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Module 1: Units and Significant Figures
1.1 The Speed of Light
When we observe and measure phenomena in the world, we try to
assign numbers to the physical quantities with as much accuracy as
we can possibly obtain from our measuring equipment. For example,
we may want to determine the speed of light, which we can calculate
by dividing the distance a known ray of light propagates over its
travel time,
speed of light = distancetime
. (1.1.1)
In 1983 the General Conference on Weights and Measures defined
the speed of light to be
c = 299, 792, 458 meters/second . (1.1.2)
This number was chosen to correspond to the most accurately
measured value of the speed of light and is well within the
experimental uncertainty.
1.2 International System of System of Units
The three quantities time, length, and the speed of light are
directly intertwined. Which quantities should we consider as base
and which ones as derived from the base quantities? For example,
are length and time base quantities while speed is a derived
quantity? This question is answered by convention. The basic system
of units used throughout science and technology today is the
internationally accepted Systme International (SI). It consists of
seven base quantities and their corresponding base units:
Mechanics is based on just the first three of these quantities,
the MKS or meter-kilogram-second system. An alternative metric
system to this, still widely used, is the so-called CGS system
(centimeter-gram-second). So far as distance and time measurements
are concerned, there is also wide use of British Imperial units
(especially in the USA) based on the foot (ft), the mile (mi),
etc., as units of length, and also making use of the minute, hour,
day and year as units of time.
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Base Quantity Base Unit Length meter (m) Mass kilogram (kg) Time
second (s) Electric Current ampere (A) Temperature Kelvin (K)
Amount of Substance mole (mol) Luminous Intensity candela (cd)
We shall refer to the dimension of the base quantity by the
quantity itself, for example
dim length length L, dim mass mass M, dim time time T.
(1.2.1)
1.3 The Atomic Clock and the Definition of the Second
Isaac Newton, in the Philosophiae Naturalis Principia
Mathematica (Mathematical Principles of Natural Philosophy),
distinguished between time as duration and an absolute concept of
time,
Absolute true and mathematical time, of itself and from its own
nature, flows equably without relation to anything external, and by
another name is called duration: relative, apparent, and common
time, is some sensible and external (whether accurate or unequable)
measure of duration by means of motion, which is commonly used
instead of true time; such as an hour, a day, a month, a year.
1.
The development of clocks based on atomic oscillations allowed
measures of timing with accuracy on the order of 1 part in
1014 , corresponding to errors of less than one microsecond (one
millionth of a second) per year. Given the incredible accuracy of
this measurement, and clear evidence that the best available
timekeepers were atomic in nature, the second (s) was redefined in
1967 by the International Committee on Weights and Measures as a
certain number of cycles of electromagnetic radiation emitted by
cesium atoms as they make transitions between two designated
quantum states:
The second is the duration of 9,192,631,770 periods of the
radiation corresponding to the transition between the two hyperfine
levels of the ground state of the cesium 133 atom.
1.4 The meter
The meter was originally defined as 1/10,000,000 of the arc from
the Equator to the North Pole along the meridian passing through
Paris. To aid in calibration and ease of comparison, the meter was
redefined in terms of a length scale etched into a platinum bar
1 Isaac Newton. Mathematical Principles of Natural Philosophy.
Translated by Andrew Motte (1729).
Revised by Florian Cajori. Berkeley: University of California
Press, 1934. p. 6.
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preserved near Paris. Once laser light was engineered, the meter
was redefined by the 17th Confrence Gnrale des Poids et Msures
(CGPM) in 1983 to be a certain number of wavelengths of a
particular monochromatic laser beam.
The metre is the length of the path travelled by light in vacuum
during a time interval of 1/299 792 458 of a second.
Example 1: Light Year Astronomical distances are sometimes
described in terms of light-years (ly). A light-year is the
distance that light will travel in one year (yr). How far in meters
does light travel in one year?
Solution: Using the relationship
distance = (speed of light) (time) , one light year corresponds
to a distance. Since the speed of light is given in terms of meters
per second, we need to know how many seconds are in a year. We can
accomplish this by converting units. We know that
1 year = 365.25 days, 1 day = 24 hours, 1 hour = 60 minutes, 1
minute = 60 seconds
Putting this together we find that the number of seconds in a
year is
1 year = 365.25 day( ) 24 hours1day
60 min1 hour
60 s
1 min
=31,557,600 s . (1.4.1)
So the distance that light travels in a one year is
1 ly = 299,792,458 m1s
31,557,600 s
1 yr
1 yr( )= 9.461 1015 m . (1.4.2)
The distance to the nearest star, Alpha Centauri, is three light
years.
A standard astronomical unit is the parsec. One parsec is the
distance at which there is one arcsecond = 1/3600 degree angular
separation between two objects that are separated by the distance
of one astronomical unit, 111AU 1.50 10 m= , which is the mean
distance between the earth and sun. One astronomical unit is
roughly equivalent to eight light minutes, 1AU 8.3l-min= One parsec
is equal to 3.26 light years, where one light year is the distance
that light travels in one earth year, 51pc 3.26ly 2.06 10 AU= =
where 151ly 9.46 10 m= .
1.5 Mass
The unit of mass, the kilogram (kg), remains the only base unit
in the International System of Units (SI) that is still defined in
terms of a physical artifact, known as the International Prototype
of the Standard Kilogram. The prototype was
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made in 1879 by George Matthey (of Johnson Matthey) in the form
of a cylinder, 39 mm high and 39 mm in diameter, consisting of an
alloy of 90 % platinum and 10 % iridium. The international
prototype is kept at the Bureau International des Poids et Mesures
(BIPM) at Sevres, France under conditions specified by the 1st
Confrence Gnrale des Poids et Msures (CGPM) in 1889 when it
sanctioned the prototype and declared This prototype shall
henceforth be considered to be the unit of mass. It is stored at
atmospheric pressure in a specially designed triple bell-jar. The
prototype is kept in a vault with six official copies.
Figure 1.1 International Prototype of the Standard Kilogram
The 3rd CGPM (1901), in a declaration intended to end the
ambiguity in popular usage concerning the word weight confirmed
that:
The kilogram is the unit of mass; it is equal to the mass of the
international prototype of the kilogram.
There is a stainless steel one-kilogram standard that can travel
for comparisons. In practice it is more common to quote a
conventional mass value (or weight-in-air, as measured with the
effect of buoyancy), than the standard mass. Standard mass is
normally only used in specialized measurements wherever suitable
copies of the prototype are stored.
Example 2: The International Prototype Kilogram Determine the
type of shape and dimensions of the platinum-iridium prototype
kilogram such that it has the smallest surface area for a given
volume. The standard kilogram is an alloy of 90 % platinum and 10 %
iridium. The density of the alloy is
= 21.56 g cm3 . You may want to consider the following
questions:
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1) Is there any reason that the surface area of the standard
could be important?
2) What is the appropriate density to use?
3) What shape (that is, sphere, cube, right cylinder,
parallelepiped, etc.) has the smallest surface area for a given
volume?
4) Why was a right-circular cylinder chosen?
Solution: The standard kilogram is an alloy of 90 % platinum and
10 % iridium. The density of platinum is
21.45 g cm- 3 and the density of iridium is
22.55 g cm3 . Thus the density of the standard kilogram,
= 21.56 g cm3 , and its volume is
V = m / 1000 g / 22 g cm3 46.38 cm3 . (1.5.1)
Corrosion would affect the mass through chemical reaction;
platinum and iridium were chosen for the standards composition as
they resist corrosion.
To further minimize corrosion, the shape should be chosen to
have the least surface area. Ideally, this would be a sphere, but
as spheres roll easily they become impractical, whereas cylinders
have flat surfaces which prevent this. The volume for a cylinder or
radius
r and height
h is a constant and given by
V = pir 2h . (1.5.2)
The surface area can be expressed in terms of the radius
r as
A = 2pir 2 + 2pirh = 2pir 2 + 2Vr
. (1.5.3)
To find the smallest surface area, minimize the area with
respect to the radius
dAdr
= 4pir 2Vr
2 = 0 . (1.5.4)
Solve for the radius
r3
=
V2pi
=
pir 2h2pi
. (1.5.5)
Thus the radius is one half the height,
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r =h2
. (1.5.6)
For the standard mass, the radius is
r =V2pi
1 3
=
46.38 cm3
2pi
1 3
1.95 cm . (1.5.7)
Twice this radius is the diameter of the standard kilogram.
Alternative Definition of Mass
Since the prototype kilogram is an artifact, there are some
intrinsic problems associated with its use as a standard. It may be
damaged, or destroyed. The prototype gains atoms due to environment
wear and cleaning, at a rate of change of mass corresponding to
approximately
1 g / year (
1 g 1microgram 1 10-6 g ).
Several new approaches to defining the SI unit of mass (kg) are
currently being explored. One possibility is to define the kilogram
as a fixed number of atoms of a particular substance, thus relating
the kilogram to an atomic mass. Silicon is a good candidate for
this approach because it can be grown as a large single crystal, in
a very pure form.
Example 3: Mass of a Silicon Crystal
A given standard unit cell of silicon has a volume
V0 and contains
N0 atoms. The number of molecules in a given mole of substance
is given by Avogadros constant
N A = 6.0221415 1023
mole-1 . The molar mass of silicon is given by
Mmolar . Find the
mass
m of a volume
V in terms of
V0 ,
N0 , V ,
Mmolar , and
N A .
Solution: The mass
m0 of the unit cell is the density of silicon cell multiplied by
the volume of the cell
V0 ,
m0 = V0 . (1.5.8)
The number of moles in the unit cell is the total mass,
m0 , of the cell, divided by the molar mass
Mmolar ,
n0 = m0 / Mmolar = V0 / Mmolar . (1.5.9)
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The number of atoms in the unit cell is the number of moles
n0 times the Avogadro constant,
N A ,
N0 = n0N A =V0N A
Mmolar
(1.5.10)
The density of the crystal is related to the mass
m of the crystal divided by the volume
V of the crystal,
= m / V (1.5.11)
So the number of atoms in the unit cell can be expressed as
N0 =mV0N AVM
molar
(1.5.12)
So the mass of the crystal is
m =M
molar
N A
VV0
N0 (1.5.13)
The molar mass, unit cell volume and volume of the crystal can
all be measured directly. Notice that
Mmolar / N A is the mass of a single atom, and
(V / V0 )N0 is the number of atoms in the volume. This approach
is therefore reduced to the problem of measuring the Avogadro
constant,
N A , with a relative uncertainty of 1 part in 108, which is
equivalent
to the uncertainty in the present definition of the
kilogram.
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1.6 Radians and Steradians
Radians
Consider the triangle drawn in Figure 1.6.1
Figure 1.2 Trigonometric relations
You know the basic trigonometric functions of an angle in a
right-angled triangle
ONB :
sin( ) = y / r , (1.6.1)
cos( ) = x / r , (1.6.2)
tan( ) = y / x (1.6.3)
It is very important to become familiar with using the measure
of the angle itself as expressed in radians [rad]. Let be the angle
between two straight lines
OX and
OP . If we draw a circle of any radius
r
centered at
O , the lines
OP and
OX cut the circle at the points
A and
B where
OA = OB = r . If the length of the arc
AB is
s ,
the radian measure of is given by
= s / r ,
and is the same for circles of all radii centered at
O -- just as the ratios
y / r
and
y / x
are the same for all right triangles with the angle at
O . As approaches 360o, s approaches the complete
circumference
2pir of the circle, so that 360o = 2pi rad .
Lets compare the behavior of
sin( ) ,
tan( ) and itself for small angles. One can see from the diagram
that
s / r > y / r . It is less obvious that
y / x > . It is very instructive to plot
sin( ) ,
tan( ) , and as functions of
[rad] between
0 and
pi / 2 on the same graph (see Figure 1.3).
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Figure 1.3 Radians compared to trigonometric functions.
For small , the values of all three functions are almost equal.
But how small is small? An acceptable condition is for
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can be used almost interchangeably, within some small percentage
error. This is the basis of many useful approximations in physics
calculations.
Steradians
The steradian (sr) is the unit of solid angle that, having its
vertex in the center of a sphere, cuts off an area of the surface
of the sphere equal to that of a square with sides of length equal
to the radius of the sphere. The conventional symbol for steradian
measure is the uppercase greek Omega. The total solid angle sphere
of a sphere is then found by dividing the surface area of the
sphere by the square of the radius,
sphere = 4pir2 / r2 = 4pi (1.6.5)
Note that this result is independent of the radius of the
sphere. Note also that it was implied that the solid angle was
measured from the center of the sphere (the radius r is constant).
It turns out that the above result does not depend on the position
of the vertex as long as the vertex is inside the sphere.
The SI unit, candela, is the luminous intensity of a source that
emits monochromatic radiation of frequency
540 1012 s-1 , in a given direction, and that has a radiant
intensity in that direction of 1/683 watts per steradian. Note that
"in a given direction" cannot be taken too literally. The intensity
is measured per steradian of spread, so if the radiation has no
spread of directions, the luminous intensity would be infinite.
1.7 Dimensions of Commonly Encountered Quantities
Many physical quantities are derived from the base quantities by
set of algebraic relations defining the physical relation between
these quantities. The dimension of the derived quantity is written
as a power of the dimensions of the base quantitites,
For example velocity is a derived quantity and the dimension is
given by the relationship
dim velocity = (length)/(time) = L T-1 . (1.6.6)
where
L length ,
T time .
Force is also a derived quantity and has dimension
dim force = (mass)(dim velocity)(time) . (1.6.7)
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where
M mass .We could express force in terms of mass, length, and
time by the relationship
dim force = (mass)(length)(time)2 = M L T-2
. (1.6.8)
The derived dimension of kinetic energy is
dim kineticenergy = (mass)(dim velocity)2 , (1.6.9)
which in terms of mass, length, and time is
dim kineticenergy = (mass)(length)2
(time)2 = M L2
T-2 (1.6.10)
The derived dimension of work is
dim work = (dim force)(length) , (1.6.11)
which in terms of our fundamental dimensions is
dim work = (mass)(length)2
(time)2 = M L2
T-2 (1.6.12)
So work and kinetic energy have the same dimensions.
Power is defined to be the rate of change in time of work so the
dimensions are
dim power = dim worktime
=(dim force)(length)
time=
(mass)(length)2(time)3 = M L
2 T-3 (1.6.13)
In Table 1.1 we include the derived dimensions of some common
mechanical quantities in terms of mass, length, and time.
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Table 1.1 Dimensions of Some Common Mechanical Quantities
M mass ,
L length ,
T time
Quantity Dimension MKS unit Angle dimensionless Dimensionless =
radian Steradian dimensionless Dimensionless = radian2 Area
L2
m2 Volume
L3
m3 Frequency
T-1
s1 = hertz = Hz Velocity
L T-1
m s1
Acceleration
L T-2
m s2
Angular Velocity
T-1
rad s1 Angular Acceleration
T-2
rad s2 Density
M L-3
kg m3 Momentum
M L T-1
kg m s1 Angular Momentum
M L2 T-1
kg m2 s1 Force
M L T-2
kg m s- 2 = newton = N Work, Energy
M L2 T-2
kg m2 s2 = joule =J Torque
M L2 T-2
kg m2 s2 Power
M L2 T-3
kg m2 s3= watt = W Pressure
M L-1 T-2
kg m1 s2 = pascal= Pa
Dimensional Analysis
There are many phenomena in nature that can be explained by
simple relationships between the observed phenomena.
Example 5: Period of a Pendulum
Consider a simple pendulum consisting of a massive bob suspended
from a fixed point by a string. Let
Tperiod denote the time (period of the pendulum) that it takes
the bob to complete one cycle of oscillation. How does the period
of the simple pendulum depend on the quantities that define the
pendulum and the quantities that determine the motion?
Solution:
What possible quantities are involved? The length of the
pendulum
l , the mass of the pendulum bob
m , the gravitational acceleration
g , and the angular amplitude of the bob
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0 are all possible quantities that may enter into a relationship
for the period of the swing. Have we included every possible
quantity? We can never be sure but lets first work with this set
and if we need more than we will have to think harder!
Our problem is then to find a function
f such that
Tperiod = f l,m, g,0( ) (1.6.14)
We first make a list of the dimensions of our quantities as
shown in Table 1.2. Choose the set: mass, length, and time, to use
as the base dimensions.
Table 1.2 Dimensions of quantities that may describe the period
of pendulum
Name of Quantity Symbol Dimensional Formula Time of swing
t
T Length of pendulum
l
L Mass of pendulum
m
M Gravitational acceleration
g
L T-2 Angular amplitude of swing
0 No dimension
Our first observation is that the mass of the bob cannot enter
into our relationship, as our final quantity has no dimensions of
mass and no other quantity can remove the dimension of the pendulum
mass. Lets focus on the length of the string and the gravitational
acceleration. In order to eliminate length, these quantities must
divide each other in the above expression for
Tperiod must divide each other. If we choose the combination
l / g , the dimensions are
dim[l / g] = lengthlength/(time)2 = (time)
2 (1.6.15)
It appears that the time of swing is proportional to the square
root of this ratio. We have an argument that works for our choice
of constants, which depend on the units we choose for our
fundamental quantities. Thus we have a candidate formula
Tperiod :lg
1/ 2
(1.6.16)
(in the above expression, the symbol
: represents a proportionality, not an approximation).
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Since the angular amplitude
0 is dimensionless, it may or may not appear. We can account for
this by introducing some function
y 0( ) into our relationship, which is beyond the limits of this
type of analysis. Then the time of swing is
Tperiod = y ( ) lg
1/ 2
(1.6.17)
We shall discover later on that
y 0( ) is nearly independent of the angular amplitude
0 for very small amplitudes and is equal to
y 0( )= 2pi ,
Tperiod = 2pilg
1/ 2
(1.6.18)
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Module 2: Problem Solving Strategies and Estimation
2.1 Problem Solving
Solving problems is the most common task used to measure
understanding in technical and scientific courses, and in many
aspects of life as well. In general, problem solving requires
factual and procedural knowledge in the area of the problem, plus
knowledge of numerous schema, plus skill in overall problem
solving. Schema is loosely defined as a specific type of problem
such as principal, rate, and interest problems, one-dimensional
kinematic problems with constant acceleration, etc. In most
introductory university courses, improving problem solving relies
on three things:
1. increasing domain knowledge, particularly definitions and
procedures 2. learning schema for various types of problems and how
to recognize that a
particular problem belongs to a known schema 3. becoming more
conscious of and insightful about the process of problem
solving.
To improve your problem solving ability in a course, the most
essential change of attitude is to focus more on the process of
solution rather than on obtaining the answer. For homework problems
there is frequently a simple way to obtain the answer, often
involving some specific insight. This will quickly get you the
answer, but you will not build schema that will help solve related
problems further down the road. Moreover, if you rely on insight,
when you get stuck on a problem, youre stuck with no plan or
fallback position. At MIT you will see very few exam problems that
are exactly the same as problems you have seen before, but most
will use the same schema.
General Approach to Problem Solving
A great many physics textbook authors (e.g. Young and Freedman,
Knight, Halliday, Resnick and Cartwright) recommend overall problem
solving strategies. These are typically four-step procedures that
descend from George Polyas influential book, How to Solve It, on
problem solving2. Here are his four steps:
I. Understand get a conceptual grasp of the problem
What is the problem asking? What are the given conditions and
assumptions? What domain of knowledge is involved? What is to be
found and how is this determined or constrained by the given
conditions?
What knowledge is relevant? E.g. in physics, does this problem
involve kinematics, forces, energy, momentum, angular momentum,
equilibrium? If the problem involves two different areas of
knowledge, try to separate the problem into parts. Is there motion
or is it static? If the problem involves vector quantities such as
velocity or momentum, think of these geometrically
2 G. Polya, How to Solve It, 2nd ed., Princeton University
Press, 1957.
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(as arrows that add vectorially). Get conceptual understanding:
is some physical quantity (energy, momentum, angular momentum,
etc.) conserved? Have you done problems that involve the same
concepts in roughly the same way?
Model: Real life contains great complexity, so in physics
(chemistry, economics) you actually solve a model problem that
contains the essential elements of the real problem. The bike and
rider become a point mass (unless angular momentum is involved),
the ladders mass is regarded as being uniformly distributed along
its length, the car is assumed to have constant acceleration or
constant power (obviously not true when it shifts gears), etc.
Become sensitive to information that is implicitly assumed
(Presence of gravity? No friction? That the collision is of short
duration relative to the timescale of the subsequent motion? ).
Advice: Write your own representation of the problems stated
data; redraw the picture with your labeling and comments. Get the
problem into your brain! Go systematically down the list of topics
in the course or for that week if you are stuck.
II. Devise a Plan - set up a procedure to obtain the desired
solution
General - Have you seen a problem like this i.e., does the
problem fit in a schema you already know? Is a part of this problem
a known schema? Could you simplify this problem so that it is? Can
you find any useful results for the given problem and data even if
it is not the solution (e.g. in the special case of motion on an
incline when the plane is at
q = 0 )? Can you imagine a route to the solution if only you
knew some apparently not given information? If your solution plan
involves equations, count the unknowns and check that you have that
many independent equations.
In Physics, exploit the freedoms you have: use a particular type
of coordinate system (e.g. polar) to simplify the problem, pick the
orientation of a coordinate system to get the unknowns in one
equation only (e.g. only the
x -
direction), pick the position of the origin to eliminate torques
from forces you dont know, pick a system so that an unknown force
acts entirely within it and hence does not change the systems
momentum Given that the problem involves some particular thing
(constant acceleration, momentum) think over all the equations that
involve this concept.
III. Carry our your plan solve the problem!
This generally involves mathematical manipulations. Try to keep
them as simple as possible by not substituting in lengthy algebraic
expressions until the end is in sight, make your work as neat as
you can to ease checking and
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reduce careless mistakes. Keep a clear idea of where you are
going and have been (label the equations and what you have now
found), if possible, check each step as you proceed.
IV. Look Back check your solution and method of solution
Can you see that the answer is correct now that you have it
often simply by retrospective inspection? Can you solve it a
different way? Is the problem equivalent to one youve solved before
if the variables have some specific values?
In physics: Check dimensions if analytic, units if numerical.
Check special cases (for instance, for a problem involving two
massive objects moving on an inclined plane, if
m1 = m2 or q = 0 does the solution reduce to a simple expression
that you can easily derive by inspection or a simple argument?) Is
the scaling what youd expect (an energy should vary as the velocity
squared, or linearly with the height). Does it depend sensibly on
the various quantities (e.g. is the acceleration less if the masses
are larger, more if the spring has a larger
k )? Is the answer physically reasonable (especially if numbers
are given or reasonable ones substituted).
Review the schema of your solution: Review and try to remember
the outline of the solution what is the model, the physical
approximations, the concepts needed, and any tricky math
manipulation.
2.2 Significant Figures, Scientific Notation, and Rounding
Significant Figures
We shall define significant figures by the following rules.3
1. The leftmost nonzero digit is the most significant digit.
2. If there is no decimal place, the rightmost nonzero digit is
the least significant digit.
3. If there is a decimal point then the right most digit is the
least significant digit even if it is a zero.
4. All digits between the least and most significant digits are
counted as significant digits.
3 Philip R Bevington and D. Keith Robinson, Data Reduction and
Error Analysis for the Physical Sciences,
2nd Edition, McGraw-Hill, Inc., New York, 1992.
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When reporting the results of an experiment, the number of
significant digits used in reporting the result is the number of
digits needed to state the result of that measurement (or a
calculation based on that measurement) without any loss of
precision.
There are exceptions to these rules, so you may want to carry
around one extra significant digit until you report your result.
For example if you multiply
2 0.56 = 1.12 , not
1.1 .
There is some ambiguity about the number of significant figure
when the rightmost digit is 0, for example 1050, with no terminal
decimal point. This has only three significant digits. If all the
digits are significant the number should be written as 1050., with
a terminal decimal point. To avoid this ambiguity it is wiser to
use scientific notation.
Scientific Notation
Careless use of significant digits can be easily avoided by the
use of decimal notation times the appropriate power of ten for the
number. Then all the significant digits are manifestly evident in
the decimal number. So the number
1050 = 1.05 103 while the number
1050. = 1.050 103 .
Rounding
To round off a number by eliminating insignificant digits we
have three rules. For practical purposes, rounding will be done
automatically by a calculator or computer, and all we need do is
set the desired number of significant figures for whichever tool is
used.
1. If the fraction is greater than 1/2, increment the new least
significant digit.
2. If the fraction is less than 1/2, do not increment.
3. If the fraction equals 1/2, increment the least significant
digit only if it is odd.
The reason for Rule 3 is that a fractional value of 1/2 may
result from a previous rounding up of a fraction that was slightly
less than 1/2 or a rounding down of a fraction that was slightly
greater than 1/2. For example, 1.249 and 1.251 both round to three
significant digits 1.25. If we were to round again to two
significant digits, both would yield the same value, either 1.2 or
1.3 depending on our convention in Rule 3. Choosing to round up if
the resulting last digit is odd and to round down if the resulting
last digit is even reduces the systematic errors that would
otherwise be introduced into the average of a group of such
numbers.
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2.3 Order of Magnitude Estimates - Fermi Problems
Counting is the first mathematical skill we learn. We came to
use this skill by distinguishing elements into groups of similar
objects, but we run into problems when our desired objects are not
easily identified, or there are too many to count.
Rather than spending a huge amount of effort to attempt an exact
count, we can try to estimate the number of objects in a
collection. For example, we can try to estimate the total number of
grains of sand contained in a bucket of sand. Since we can see
individual grains of sand, we expect the number to be very large
but finite. Sometimes we can try to estimate a number which we are
fairly sure but not certain is finite, such as the number of
particles in the universe (See Chapter 20).
We can also assign numbers to quantities that carry dimensions,
such as mass, length, time, or charge, which may be difficult to
measure exactly. We may be interested in estimating the mass of the
air inside a room, or the length of telephone wire in the United
States, or the amount of time that we have slept in our lives, or
the number of electrons inside our body. So we choose some set of
units, such as kilograms, miles, hours, and coulombs, and then we
can attempt to estimate the number with respect to our standard
quantity.
Often we are interested in estimating quantities such as speed,
force, energy, or power. We may want to estimate our natural
walking speed, or the force of wind acting against a bicycle rider,
or the total energy consumption of a country, or the electrical
power necessary to operate this institute. All of these quantities
have no exact, well-defined value; they instead lie within some
range of values.
When we make these types of estimates, we should be satisfied if
our estimate is reasonably close to the middle of the range of
possible values. But what does reasonably close mean? Once again,
this depends on what quantities we are estimating. If we are
describing a quantity that has a very large number associated with
it, then an estimate within an order of magnitude should be
satisfactory. The number of molecules in a breath of air is close
to
1022 ; an estimate anywhere between
1021 and
1023 molecules is close enough. If we are trying to win a
contest by estimating the number of marbles in a glass container,
we cannot be so imprecise; we must hope that our estimate is within
1% of the real quantity.
These types of estimations are called Fermi Problems. The
technique is named after the physicist Enrico Fermi, who was famous
for making these sorts of back of the envelope calculations.
Methodology for Estimation Problems
Estimating is a skill that improves with practice. Here are two
guiding principles that may help you get started.
-
20
(1) You must identify a set of quantities that can be estimated
or calculated.
(2) You must establish an approximate or exact relationship
between these quantities and the quantity to be estimated in the
problem.
Estimations may be characterized by a precise relationship
between an estimated quantity and the quantity of interest in the
problem. When we estimate, we are drawing upon what we know. But
different people are more familiar with certain things than others.
If you are basing your estimate on a fact that you already know,
the accuracy of your estimate will depend on the accuracy of your
previous knowledge. When there is no precise relationship between
estimated quantities and the quantity to be estimated in the
problem, then the accuracy of the result will depend on the type of
relationships you decide upon. There are often many approaches to
an estimation problem leading to a reasonably accurate estimate. So
use your creativity and imagination!
Example: Lining Up Pennies
Suppose you want to line pennies up, diameter to diameter, until
the total length is
1 kilometer . How many pennies will you need? How accurate is
this estimation?
Solution: The first step is to consider what type of quantity is
being estimated. In this example we are estimating a dimensionless
scalar quantity, the number of pennies. We can now give a precise
relationship for the number of pennies needed to mark off 1
kilometer
# of pennies = totaldistancediameter of penny
. (2.3.1)
We can estimate a penny to be approximately 2 centimeters wide.
Therefore the number of pennies is
# of pennies= totaldistancelengthof a penny
=
(1 km)(2 cm)(1 km / 105 cm) = 5 10
4 pennies . (2.3.2)
When applying numbers to relationships we must be careful to
convert units whenever necessary.
How accurate is this estimation? If you measure the size of a
penny, you will find out that the width is
1.9 cm , so our estimate was accurate to within 5%. This
accuracy was fortuitous. Suppose we estimated the length of a penny
to be 1 cm. Then our estimate for the total number of pennies would
be within a factor of 2, a margin of error we can live with for
this type of problem.
-
1
Example: Estimate the total mass of all the water in the earth's
oceans.
Solution: In this example we are estimating mass, a quantity
that is a fundamental in SI units, and is measured in kg. Initially
we will try to estimate two quantities: the density of water and
the volume of water contained in the oceans. Then the relationship
we want is
(mass)ocean
=(density)water (volume)ocean . (2.3.3)
One of the hardest aspects of estimation problems is to decide
which relationship applies. One way to check your work is to check
dimensions. Density has dimensions of mass/volume, so our
relationship is
massocean( )= massvolume
volume
ocean( ). (2.3.4)
The density of fresh water is
water = 1.0 g cm
3 ; the density of seawater is slightly higher, but the
difference wont matter for this estimate. You could estimate this
density by envisioning how much mass is contained in a one-liter
bottle of water. (The density of water is a point of reference for
all density problems. Suppose we need to estimate the density of
iron. If we compare iron to water, we estimate that iron is 5 to 10
times denser than water. The actual density of iron is
iron = 7.8 g cm-3 ).
Since there is no precise relationship, estimating the volume of
water in the oceans is much harder. Lets model the volume occupied
by the oceans as if they completely cover the earth, forming a
spherical shell (Figure 1.5, which is decidedly not to scale). The
volume of a spherical shell of radius
Rearth and thickness d is
( )2shell earthvolume 4piR d , (2.3.5)
where
Rearth is the radius of the earth and d is the average depth of
the ocean.
Figure 1.5 A model for estimating the mass of the oceans.
-
2
We first estimate that the oceans cover about 75% of the surface
of the earth. So the volume of the oceans is
volumeocean
0.75( ) 4pi Rearth2 d( ). (2.3.6)
We therefore have two more quantities to estimate, the average
depth of the ocean, which we can estimate the order of magnitude
as
d 1km , and the radius of the earth, which is approximately
Rearth 6 10
3km . (The quantity that you may remember is the circumference
of the earth, about
25,000 miles . Historically the circumference of the earth was
defined to be
4 107 m ). The radius
Rearth and the circumference s are exactly
related by
s = 2pi Rearth . (2.3.7)
Thus
Rearth =
s
2pi=
2.5 104 mi( )1.6 km mi-1( )2pi
= 6.4 103 km (2.3.8)
We will use
Rearth 6 10
3km ; additional accuracy is not necessary for this problem,
since the ocean depth estimate is clearly less accurate. In fact,
the factor of
75% is not needed, but included more or less from habit.
Altogether, our estimate for the mass of the oceans is
(mass)ocean
=(density)water (volume)ocean water 0.75( ) 4pi Rearth2 d( ),
(2.3.9)
(mass)ocean
1g
cm3
1 kg103 g
(105 cm)3(1 km)3
(0.75)(4 )(6 103 km)2(1km) , (2.3.10)
(mass)ocean
3 1020 kg 1020 kg
-
3
Module 3: Cartesian Coordinates and Vectors
Philosophy is written in this grand book, the universe which
stands continually open to our gaze. But the book cannot be
understood unless one first learns to comprehend the language and
read the letters in which it is composed. It is written in the
language of mathematics, and its characters are triangles, circles
and other geometric figures without which it is humanly impossible
to understand a single word of it; without these, one wanders about
in a dark labyrinth.
Galileo Galilee in Assayer
3.1 Introduction
Mathematics and physics have been historically interwoven since
the time of the ancient Greeks.
The calculus had its origins in the logical difficulties
encountered by the ancient Greek mathematicians in their attempt to
express their intuitive ideas on the ratios and proportionalities
of lines, which they vaguely recognized as continuous, in terms of
numbers, which they regarded as discrete.4
The modern science of kinematics began in the 16th century,
culminating with Galileos description of the motion of bodies,
published in at the start of the 17th century. Galileo used
geometric techniques derived from Euclids Elements to introduce the
concepts of velocity and acceleration. The reliance on mathematics
to describe nature is the foundation on which science is built.
With the introduction of analytic geometry, the systematic use
of variables, and a use of the infinitesimal, Leibnitz and Newton
developed algorithms for introducing and calculating the derivative
and the integral, the core concepts of calculus. Although the
calculus was not used by Newton in his development of the
Principles of the Mechanics, its use was found to be indispensable
in solving countless problems.
In physics, there are well-defined problem solving methodologies
that provide step-by-step instructions on applying physics concepts
but the key question is: How do these methodologies fit into the
larger context of expert problem solving?
Although algorithmic steps can be articulated (and memorized),
it is essential for students to understand the meaning of these
methodologies. In particular how the methodologies are related to
the core physics concepts. The process of understanding and
mastering these methodologies is slow because it is built on
practice. It is also difficult because it requires a simultaneous
understanding of physical and mathematical concepts. The
methodologies can bring out the subtleties of the physical concepts
while the concepts illustrate the need for the mathematics. This is
a non-algorithmic learning
4 Carl B. Boyer,The History of the Calculus and its Conceptual
Development, Dover, New York, 1949, p. 4
-
4
process of synthesizing two difficult abstract knowledge systems
simultaneously, mathematical physics.
The conceptual foundation of the methodologies provides the
framework for thinking about the physics problem. Thinking about
physics problems and designing problem solving strategies is the
fusion that turns students into expert problem solvers. Learning to
think about a physics problem using force diagrams is an example of
this type of fusion.
An emphasis on modeling of physical systems is an effective way
to mitigate against two of the main hazards of mathematics
classes:
1. The students recognize certain formulas as relevant to other
subjects, but view mathematical reasoning as an isolated game with
fixed rules that they are forced to master, but with no connection
to other subjects.
2. Because they can pass the course by memorizing some templated
techniques (and because of point 1) the mathematics they learn is
not truly available to them when solving problems in other
domains.
Today, the first-year student at MIT is expected to understand
single variable calculus (18.01), multiple variable calculus
(18.02), along with Newtonian Mechanics (8.01), and Electricity and
Magnetism (8.02). These General Institute Requirements are in fact
a rite of passage for the modern MIT student into the world of
science and engineering.
The understanding of physics will deepen because students can
solve more mathematically challenging homework problems. This will
reinforce the idea that calculus is an important part of physics.
The understanding of mathematics will deepen because physics
provides a rich source of problems
3.2 Cartesian Coordinate System
Physics involve the study of phenomena that we observe in the
world. In order to connect the phenomena to mathematics we begin by
introducing the concept of a coordinate system. A coordinate system
consists of four basic elements:
(1) Choice of origin (2) Choice of axes (3) Choice of positive
direction for each axis (4) Choice of unit vectors for each
axis
There are three commonly used coordinate systems: Cartesian,
cylindrical and spherical. What makes these systems extremely
useful is the associated set of infinitesimal line, area, and
volume elements that are key to making many integration
-
5
calculations in classical mechanics, such as finding the center
of mass and moment of inertia.
Cartesian Coordinates
Cartesian coordinates consist of a set of mutually perpendicular
axes, which intersect at a common point, the origin O . We live in
a three-dimensional environment; for that reason, the most common
system we will use has three axes, for which we choose the
directions of the axes and position of the origin are.
(1) Choice of Origin
Choose an originO . If you are given an object, then your choice
of origin may coincide with a special point in the body. For
example, you may choose the mid-point of a straight piece of
wire.
(2) Choice of Axis
Now we shall choose a set of axes. The simplest set of axes are
known as the Cartesian axes, x -axis, y -axis, and the z -axis.
Once again, we adapt our choices to the physical object. For
example, we select the x -axis so that the wire lies on the x
-axis, as shown in Figure 3.1
Figure 3.1 A segment of wire of length a lying along the x -axis
of a Cartesian coordinate system.
Then each point P in space our S can be assigned a triplet of
values ( , , )P P Px y z , the Cartesian coordinates of the point P
. The ranges of these values are: Px < < + ,
Py < < + , Pz < < + .
The collection of points that have the same the coordinate Py is
called a level surface. Suppose we ask what collection of points in
our space S have the same value of Py y= . This is the set of
points { }( , , ) such thatPy PS x y z S y y= = . This set PyS is
a
-
6
plane, the x z
plane (Figure 3.2), called a level set for constant Py . Thus,
the y -coordinate of any point actually describes a plane of points
perpendicular to the y -axis.
Figure 3.2 Level surface set for constant value Py .
(3) Choice of Positive Direction
Our third choice is an assignment of positive direction for each
coordinate axis. We shall denote this choice by the symbol + along
the positive axis. Conventionally, Cartesian coordinates are drawn
with the y z
plane corresponding to the plane of the paper. The horizontal
direction from left to right is taken as the positive y -axis, and
the vertical direction from bottom to top is taken as the positive
z -axis. In physics problems we are free to choose our axes and
positive directions any way that we decide best fits a given
problem. Problems that are very difficult using the conventional
choices may turn out to be much easier to solve by making a
thoughtful choice of axes. The endpoints of the wire now have
coordinates ( / 2,0,0)a and ( / 2,0,0)a .
(4) Choice of Unit Vectors
We now associate to each point P in space, a set of three unit
directions vectors ( , , )P P Pi j k . A unit vector has magnitude
one: 1P =i , 1P =j , and 1P =k . We assign the direction of Pi to
point in the direction of the increasing x -coordinate at the point
P . We define the directions for Pj and Pk in the direction of the
increasing y -coordinate and z -coordinate respectively. (Figure
3.3).
-
7
Figure 3.3 Choice of unit vectors.
3.3 Vector Analysis
Introduction to Vectors
Certain physical quantities such as mass or the absolute
temperature at some point only have magnitude. These quantities can
be represented by numbers alone, with the
appropriate units, and they are called scalars. There are,
however, other physical quantities which have both magnitude and
direction; the magnitude can stretch or shrink, and the direction
can reverse. These quantities can be added in such a way that takes
into
account both direction and magnitude. Force is an example of a
quantity that acts in a certain direction with some magnitude that
we measure in newtons. When two forces act on an object, the sum of
the forces depends on both the direction and magnitude of the two
forces. Position, displacement, velocity, acceleration, force,
momentum and torque are all physical quantities that can be
represented mathematically by vectors. We shall
begin by defining precisely what we mean by a vector.
Properties of a Vector
A vector is a quantity that has both direction and magnitude.
Let a vector be denoted by the symbol A
r. The magnitude of A
r is | AA |r . We can represent vectors as geometric
objects using arrows. The length of the arrow corresponds to the
magnitude of the vector. The arrow points in the direction of the
vector (Figure 3.4).
-
8
Figure 3.4 Vectors as arrows.
There are two defining operations for vectors:
(1) Vector Addition:
Vectors can be added. Let Ar
and Br
be two vectors. We define a new vector, = +C A Br r r
,
the vector addition of Ar
and Br
, by a geometric construction. Draw the arrow that represents
A
r. Place the tail of the arrow that represents B
r at the tip of the arrow for A
r as
shown in Figure 3.5(a). The arrow that starts at the tail of Ar
and goes to the tip of Br is defined to be the vector addition = +C
A B
r r r. There is an equivalent construction for the
law of vector addition. The vectors Ar
and Br
can be drawn with their tails at the same point. The two vectors
form the sides of a parallelogram. The diagonal of the
parallelogram corresponds to the vector = +C A B
r r r, as shown in Figure 3.5(b).
Figure 3.5(a) Geometric sum of vectors. Figure 3.5 (b) Geometric
sum of vectors.
Vector addition satisfies the following four properties:
(i) Commutivity:
The order of adding vectors does not matter;
-
9
+ = +A B B Ar rr r
. (3.3.1)
Our geometric definition for vector addition satisfies the
commutivity property (i) since in the parallelogram representation
for the addition of vectors, it doesnt matter which side you start
with, as seen in Figure 3.6.
Figure 3.6 Commutative property of vector addition
(ii) Associativity:
When adding three vectors, it doesnt matter which two you start
with
( ) ( )+ + = + +A B C A B Cr r r rr r (3.3.2)
In Figure 3.7, we add ( )+ +A B Cr rr , and ( )+ +A B Cr rr to
arrive at the same vector sum in either case.
Figure 3.7 Associative law.
(iii) Identity Element for Vector Addition:
There is a unique vector, 0r
, that acts as an identity element for vector addition. For all
vectors A
r,
-
10
+ = + =A 0 0 A Ar r r r r
(3.3.3)
(iv) Inverse Element for Vector Addition:
For every vector Ar
, there is a unique inverse vector
( 1) A Ar r (3.3.4)
such that ( )+ =A A 0r r r
The vector Ar
has the same magnitude as Ar
, | | | | A= =A Aur ur , but they point in opposite directions
(Figure 3.8).
Figure 3.8 additive inverse.
(2) Scalar Multiplication of Vectors:
Vectors can be multiplied by real numbers. Let Ar
be a vector. Let c be a real positive number. Then the
multiplication of A
r by c is a new vector, which we denote by the
symbol c Ar
. The magnitude of c Ar
is c times the magnitude of Ar
(Figure 3.9a),
c A Ac= (3.3.5)
Since 0c > , the direction of c Ar
is the same as the direction of Ar
. However, the direction of c A
r is opposite of A
r (Figure 3.9b).
-
11
Figure 3.9 Multiplication of vector Ar
by (a) 0c > , and (b) 0c < .
Scalar multiplication of vectors satisfies the following
properties:
(i) Associative Law for Scalar Multiplication:
The order of multiplying numbers is doesnt matter. Let b and c
be real numbers. Then
( ) ( ) ( ) ( )b c bc cb c b= = =A A A Ar r r r
(3.3.6)
(ii) Distributive Law for Vector Addition:
Vector addition satisfies a distributive law for multiplication
by a number. Let c be a real number. Then
( )c c c+ = +A B A Br rr r
(3.3.7)
Figure 3.10 illustrates this property.
Figure 3.10 Distributive Law for vector addition.
(iii) Distributive Law for Scalar Addition:
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12
The multiplication operation also satisfies a distributive law
for the addition of numbers. Let b and c be real numbers. Then
( )b c b c+ = +A A Ar r r
(3.3.8)
Our geometric definition of vector addition satisfies this
condition as seen in Figure 3.11.
Figure 3.11 Distributive law for scalar multiplication
(iv) Identity Element for Scalar Multiplication:
The number 1 acts as an identity element for multiplication,
1 =A Ar r
(3.3.9)
3.4 Application of Vectors
When we apply vectors to physical quantities its nice to keep in
the back of our minds all these formal properties. However from the
physicists point of view, we are interested in representing
physical quantities such as displacement, velocity, acceleration,
force, impulse, momentum, torque, and angular momentum as vectors.
We cant add force to velocity or subtract momentum from torque. We
must always understand the physical context for the vector
quantity. Thus, instead of approaching vectors as formal
mathematical objects we shall instead consider the following
essential properties that enable us to represent physical
quantities as vectors.
(1) Vectors can exist at any point P in space.
(2) Vectors have direction and magnitude.
(3) Vector Equality: Any two vectors that have the same
direction and magnitude are equal no matter where in space they are
located.
-
13
(4) Vector Decomposition: Choose a coordinate system with an
origin and axes. We can decompose a vector into component vectors
along each coordinate axis. In Figure 3.12 we choose Cartesian
coordinates for the -x y plane (we ignore the z -direction for
simplicity but we can extend our results when we need to). A vector
A
r at P can be
decomposed into the vector sum,
x y= +A A Ar r r
, (3.4.1)
where xAr
is the x -component vector pointing in the positive or negative
x -direction, and yA
r is the y -component vector pointing in the positive or
negative y -direction
(Figure 3.12).
Figure 3.12 Vector decomposition
(5) Unit vectors: The idea of multiplication by real numbers
allows us to define a set of unit vectors at each point in space.
We associate to each point P in space, a set of three unit vectors
( , , )i j k . A unit vector means that the magnitude is one: | |
1=i , | | 1=j , and
| | 1=k . We assign the direction of i to point in the direction
of the increasing x -coordinate at the point P . We call i the unit
vector at P pointing in the + x -direction. Unit vectors j and k
can be defined in a similar manner (Figure 3.13).
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14
Figure 3.13 Choice of unit vectors in Cartesian coordinates.
(6) Vector Components: Once we have defined unit vectors, we can
then define the x -component and y -component of a vector. Recall
our vector decomposition,
x y= +A A Ar r r
. We can write the x-component vector, xAr
, as
x xA=A ir
(3.4.2)
In this expression the term xA , (without the arrow above) is
called the x-component of the vector A
r. The x -component xA can be positive, zero, or negative. It is
not the
magnitude of xAr
which is given by 2 1/ 2( )xA . Note the difference between the
x -component, xA , and the x -component vector, xA
r.
In a similar fashion we define the y -component, yA , and the z
-component, zA , of the vector A
r
,y y z zA A= =A j A k
r r (3.4.3)
A vector Ar
can be represented by its three components ( , , )x y zA A
A=Ar
. We can also write the vector as
x y zA A A= + +A i j kr
(3.4.4)
(7) Magnitude: In Figure 3.13, we also show the vector
components ( , , )x y zA A A=Ar
.
Using the Pythagorean theorem, the magnitude of Ar
is,
2 2 2x y zA A A A= + + (3.4.5)
(8) Direction: Lets consider a vector ( , ,0)x yA A=Ar
. Since the z -component is zero, the
vector Ar
lies in the -x y plane. Let denote the angle that the vector
Ar
makes in the counterclockwise direction with the positive x
-axis (Figure 3.14). Then the x -component and y -component are
), )x yA A A A = =cos( sin( (3.4.6)
-
15
Figure 3.14 Components of a vector in the x-y plane.
We can now write a vector in the -x y plane as
cos( ) sin( )A A = +A i jr (3.4.7)
Once the components of a vector are known, the tangent of the
angle can be determined by
) ))y
x
A AA A
= =
sin( tan(cos( , (3.4.8)
that yields
1tan y
x
AA
=
. (3.4.9)
Clearly, the direction of the vector depends on the sign of xA
and yA . For example, if both 0xA > and 0yA > , then 0 / 2
pi< < , and the vector lies in the first quadrant. If,
however, 0xA > and 0yA < , then / 2 0pi < < , and the
vector lies in the fourth quadrant.
(9) Unit vector in the direction of Ar
: Let x y zA A A= + +A i j kr
. Let A denote a unit vector in the direction of A
r. Then
2 2 2 1/2
( )x y z
x y z
A A AA A A
+ += =
+ +
i j kAAA
r
r (3.4.10)
-
16
(10) Vector Addition: Let Ar
and Br
be two vectors in the x-y plane. Let A and B denote the angles
that the vectors A
r and B
r make (in the counterclockwise direction)
with the positive x-axis. Then
) )A AA A = +A cos( i sin( jr
(3.4.11)
) )B BB B = +B cos( i sin( jr
(3.4.12)
In Figure 3.15, the vector addition = +C A Br r r
is shown. Let C denote the angle that the vector C
r makes with the positive x-axis.
Figure 3.15 Vector addition with components
Then the components of Cr
are
,x x x y y yC A B C A B= + = + (3.4.13)
In terms of magnitudes and angles, we have
) ) )) ) )
x C A B
y C A B
C C A BC C A B
= = +
= = +
cos( cos( cos(sin( sin( sin( (3.4.14)
We can write the vector Cr
as
( ) ( ) ) )x x y y C CA B A B C C = + + + = +C i j cos( i sin(
jr
(3.4.15)
-
17
Example 1: Given two vectors, 2 3 7= + +A i j kr and 5 2= + +B i
j kr , find: (a) Ar ; (b) Br
; (c) +A Br r
; (d) A Br r
; (e) a unit vector A pointing in the direction of Ar
; (f) a unit vector B pointing in the direction of B
r.
(a) ( )1/ 22 2 22 ( 3) 7 62 7.87= + + = =Ar
(b) ( )1/ 22 2 25 1 2 30 5.48= + + = =Br
(c)
( ) ( ) ( ) (2 5) ( 3 1) (7 2)
7 2 9
x x y y z zA B A B A B+ = + + + + +
= + + + + +
= +
A B i j ki j k
i j k
r r
(d)
( ) ( ) ( ) (2 5) ( 3 1) (7 2)
3 4 5
x x y y z zA B A B A B = + +
= + +
= +
A B i j ki j k
i j k
r r
(e) A unit vector A in the direction of Ar
can be found by dividing the vector A
r by the magnitude of A
r. Therefore
( ) / 2 3 7 / 62= = + +A A A i j kr r
(f) In a similar fashion, ( ) / 5 2 / 30= = + +B B B i j kr r
.
Example 2: Consider two points located at 1rr
and 2rr
, separated by distance 12 1 2r = r rr r
.
Find a vector Ar
from the origin to the point on the line between 1rr
and 2rr
at a distance
12xr from the point at 1rr
, where x is some number.
Solution: Consider the unit vector pointing from 1rr
and 2rr
given by
12 1 2 1 2 1 2 12 / / r= = r r r r r r rr r r r r r
. The vector r in the figure connects Ar
to the point at 1rr
,
therefore we can write ( ) ( )12 12 12 1 2 12 1 2 /xr xr r x= =
= r r r r rr r r r r . The vector 1 = +r A rr r
.
Therefore ( )1 1 1 2 1 2(1 )x x x= = = +A r r r r r rr r r r r r
r r .
3.5 Dot Product
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18
We shall now introduce a new vector operation, called the dot
product or scalar product that takes any two vectors and generates
a scalar quantity (a number). We shall see that the physical
concept of work can be mathematically described by the dot product
between the force and the displacement vectors.
Let Ar
and Br
be two vectors. Since any two non-collinear vectors form a
plane, we define the angle to be the angle between the vectors
A
r and B
r as shown in Figure
3.16. Note that can vary from 0 to pi .
Figure 3.16 Dot product geometry.
Definition: Dot Product
The dot product rA
rB of the vectors A
r and B
r is defined to be product of the
magnitude of the vectors Ar
and Br
with the cosine of the angle between the two vectors:
Aur
Bur
= ABcos() (3.5.1)
Where | |A = Ar and | |B = Br represent the magnitude of Ar and
Br respectively. The dot product can be positive, zero, or
negative, depending on the value of cos . The dot product is always
a scalar quantity.
The angle formed by two vectors is therefore
= cos1rA
rB
rA
rB
(3.5.2)
The magnitude of a vector rA is given by the square root of the
dot product of the vector
rA with itself.
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19
rA = ( rA rA)1/ 2 (3.5.3)
We can give a geometric interpretation to the dot product by
writing the definition as
rA
rB = ( Acos( )) B (3.5.4)
In this formulation, the term cosA is the projection of the
vector Br in the direction of the vector B
r. This projection is shown in Figure 3.17(a). So the dot
product is the
product of the projection of the length of Ar in the direction
of Br with the length of Br . Note that we could also write the dot
product as
rA
rB = A(Bcos( )) (3.5.5)
Now the term
Bcos( ) is the projection of the vector Br in the direction of
the vector Ar as shown in Figure 3.17(b). From this perspective,
the dot product is the product of the projection of the length of
Br in the direction of Ar with the length of Ar .
Figure 3.17(a) and 2.17(b) Projection of vectors and the dot
product.
From our definition of the dot product we see that the dot
product of two vectors that are perpendicular to each other is zero
since the angle between the vectors is / 2pi and
cos(pi / 2) = 0 .
Algebraic Properties of Dot Product
The first property involves the dot product between a vector c
Ar
where c is a scalar and a vector B
r,
(1a) c
rA
rB = c( rA rB) (3.5.6)
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20
The second involves the dot product between the sum of two
vectors Ar
and Br
with a vector C
r,
(2a) ( rA + rB) rC = rA rC + rB rC (3.5.7)
Since the dot product is a commutative operation,
rA
rB =
rB
rA , (3.5.8)
similar definitions hold;
(1b)
rA c
rB = c( rA rB) (3.5.9)
(2b)
rC ( rA + rB) = rC rA + rC rB . (3.5.10)
Vector Decomposition and the Dot Product
With these properties in mind we can now develop an algebraic
expression for the dot product in terms of components. Lets choose
a Cartesian coordinate system with the vector B
r pointing along the positive x -axis with positive x -component
xB , i.e.,
Bur
= Bxi . The vector A
r can be written as
Aur
= Ax
i + Ay j + Az k (3.5.11)
We first calculate that the dot product of the unit vector i
with itself is unity:
i i =| i || i | cos(0) = 1 (3.5.12)
since the unit vector has magnitude
| i |= 1 and cos(0) 1= . We note that the same rule applies for
the unit vectors in the y and z directions:
j j = k k = 1 (3.5.13)
The dot product of the unit vector i with the unit vector j is
zero because the two unit vectors are perpendicular to each
other:
i j =| i || j | cos(pi / 2) = 0 (3.5.14)
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21
Similarly, the dot product of the unit vector i with the unit
vector k , and the unit vector j with the unit vector k are also
zero:
i k = j k = 0 (3.5.15)
The dot product of the two vectors now becomes
rA
rB = ( A
xi + Ay j + Az k) Bx i
= Ax
i Bx
i + Ay j Bx i + Az k Bx i property (2a)= A
xB
x(i i) + Ay Bx (j i) + Az Bx ( k i) property (1a) and (1b)
= AxB
x
(3.5.16)
This third step is the crucial one because it shows that it is
only the unit vectors that undergo the dot product operation.
Since we assumed that the vector Br
points along the positive x -axis with positive x -component xB
, our answer can be zero, positive, or negative depending on the x
-component of the vector A
r. In Figure 3.18, we show the three different cases.
Figure 3.18 Dot product that is (a) positive, (b) zero or (c)
negative.
The result for the dot product can be generalized easily for
arbitrary vectors
rA = A
xi + Ay j + Az k (3.5.17)
and
rB = B
xi + By j + Bz k (3.5.18)
to yield
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22
rA
rB = A
xB
x+ Ay By + Az Bz (3.5.19)
Example 3: Given two vectors, 2 3 7= + +A i j kr and 5 2= + +B i
j kr , find A Br r .
(2)(5) ( 3)(1) (7)(2) 21x x y y z zA B A B A B = + +
= + + =
A Br r
Example 4: Find the cosine of the angle between the vectors 3= +
+A i j kr and 2 3= B i j kr .
Solution: The dot product of two vectors is equal to cos =A B A
Br rr r
where is the angle between the two vectors. Therefore
2 2 2 1/ 2 2 2 2 1/ 2
2 2 2 1/ 2 2 2 2 1/ 2
1/ 2 1/ 2
cos /
( ) /( ) ( )((3)( 2) (1)( 3) (1)( 1)) /((3) (1) (1) ) ( ( 2) (
3) ( 1) )( 10) /(11) (14) 0.806
x x y y z z x y z x y zA B A B A B A A A B B B
=
= + + + + + +
= + + + + + +
= =
A B A Br rr r
Example 5: Show that if = +A B A Br rr r
, then Ar
is perpendicular to Br
.
Solution: Recall that ( ) ( )1/ 2 1/ 2( ) ( ) ( 2 ) = = + A B A
B A B A A A B B Br r r r r rr r r r r r . Similarly ( ) ( )1/ 2 1/
2( ) ( ) ( 2 )+ = + + = + + A B A B A B A A A B B Br r r r r rr r r
r r r . So if
= +A B A Br rr r
, then 0 =A Br r
hence Ar
is perpendicular to Br
.
Work:
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23
A typical physics application of the dot product between two
vectors involves the calculation of the work done by a force F
r on an object that undergoes a displacement
rr . If the force is uniform in space and constant in time
during the entire displacement,
then the work done by the force on the object is given by
WF =
rF rr (3.5.20)
3.6 Cross Product
We shall now introduce our second vector operation, called the
cross product that takes any two vectors and generates a new
vector. The cross product is a type of multiplication law that
turns our vector space (law for addition of vectors) into a vector
algebra (a vector algebra is a vector space with an additional rule
for multiplication of vectors). The first application of the cross
product will be the physical concept of torque about a point P that
can be described mathematically by the cross product between two
vectors: a vector from P to where the force acts, and the force
vector.
Definition: Cross Product
Let Ar
and Br
be two vectors. Since any two non-parallel vectors form a plane,
we define the angle to be the angle between the vectors A
r and B
r as shown in
Figure 3.19. The magnitude of the cross product rA
rB of the vectors A
r and B
r
is defined to be product of the magnitude of the vectors Ar
and Br
with the sine of the angle between the two vectors,
rA
rB = ABsin( ) , (3.6.1)
where A and B denote the magnitudes of Ar
and Br
, respectively. The angle between the vectors is limited to the
values
0 pi insuring that
sin( ) 0 .
Figure 3.19 Cross product geometry.
The direction of the cross product is defined as follows. The
vectors Ar
and Br
form a plane. Consider the direction perpendicular to this
plane. There are two possibilities: we
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24
shall choose one of these two (shown in Figure 3.19) for the
direction of the cross product
rA
rB using a convention that is commonly called the right-hand
rule.
Right-hand Rule for the Direction of Cross Product
The first step is to redraw the vectors Ar
and Br
so that their tails are touching. Then draw an arc starting from
the vector A
r and finishing on the vector B
r. Curl your right
fingers the same way as the arc. Your right thumb points in the
direction of the cross product
rA
rB (Figure 3.20).
Figure 3.20 Right-Hand Rule.
You should remember that the direction of the cross product
rA
rB is perpendicular to
the plane formed by Ar
and Br
.
We can give a geometric interpretation to the magnitude of the
cross product by writing the magnitude as
rA
rB = A(Bsin ) (3.6.2)
The vectors Ar
and Br
form a parallelogram. The area of the parallelogram is equal to
the height times the base, which is the magnitude of the cross
product. In Figure 3.22, two different representations of the
height and base of a parallelogram are illustrated. As depicted in
Figure 3.21(a), the term
Bsin( ) is the projection of the vector Br in the direction
perpendicular to the vector B
r. We could also write the magnitude of the cross
product as
rA
rB = ( Asin( )) B (3.6.3)
Now the term
Asin( ) is the projection of the vector Ar in the direction
perpendicular to the vector B
r as shown in Figure 3.21(b).
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25
Figure 3.21(a) and 3.21(b) Projection of vectors and the cross
product
The cross product of two vectors that are parallel (or
anti-parallel) to each other is zero since the angle between the
vectors is 0 (or pi ) and
sin(0) = 0 (or
sin(pi ) = 0 ). Geometrically, two parallel vectors do not have
a unique component perpendicular to their common direction.
Properties of the Cross Product
(1) The cross product is anti-commutative since changing the
order of the vectors cross product changes the direction of the
cross product vector by the right hand rule:
rA
rB =
rB
rA (3.6.4)
(2) The cross product between a vector c Ar where c is a scalar
and a vector Br is
c
rA
rB = c( rA rB) (3.6.5)
Similarly,
rA c
rB = c( rA rB) (3.6.6)
(3) The cross product between the sum of two vectors Ar and Br
with a vector Cr
is
( rA + rB) rC = rA rC + rB rC (3.6.7)
Similarly,
rA ( rB + rC) = rA rB + rA rC (3.6.8)
Vector Decomposition and the Cross Product
We first calculate that the magnitude of cross product of the
unit vector i with j :
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26
| i j |=| i || j |sin(pi / 2) = 1 (3.6.9)
since the unit vector has magnitude
| i |=| j |= 1 and
sin(pi / 2) = 1. By the right hand rule, the direction of
i j is in the +k as shown in Figure 3.22. Thus
i j = k .
Figure 3.22 Cross product of
i j
We note that the same rule applies for the unit vectors in the y
and z directions,
j k = i, k i = j (3.6.10)
Note that by the anti-commutatively property (1) of the cross
product,
j i = k, i k = j (3.6.11)
The cross product of the unit vector i with itself is zero
because the two unit vectors are parallel to each other, (
sin(0) = 0 ),
| i i |=| i || i | sin(0) = 0 (3.6.12)
The cross product of the unit vector j with itself and the unit
vector k with itself are also zero for the same reason.
j j = 0, k k = 0 (3.6.13)
With these properties in mind we can now develop an algebraic
expression for the cross product in terms of components. Lets
choose a Cartesian coordinate system with the vector B
r pointing along the positive x-axis with positive x-component
xB . Then the
vectors Ar
and Br
can be written as
rA = A
xi + Ay j + Az k (3.6.14)
and
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27
rB = B
xi (3.6.15)
respectively. The cross product in vector components is
rA
rB = ( A
xi + Ay j + Az k) Bx i (3.6.16)
This becomes, using properties (3) and (2),
rA
rB = ( A
xi B
xi) + ( Ay j Bx i) + ( Az k Bx i)
= AxB
x(i i) + Ay Bx (j i) + Az Bx ( k i)
= Ay Bx k + Az Bx j (3.6.17)
The vector component expression for the cross product easily
generalizes for arbitrary vectors
rA = A
xi + Ay j + Az k (3.6.18)
and
rB = B
xi + By j + Bz k
(3.6.19)
to yield
rA
rB = ( Ay Bz Az By ) i + ( Az Bx Ax Bz ) j + ( Ax By Ay Bx ) k .
(3.6.20)
Example 6: Given two vectors, 2 3 7= + +A i j kr and 5 2= + +B i
j kr , find ABr r .
Solution:
( ) ( ) ( ) (( 3)(2) (7)(1)) ((7)(5) (2)(2)) ((2)(1) (
3)(5))
13 31 17
y z z y z x x z x y y xA B A B A B A B A B A B = + +
= + +
= + +
A B i j ki j k
i j k
r r
Example 6: Law of Sines: Prove that / sin / sin =B Arr and / sin
/ sin =B Crr using the cross product. (Hint: Consider the area of a
triangle formed by three vectors Ar , Br
, and Cr
, where 0+ + =A B Cr rr
.)
Solution: Since 0+ + =A B Cr rr
, we have that 0 ( )= + + = + A A B C A B A Cr r r r r rr r
. Thus
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28
= A B A Cr r rr
or = A B A Cr r rr
.
From the figure we see that sin =A B A Br rr r
and sin =A C A Cr r r r . Therefore sin sin =A B A Cr r rr , and
hence / sin / sin =B Crr . A similar argument shows that
/ sin / sin =B Arr proving the law of sines.
Example 8: Show that the volume of a parallelpiped with edges
formed by the vectors Ar
, Br
, and Cr
is given by ( ) A B Cr rr
.
Solution: The volume of a parallelpiped is given by area of the
base times height. If the base is formed by the vectors B
rand C
r, then the area of the base is given by the
magnitude of B Crr
. The vector = B C B C nr rr r
where n is a unit vector perpendicular
to the base. The projection of the vector Ar along the direction
n gives the height of the parallelpiped. This projection is given
by taking the dot product of Ar with a unit vector and is equal to
height =A n
r. Therefore
rA ( rB rC) = rA ( rB rC ) n = ( rB rC ) rA n = (area)(height) =
(volume) .
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29
Module 4: One-Dimensional Kinematics
4.1 Introduction
Kinematics is the mathematical description of motion. The term
is derived from the Greek word kinema, meaning movement. In order
to quantify motion, a mathematical coordinate system, called a
reference frame, is used to describe space and time. Once a
reference frame has been chosen, we can introduce the physical
concepts of position, velocity and acceleration in a mathematically
precise manner. Figure 4.1 shows a Cartesian coordinate system in
one dimension with unit vector i pointing in the direction of
increasing
x -coordinate.
Figure 4.1 A one-dimensional Cartesian coordinate system.
4.2 Position, Time Interval, Displacement
Position
Consider an object moving in one dimension. We denote the
position coordinate of the center of mass of the object with
respect to the choice of origin by ( )x t . The position coordinate
is a function of time and can be positive, zero, or negative,
depending on the location of the object. The position has both
direction and magnitude, and hence is a vector (Figure 4.2),
( ) ( )t x t=x ir . (4.2.1)
We denote the position coordinate of the center of the mass at
0t = by the symbol 0 ( 0)x x t = . The SI unit for position is the
meter [m] (see Section 1.3).
Figure 4.2 The position vector, with reference to a chosen
origin.
-
30
Time Interval
Consider a closed interval of time 1 2[ , ]t t . We characterize
this time interval by the difference in endpoints of the interval
such that
2 1t t t = . (4.2.2)
The SI units for time intervals are seconds [s].
Definition: Displacement
The change in position coordinate of the mass between the times
1t and 2t is
2 1 ( ( ) ( )) ( )x t x t x t x i ir . (4.2.3)
This is called the displacement between the times 1t and 2t
(Figure 4.3). Displacement is a vector quantity.
Figure 4.3 The displacement vector of an object over a time
interval is the vector difference between the two position
vectors
4.3 Velocity
When describing the motion of objects, words like speed and
velocity are used in common language; however when introducing a
mathematical description of motion, we need to define these terms
precisely. Our procedure will be to define average quantities for
finite intervals of time and then examine what happens in the limit
as the time interval becomes infinitesimally small. This will lead
us to the mathematical concept that velocity at an instant in time
is the derivative of the position with respect to time.
Definition: Average Velocity
The component of the average velocity, xv , for a time interval
t is defined to be the displacement x divided by the time interval
t ,
-
31
x
xv
t
. (4.3.1)
The average velocity vector is then
( ) ( )xx
t v tt
=
v i ir . (4.3.2)
The SI units for average velocity are meters per second 1m s
.
Instantaneous Velocity
Consider a body moving in one direction. We denote the position
coordinate of the body by ( )x t , with initial position 0x at time
0t = . Consider the time interval [ , ]t t t+ . The average
velocity for the interval t is the slope of the line connecting the
points ( , ( ))t x t and ( , ( ))t x t t+ . The slope, the rise
over the run, is the change in position over the change in time,
and is given by
rise ( ) ( )run
x
x x t t x tv
t t
+ = =
. (4.3.3)
Lets see what happens to the average velocity as we shrink the
size of the time interval. The slope of the line connecting the
points ( , ( ))t x t and ( , ( ))t x t t+ approaches the slope of
the tangent line to the curve ( )x t at the time t (Figure
4.4).
Figure 4.4 Graph of position vs. time showing the tangent line
at time t .
In order to define the limiting value for the slope at any time,
we choose a time interval [ , ]t t t+ . For each value of t , we
calculate the average velocity. As 0t ,
-
32
we generate a sequence of average velocities. The limiting value
of this sequence is defined to be the x -component of the
instantaneous velocity at the time t .
Definition: Instantaneous Velocity
The x -component of instantaneous velocity at time t is given by
the slope of the tangent line to the curve of position vs. time
curve at time t :
0 0 0
( ) ( )( ) lim lim limx xt t tx x t t x t dx
v t vt t dt
+ = =
. (4.3.4)
The instantaneous velocity vector is then
( ) ( )xt v t=v ir
. (4.3.5)
Example 1: Determining Velocity from Position
Consider an object that is moving along the x -coordinate axis
represented by the equation
20
1( )2
x t x bt= + (4.3.6)
where 0x is the initial position of the object at 0t = .
We can explicitly calculate the x -component of instantaneous
velocity from Equation (4.3.4) by first calculating the
displacement in the x -direction, ( ) ( )x x t t x t = + . We need
to calculate the position at time t t+ ,
( )2 2 20 01 1( ) ( ) 22 2x t t x b t t x b t t t t+ = + + = + +
+ . (4.3.7)
Then the instantaneous velocity is
2 2 20 0
0 0
1 1( 2 )( ) ( ) 2 2( ) lim limx t tx b t t t t x bt
x t t x tv t
t t
+ + + + + = =
. (4.3.8)
This expression reduces to
0
1( ) lim2x t
v t bt b t
= +
. (4.3.9)
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33
The first term is independent of the interval t and the second
term vanishes because the limit as 0t of t is zero. Thus the
instantaneous velocity at time t is
( )xv t bt= . (4.3.10)
In Figure 4.5 we graph the instantaneous velocity, ( )xv t , as
a function of time t .
Figure 4.5 A graph of instantaneous velocity as a function of
time.
4.4 Acceleration
We shall apply the same physical and mathematical procedure for
defining acceleration, the rate of change of velocity. We first
consider how the instantaneous velocity changes over an interval of
time and then take the limit as the time interval approaches
zero.
Average Acceleration
Acceleration is the quantity that measures a change in velocity
over a particular time interval. Suppose during a time interval t a
body undergoes a change in velocity
( ) ( )t t t = + v v vr r r . (4.4.1)
The change in the x -component of the velocity, xv , for the
time interval [ , ]t t t+ is then
( ) ( )x x xv v t t v t = + . (4.4.2)
Definition: Average Acceleration
The x -component of the average acceleration for the time
interval t is defined to be
( ( ) ( )) x x x x
x
v v t t v t va
t t t
+ = = =
a i i i ir . (4.4.3)
The SI units for average acceleration are meters per second
squared, 2[m s ] .
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34
Instantaneous Acceleration
On a graph of the x -component of velocity vs. time, the average
acceleration for a time interval t is the slope of the straight
line connecting the two points ( , ( ))xt v t and ( , ( ))xt t v t
t+ + . In order to define the x -component of the instantaneous
acceleration at time t , we employ the same limiting argument as we
did when we defined the instantaneous velocity in terms of the
slope of the tangent line.
Definition: Instantaneous Acceleration.
The x -component of the instantaneous acceleration at time t is
the limit of the slope of the tangent line at time t of the graph
of the x -component of the velocity as a function of time,
0 0 0
( ( ) ( ))( ) lim lim limx x x xx xt t tv t t v t v dv
a t at t dt
+ = =
. (4.4.4)
The instantaneous acceleration vector is then
( ) ( )xt a t=a ir
. (4.4.5)
In Figure 4.6 we illustrate this geometrical construction.
Figure 4.6 Graph of velocity vs. time showing the tangent line
at time t .
Since velocity is the derivative of position with respect to
time, the x -component of the acceleration is the second derivative
of the position function,
-
35
2
2x
x
dv d xa
dt dt= = . (4.4.6)
Example 2: Determining Acceleration from Velocity
Lets continue Example 1, in which the position function for the
body is given by 2
0 (1/ 2)x x bt= + , and the x -component of the velocity is xv
bt= . The x -component of the instantaneous acceleration at time t
is the limit of the slope of the tangent line at time t of the
graph of the x -component of the velocity as a function of time
(Figure 4.5)
0 0
( ) ( )lim limx x xx t tdv v t t v t bt b t bt
a bdt t t
+ + = = = =
. (4.4.7)
Note that in Equation (4.4.7), the ratio /v t is independent of
t , consistent with the constant slope of the graph in Figure
4.4.
4.5 Constant Acceleration
Lets consider a body undergoing constant acceleration for a time
interval [0, ]t t = . When the acceleration xa is a constant, the
average acceleration is equal to the instantaneous acceleration.
Denote the x -component of the velocity at time 0t = by
,0 ( 0)x xv v t = . Therefore the x -component of the
acceleration is given by
,0( )x xxx x
v t vva a
t t
= = =
. (4.5.1)
Thus the velocity as a function of time is given by
,0( )x x xv t v a t= + . (4.5.2)
When the acceleration is constant, the velocity is a linear
function of time.
Velocity: Area Under the Acceleration vs. Time Graph
In Figure 4.7, the x -component of the acceleration is graphed
as a function of time.
-
36
Figure 4.7 Graph of the x -component of the acceleration for xa
constant as a function of time.
The area under the acceleration vs. time graph, for the time
interval 0t t t = = , is
Area( , )x xa t a t . (4.5.3)
Using the definition of average acceleration given above,
,0Area( , ) ( )x x x x xa t a t v v t v = = . (4.5.4)
Displacement: Area Under the Velocity vs. Time Graph
In Figure 4.8, we graph the x -component of the velocity vs.
time curve.
Figure 4.8 Graph of velocity as a function of time for xa
constant.
The region under the velocity vs. time curve is a trapezoid,
formed from a rectangle and a triangle and the area of the
trapezoid is given by
( ),0 ,01Area( , ) ( )2x x x xv t v t v t v t= + . (4.5.5)
Substituting for the velocity (Equation (4.5.2)) yields
-
37
2,0 ,0 ,0 ,0
1 1Area( , ) ( )2 2x x x x x x x
v t v t v a t v t v t a t= + + = + . (4.5.6)
Figure 4.9 The average velocity over a time interval.
We can then determine the average velocity by adding the initial
and final velocities and dividing by a factor of two (see Figure
4.9).
,01 ( ( ) )2x x x
v v t v= + . (4.5.7)
The above method for determining the average velocity differs
from the definition of average velocity in Equation (4.3.1). When
the acceleration is constant over a time interval, the two methods
will give identical results. Substitute into Equation (4.5.7) the x
-component of the velocity, Equation (4.5.2), to yield
vx
=
12
(vx(t) + v
x ,0 ) =12
((vx ,0 + ax t) + vx ,0 ) = vx ,0 +
12
axt . (4.5.8)
Recall Equation (4.3.1); the average velocity is the
displacement divided by the time interval (note we are now using
the definition of average velocity that always holds, for
non-constant as well as constant acceleration). The displacement is
equal to
0( ) xx x t x v t = . (4.5.9)
Substituting Equation (4.5.8) into Equation (4.5.9) shows that
displacement is given by
20 ,0
1( )2x x x
x x t x v t v t a t = = + . (4.5.10)
Now compare Equation (4.5.10) to Equation (4.5.6) to conclude
that the displacement is equal to the area under the graph of the x
-compone