, the speed of a body does not remain uniform over a certain time interval we are considering. For instance, a bus that carry passengers between Addis Ababa and Ambo, with a half dozen intermediate stops, gains speed when it starts from a station and loses speed when it is approaching a station. Again, it slows down in motion while passing over the bridges etc. It also changes its direction quit often while proceeding along its journey. Thus, the speed of the bus is not uniform but variable. When the speed of a body varies, then we should use the term average speed since we are determining the average value of the spree over the time interval we are considering. Thus, the average speed of a body between two points is measured by the total distance covered by the body between those points divided by the total time taken by the body to travel that distance. Mathematically, this is expressed as ∆𝒙 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝒔𝒑𝒆𝒆𝒅 = ∆𝒕 (2-2) Where ∆𝒙 is the total distance covered by the body in the corresponding time interval ∆𝒕. Remark: Speed is a scalar physical quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. Contrast this to a slow-moving object that has a low speed; it covers a relatively small amount of distance in the same amount of time. An object with no movement at all has a zero speed. Speed is represented by “v” usually, and hence, ∆𝒙 𝒗= ∆𝒕 Kinematics in one Dimensions 47 General Physics Module Phys 1011 AAU 2.2.2 Velocity Average Velocity Velocity is related to speed in the same way that displacement is related to distance. It is the rate of displacement i.e., it is the ratio of the displacement which takes account of direction as well distance to time interval. In other words, velocity is the rate of change of position of a body in a particular direction. The velocity of a body is also expressed in meters per second (ms-1) or kilometres per sec or kilometres per hour (km/h) like speed. Its dimensions are M0L1T-1. Consider a particle moving along the x-axis, as shown in Figure 2.1. y A(t1) B(t1) O x x(t1) x(t2)-x(t1) x(t2) Figure 2-2: Variation of position with time. The instantaneous positions of the particle are uniquely determined by its distance from the origin O, i.e., its x coordinate. The distance travelled during the time interval 𝑡2 − 𝑡1 is 𝑥(𝑡2 ) − 𝑥(𝑡1 ). Kinematics in one Dimensions 48 General Physics Module Phys 1011 AAU y x(t2) x(t2)-x(t1) = Δx Distance (x) x(t1) t2 – t1 O t1 T2 x Time (t) Figure 2-3: Distance – time graph The ratio of the displacement of the particle to this time interval gives the average velocity between points A and B. i.e., on a graph of x versus t, vav is the slope of the straight line that connects two particular points on the x (t) curve: one is the point that corresponds to x(t2) and t2, and the other is the point that corresponds to x(t1) and t1. 𝑫𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 ̅= 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = 𝒗𝒂𝒗 = 𝒗 , 𝑻𝒊𝒎𝒆 Or 𝒙(𝒕𝟐 )−𝒙(𝒕𝟏 ) ∆𝒙 𝒗𝒂𝒗 = = (2-3) 𝒕𝟐 −𝒕𝟏 ∆𝒕 Remark: When the velocity of a body is constant, the velocity is very simply defined as the displacement travelled divided by time taken; but when the velocity changes with time, (i.e., either its speed changes or direction of motion changes or both changes), then a more careful definition is required. 2.2.3 Instantaneous velocity: We have now seen two ways to describe how fast something moves: average velocity and average speed, both of which are measured over a time interval Δt. However, the phrase “how fast” more Kinematics in one Dimensions 49 General Physics Module Phys 1011 AAU commonly refers to how fast a particle is moving at a given instant—its instantaneous velocity (or simply velocity) v. The velocity at any instant is obtained from the average velocity by shrinking the time interval Δt closer and closer to 0. 𝚫𝒙 𝒗𝒊𝒏𝒔𝒕 = 𝐥𝐢𝐦 𝚫𝒕→𝟎 𝚫𝒕 For uniform motion 𝑣𝑖𝑛𝑠𝑡 = 𝑣𝑎𝑣 Figure 2-4: x-t curve for non-uniform motion. Examples: 1. A car travelled a distance of 150 km in 3 hours on straight level road. What is the speed of the car? Solution: 𝒔 𝟏𝟓𝟎𝒌𝒎 𝒗= = = 𝟓𝒌𝒎/𝒉𝒓 𝒕 𝟑𝒉𝒓 2. A bus travelling at a speed of 120km/hr east wards continues its journey steadily on a straight level road for 3 hours. What is the actual displacement of the bus after 3 hours? Solution: 𝟏𝟐𝟎𝒌𝒎 ⃗⃗ = 𝒗 𝒔 ⃗⃗𝒂𝒗 𝒕 = ( 𝒆𝒂𝒔𝒕) (𝟑𝒉𝒓𝒔) = 𝟑𝟔𝟎 𝒌𝒎 𝒆𝒂𝒔𝒕 𝒉𝒓 Kinematics in one Dimensions 50 General Physics Module Phys 1011 AAU 3. A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0-min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a total time of 1.75 h, the rabbit wins the race. (a) Calculate the average speed of the rabbit. (b) What was his average speed before he stopped for a nap? Solution: Finding the overall average speed in part (a) is just a matter of dividing the total distance by the total time. Part (b) requires two equations and two unknowns, the latter turning out to be the two different average speeds: v1 before the nap and v2 after the nap. One equation is given in the statement of the problem (v2 _ 2v1), whereas the other comes from the fact that the rabbit ran for only 15 minutes because he napped for 90 minutes. (a) Find the rabbit’s overall average speed. 𝐭𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐬𝐩𝐞𝐞𝐝 = 𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝟒. 𝟎𝒌𝒎 = 𝟏. 𝟕𝟓 𝒉𝒓 = 𝟐. 𝟐𝟗 𝒌𝒎⁄𝒉𝒓 (b) Find the rabbit’s average speed before his nap. Sum the running times, and set the sum equal to 0.25 h: 𝒕𝟏 + 𝒕𝟐 = 𝟎. 𝟐𝟓𝟎 𝒉. Substitute 𝑡1 = 𝑑1 /𝑣1 𝑎𝑛𝑑 𝑡2 = 𝑑2 /𝑣2 : 𝒅𝟏 𝒅 𝒗𝟏 + 𝒗𝟐 = 𝟎. 𝟐𝟓𝟎𝒉 (1) 𝟐 Substitute 𝑣2 = 2𝑣1 and the values of 𝑑1 and 𝑑2 into Equation (1) 𝟎.𝟓𝟎𝟎 𝒌𝒎 𝟑.𝟓𝟎 𝒌𝒎 + = 𝟎. 𝟐𝟓𝟎 𝒉 (2) 𝒗𝟏 𝟐𝒗𝟏 Solve Equation (2) for 𝑣1 : 𝒗𝟏 = 𝟗. 𝟎𝟎 𝒌𝒎/𝒉 Kinematics in one Dimensions 51 General Physics Module Phys 1011 AAU 4. A car travelled 40km east in 1hr and then travelled 80km north in 2hrs. Calculate (a) its average speed, and (b) its average velocity 80km θ 40km Solution: (a) average speed, 𝒕𝒐𝒕𝒂𝒍 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒗𝒂𝒗 = 𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝟒𝟎𝒌𝒎 + 𝟖𝟎𝒌𝒎 = = 𝟒𝟎𝒌𝒎/𝒉𝒓 𝟏𝒉𝒓 + 𝟐𝒉𝒓 (b) Average velocity, 𝒕𝒐𝒕𝒂𝒍 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 ⃗⃗𝒂𝒗 = 𝒗 ⃗⃗𝒆𝒂𝒔𝒕 + 𝒗 =𝒗 ⃗⃗𝒏𝒐𝒓𝒕𝒉 𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 Magnitude of the average velocity, 𝒗𝒂𝒗 = √𝒗𝟐 𝒆𝒂𝒔𝒕 + 𝒗𝟐 𝒏𝒐𝒓𝒕𝒉 𝟖𝟎 𝟐 𝟐 [ ]𝒌𝒎 𝟒𝟎√𝟐𝒌𝒎 𝒗𝒂𝒗 = √(𝟒𝟎𝒌𝒎) +( 𝟐 ) = = 𝟓𝟔. 𝟓𝟔 𝒌𝒎/𝒉𝒓 𝒉𝒓 𝒉𝒓 𝒉𝒓 And its direction, 𝟖𝟎𝒌𝒎 𝜽 = 𝐭𝐚𝐧−𝟏 = 𝐭𝐚𝐧−𝟏 𝟐 = 𝟔𝟒𝟎 𝟒𝟎𝒌𝒎 Therefore, ⃗⃗𝒂𝒗 = 𝟓𝟔. 𝟓𝟔 𝒌𝒎⁄𝒉𝒓 , 𝟔𝟒° with respect to the east direction. 𝒗 5. The position vector of a body moving along the x-axis is given by 𝑥 = 10 𝑐𝑚⁄𝑠 2 𝑡 2. Compute its instantaneous velocity at time 𝑡 = 2𝑠. Kinematics in one Dimensions 52 General Physics Module Phys 1011 AAU Solution: ⃗⃗(𝒕 + ∆𝒕) − 𝒙 𝒙 ⃗⃗(𝒕) ⃗⃗𝒊𝒏𝒔 (𝒕) = 𝐥𝐢𝐦 𝒗 ∆𝒕→𝟎 ∆𝒕 𝟏𝟎 𝒄𝒎⁄𝒔𝟐 (𝒕 + ∆𝒕)𝟐 − 𝟏𝟎 𝒄𝒎⁄𝒔𝟐 𝒕𝟐 = 𝐥𝐢𝐦 ∆𝒕→𝟎 ∆𝒕 𝟏𝟎 𝒄𝒎⁄𝒔𝟐 𝒕𝟐 + 𝟏𝟎 𝒄𝒎⁄𝒔𝟐 (𝟐𝒕∆𝒕) + 𝟏𝟎 𝒄𝒎⁄𝒔𝟐 (∆𝒕)𝟐 − 𝟏𝟎 𝒄𝒎⁄𝒔𝟐 𝒕𝟐 = 𝐥𝐢𝐦 ∆𝒕→𝟎 ∆𝒕 𝟏𝟎𝒄𝒎⁄𝒔𝟐 (𝟐𝒕∆𝒕) 𝟏𝟎𝒄𝒎⁄𝒔𝟐 (∆𝒕)𝟐 = 𝐥𝐢𝐦 [ + ] = (𝟐𝟎𝒕) 𝒄𝒎⁄𝒔𝟐 ; ∆𝒕→𝟎 ∆𝒕 ∆𝒕 Hence at 𝒕 = 𝟐𝒔𝒆𝒄, 𝒗𝒊𝒏𝒔 = 𝟒𝟎𝒄𝒎/𝒔 2.3 Acceleration When velocity of a particle changes with time, the particle is said to be accelerating. For example, the magnitude of the velocity of a car increases when you step on the gas and decreases when you apply the brakes. Let us see how to quantify acceleration. Suppose an object that can be modelled as a particle moving along the x- axis has an initial velocity 𝑣𝑥𝑖 at time 𝑡𝑖 and a final velocity 𝑣𝑥𝑓 at time 𝑡𝑥𝑓 , The average acceleration 𝑎̅𝑥 = 𝑎𝑎𝑣 of the particle is defined as the change in velocity Δvx divided by the time interval Δt during which that changes occur: ∆𝒗𝒙 𝒗𝒙𝒇 −𝒗𝒙𝒊 ̅𝒙 = 𝒂 = (2-4) ∆𝒕 𝒕𝒇 −𝒕𝒊 As with velocity, when the motion being analysed is one-dimensional, we can use positive and negative signs to indicate the direction of the acceleration. Because the dimensions of velocity are L/T and the dimension of time is T, acceleration has dimensions of length divided by time squared, or L/T2. The SI unit of acceleration is meters per second squared (m/s2). There are other non SI unit, such as (cm/s2, ft/s2). In some situations, the value of the average acceleration may be different over different time intervals. It is therefore useful to define the instantaneous acceleration as the limit of the average acceleration as Δt approaches zero. ⃗⃗(𝒕+∆𝒕)−𝒗 𝒗 ⃗⃗(𝒕) ⃗⃗𝒊𝒏𝒔 (𝒕) = 𝒍𝒊𝒎 𝒂 (2-5) ∆𝒕→𝟎 ∆𝒕 Example: Kinematics in one Dimensions 53 General Physics Module Phys 1011 AAU 1. A particle is in motion and is accelerating. The functional form of the velocity is 𝑣(𝑡) = 20𝑡 − 5𝑡 2 𝑚⁄𝑠. Find the instantaneous acceleration at t = 1, 2, 3, and 5 s. Solution: ⃗⃗(𝒕 + ∆𝒕) − 𝒗 𝒗 ⃗⃗(𝒕) ⃗⃗𝒊𝒏𝒔 (𝒕) = 𝐥𝐢𝐦 𝒂 ∆𝒕→𝟎 ∆𝒕 𝟐𝟎(𝒕 + ∆𝒕) − 𝟓(𝒕 + ∆𝒕)𝟐 − (𝟐𝟎𝒕 − 𝟓𝒕𝟐 ) ⃗⃗𝒊𝒏𝒔 (𝒕) = 𝐥𝐢𝐦 𝒂 ∆𝒕→𝟎 ∆𝒕 ⃗⃗𝒊𝒏𝒔 (𝒕) = 𝐥𝐢𝐦 (𝟐𝟎 − 𝟓∆𝒕 − 𝟏𝟎𝒕) = 𝟐𝟎 − 𝟏𝟎𝒕 𝒂 ∆𝒕→𝟎 i) ⃗𝒂⃗𝒊𝒏𝒔 (𝒕 = 𝟏) = 𝟐𝟎 − 𝟏𝟎𝒕 = 𝟏𝟎𝒎/𝒔𝟐 ii) ⃗𝒂⃗𝒊𝒏𝒔 (𝒕 = 𝟐) = 𝟐𝟎 − 𝟏𝟎𝒕 = 𝟎𝒎/𝒔𝟐 iii) ⃗⃗𝒊𝒏𝒔 (𝒕 = 𝟑) = 𝟐𝟎 − 𝟏𝟎𝒕 = −𝟏𝟎𝒎/𝒔𝟐 𝒂 iv) ⃗⃗𝒊𝒏𝒔 (𝒕 = 𝟓) = 𝟐𝟎 − 𝟏𝟎𝒕 = −𝟑𝟎𝒎/𝒔𝟐 𝒂 Remark: When the object’s velocity and acceleration are in the same direction, the object is speeding up. On the other hand, when the object’s velocity and acceleration are in opposite directions, the object is slowing down (which we call deceleration). Average acceleration and instantaneous acceleration are equal during uniformly accelerated motion. Acceleration of a body is 2 𝑚⁄𝑠 2 means that the velocity increases by 2 𝑚⁄𝑠 every one second. 2.3.1 Motion with constant acceleration The simplest kind of accelerated motion is straight-line motion in which the acceleration is constant. This means that the velocity changes at the same rate throughout the motion. As can be seen from the velocity-time graph, the velocity is increasing by equal amounts in equal intervals of time. The slope of a chord between any two points on the line is the same as the slope of a tangent at any point, and the average and instantaneous accelerations are equal. Kinematics in one Dimensions 54 General Physics Module Phys 1011 AAU