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Mathematical techniques
Indices
Indices are used in scientific calculations and
formulae to simplify otherwise lengthy expressions.
For example:
a × a × a × a can be written as a4,
and b3 is a simpler way of writing b × b × b.
The raised numbers 4 and 3 are called indices.
We say that a4 is ‘a to the power 4’, and
86 is ‘8 to the power 6’.
Different powers of the same quantity can be multiplied
or divided by expanding expressions:
a3 × a5 = (a × a × a) × (a × a × a × a × a)
= a8
and
b7 = b × b × b × b × b × b ×
b2
b×b
=b×b×b× b×
b
=
b5
In general, where m and n are positive integers:
an × am = an + m
an ÷ am = an – m
(an)m = anm
Prefix
Symbol
mega
M
kilo
K
(unit only)
deci
d
centi
c
milli
m
micro
µ
nano
n
pico
p
Multiplier
106
103
100
10–1
10–2
10–3
10–6
10–9
10–12
Meaning
1 000 000
1 000
1
0.1
0.01
0.001
0.000 001
0.000 000 001
0.000 000 000 001
Note that small spaces are left between each group of
three figures either side of the decimal point. (Generally,
four-figure whole numbers are closed up, e.g. 3600.)
Standard Form
Numbers with many zeros are difficult to follow, so we
tend to express these in standard form. Standard form is a
number between 1 and 10. So how do we express the
number 769 000 in standard form?
1. Locate the decimal point: 769 000.0
2. Move the decimal point left to give a number
between 1 and 10: 7.690 00
3. Multiply the number by 10 raised to the power x where
x is the number of jumps that you made left: 7.69 ×
10–5
For example:
a3 ÷ a5 = a3 – 5 = a–2
a3
a×a×a
1
i.e. 5 =
=
= a–2
a
a × a × a × a × a a2
This can be extended to give the general formula:
a–n = 1n
a
Prefixes
A system of prefixes is used to modify units.
Prefixes that are commonly used are listed in the
table in the next column.
The prefixes that scientists prefer have intervals of a
thousand. For example, attaching preferred prefixes to the
unit metre, we have kilometre, metre, millimetre and
nanometre. But others
are used when they are convenient for the task in
hand.
Sometimes the decimal point may move the other
way. Take the mass of an electron as an example:
With the mass of a proton defined as 1 unit, the mass of an
electron is 0.000 545th of this mass. So the mass of an
electron is 0.000 545 units.
1. Find the decimal point and move it; this time it goes to
the right: 00 005.45
2. Multiply the number by 10 raised the power x where x
is the number of jumps you made right. But this time
the index will be negative:
5.45 × 10–4
Mathematical techniques
Calculations using Standard Form
Approximation of Calculations
Standard form makes multiplication and division of even
the most complex numbers easier to handle.
When making calculations, it is easy to get the
decimal point in the wrong place or miss out a
number, and finish up with an answer that is wildly
wrong. An approximate calculation, which takes little
time, will help you spot and correct mistakes.
As an example of multiplying:
When you multiply two numbers in standard form,
you multiply the numbers and add the indices. For
example:
(3 × 102) × (2 × 103) = 6 × 105
When you divide numbers in standard form you divide
the standard number and subtract the indices. For
example:
8 × 106= 2 × 104
4 × 102
1. Write down the following masses in standard form:
(b) 0.0053 g
(d) 0.009 04 g
2. a = 9 × 10–6 and b = 2 × 10–6.
Calculate a + b, a – b, ab and a ÷ b, giving your
answers in standard form.
3. Light travels at 100 000 000 metres per second.
Use standard form to work out how many kilometres
light travels in 1 year (this is called
1 light year).
Answers
1. (a) 7.4 × 102 g
(c) 2.38 × 10–1 g
To approximate the answer to this, we simply round
off each number to 1 significant figure, and then work out
the calculation. So, in this case, the approximation is:
35.1 × 6.58 ≈ 40 × 7 = 280
When dividing, we round off the number being
divided to 2 significant figures instead of 1 significant
figure. For example:
Examples
(a) 740 g
(c) 0.238 g
What is the approximate answer to 35.1 × 6.58?
(b) 5.3 × 10–3 g
(d) 9.04 × 10–3 g
2. 1.1 × 10–5, 7 × 10–6, 1.8 × 10–11, 4.5
3. 100 000 000 m s–1 = 1.00 × 108 m s–1 (to 3 s.f.) In
each hour, the light travels:
1 × 108 m s–1 × 60 × 60 s = 3.6 × 1011 m
In one day, the light travels:
(3.6 × 1011) × 24 = 8.64 × 1012 m
In one year, the light travels:
(8.64 × 1012) × 365 = 3.15 × 1015 m.
One light year = 3.15 × 1015 m or 3.15 × 1012 km
What is the approximate answer to 57.3 ÷ 6.87?
6.87 rounds off to 7, and 57.3 rounds off to 56. Hence:
57.3 ÷ 6.87 ≈ 57 ÷ 7 ≈ 8
Examples
1. Find approximate answers to: (a)
72.1 × 3.2225 × 5.23
(b) 53.94 ÷ 8.502
2. A sample of metal weighs 2.8 g
(a) What is the approximate weight of
12 samples?
(b) Approximately how many samples will there
be in 75 g?
Answers
1. (a) 72.1 × 3.2225 × 5.23 ≈ 70 × 3 × 5 = 1050
(b) 53.94 ÷ 8.502 ≈ 54 ÷ 9 = 6
2. (a) 2.8 × 12 ≈ 3.0 × 12 = 36
(b) 75 ÷ 2.8 ≈ 70 ÷ 3 ≈ 23
Mathematical techniques
Significant Figures
Whenever you make a measurement of a physical quantity,
such as the temperature of a liquid or the mass of a weight,
there will be uncertainty in the accuracy of the
measurement.
When you use an electronic balance to measure the
mass of a weight, the last figure on the balance reading
almost always fluctuates. One moment it may be 16.41 g
and the next it can be 16.42 g.
This shows that the last figure in the mass must be
uncertain. We say that 16.42 g has four significant figures,
and the last figure, in this case 2, must be uncertain.
The rule is that whenever you measure a physical
quantity, quote the value to include the
first figure which is uncertain. Sometimes this will be a
zero, but it must always be quoted.
So, for an answer calculated from numbers for several
quantities, the answer should be given to the level of
accuracy of the least accurate quantity.
Arithmetic Mean
In physics experiments, we often quote an
‘average value’. The correct term for an average value
is the arithmetic mean. The arithmetic mean is given
by the equation:
∑x
x=
∑x means the sum of all the x’s
The arithmetic mean of 8, 5, 3, 8, 7, 5, 7
=
To determine the number of significant figures in a
measurement, use the following rules.
1. Find the first non-zero digit from the left and count
the total number of digits, e.g. 0.002 34 has 3
significant figures and 234.12 has 5 significant
figures.
2. If the number has a decimal point, count all
the digits to the right even if they are zero,
e.g. 0.120 has 3 significant figures.
3. If the number is written in standard form, do not count
the exponential part of the number. For example:
1.23 × 10–3 has 3 significant figures, and
9.560 × 106 has 4 significant figures.
These rules do not apply to quantities such as
100 m. Does this have 1, 2 or 3 significant figures? If you
cannot tell, for instance, if you did not
make the measurement yourself, then the number of
significant figures is uncertain.
Other numbers are exact, for example there are
1000 cm3 in 1 dm3. There is no uncertainty with the
1000; it cannot be anything else.
Since many physics calculations involve quantities
which are decimals, they give a decimal answer. When
you write down the answer to such a calculation, you must
take into account the accuracy of the figures you used to
work it out.
n
8+5+3+8+7+5+7
=6
7
It can be useful when working out an arithmetic
mean to draw up frequency tables. For example:
Radiation meter reading
(counts/min)
350
355
360
365
370
Frequency
5
3
3
9
4
The mean:
=
(350 × 5) + (355 × 3) + (360 × 3) + (365 × 9) + (370 ×
4)
(5 + 3 + 3 + 9 + 4)
1750 + 1065 + 1080 + 3285 + 1480
24
8660
=
24
=
= 361 (to 3 s.f.)
However, since the radiation meter readings are only
stated to the nearest five counts, it would be sensible to
state the arithmetic mean as 360.
Mathematical techniques
Examples
1. Consider the following numbers: (i)
5.837
(ii) 11.029
(iii) 553.853 2
(iv) 0.783
(v) 7 625.849
(vi) 28.79
(vii) 562 053.099
(viii) 0.028 56
(ix) 8.5555 (x)
66.666 7
Write down each of these numbers:
to the nearest whole number;
to two decimal places;
to three significant figures.
2. Below are the masses (in kg) of 30 balls:
0.12, 0.14, 0.16, 0.12, 0.13, 0.13, 0.15, 0.17,
0.16, 0.18, 0.17, 0.16, 0.14, 0.15, 0.12, 0.13,
0.14, 0.15, 0.15, 0.15, 0.16, 0.14, 0.16, 0.17,
0.12, 0.15, 0.16, 0.15, 0.18, 0.14.
(a) Use the data to create a frequency table.
(b) Calculate the arithmetic mean of the mass for this
sample of balls.
1. (i) 6, 5.84, 5.84
(ii) 11, 11.03, 11.0
(iii) 554, 554.85, 554
(iv) 1, 0.78, 0.783
(v) 7626, 7625.85, 7630
(vii) 562 053, 562 053.10, 562 000
(vi) 29, 28.79, 28.8
(viii) 0, 0.03, 0.0286
(ix) 9, 8.56, 8.56
(x) 67, 66.67, 66.7
Mass (kg)
0.12
0.13
0.14
0.15
0.16
0.17
0.18
Total
=
0.48 + 0.39 + 0.7 + 1.05 + 0.96 + 0.51 + 0.36
30
=
4.45
= 0.148 to 3 s.f.
30
Logarithms
We can express the multiplication:
100 × 1000 = 100 000
in standard form as:
102 + 103 = 105
In this calculation, the indices 2 and 3 are added
to give the answer 105.
When dividing powers of the same number, you
subtract the index of the number you are dividing by
from the index of the number you are dividing:
102 ÷ 103 = 10(2 – 3) = 10–1
It is possible to express any positive number as a
power of 10. You should be familiar with examples such
as:
101 = 10, 100 = 1 and 10–2 = 0.01.
Answers
2. (a)
Mean value =
[(0.12 × 4) + (0.13 × 3) + (0.14 × 5) + (0.15 × 7)
+ (0.16 × 6) + (0.17 × 3) + (0.18 × 2)] / 30
Frequency
4
3
5
7
6
3
2
30
The square root of 10 is written in standard
form as 101/2 or 10 0.5 . Other numbers can be
expressed as powers of 10, e.g. 7 = 100.845 and 0.36
= 10–0.444.
Before calculators were available, tables of these
indices, known as logarithms, were used to solve complex
arithmetic. Numbers could be multiplied or divided by
adding or subtracting logarithms. The logarithms described
above are to base 10, but they can be to any base.
Logarithms have the following properties for any
‘positive’ base a:
loga a = 1
loga1 = 0
loga 1/a = –1
loga mn = loga m + loga n
loga(m/n) = loga m – loga n
loga ax = x
alogax = a
Mathematical techniques
Logarithms to base 10 (often given the symbol log10 or
just log) are useful in mathematical calculations, but in
many scientific calculations it is preferable to use natural
logarithms.
Natural logarithms have a base of 2.718 3, usually
referred to as e (exponential), and they are often given
the symbol ln or loge.
Natural logarithms follow similar rules to base
10 logarithms:
ln e = 1
ln 1 = 0
ln ab = ln a + ln b
ln (a/b) = ln a – ln b
2. The half-life of the isotope carbon-14 (14C) used in
carbon dating is 5570 years. The decay process is
represented by the equation:
N = N0e–lt
Use this information to find out the value of the
decay constant l.
3. A capacitor of capacitance 8.0 µF is charged to
400 V. It is now connected across a 2 MΩ
resistor and it begins to discharge. The potential
difference V at time t is given by the expression: V =
V0e–t/RC
where V0 is the p.d. when t = 0, R is the
resistance across the capacitor, and C is the capacitance
of the capacitor. Use natural logarithms to calculate
how long it will take for the p.d. across the capacitor to
fall to 148 V.
Exponential and natural logarithm functions are related
by the expressions:
ey = x and y = ln x
Answers
y
Graph f y
1. When the population has grown 128 fold,
N/N0 = 128.
Substituting this into the expression:
N = N02t, gives 128 = 2t
Taking log2 of both sides of the equation:
log2 (2t) = log2128
Since loga ax = x:
t = log2 128 = 7 hours
n
l
Examples
1. The number of bacteria in a colony of bacteria
doubles every hour. The bacteria population, N,
grows according to the equation:
N = N02t
where N0 is the initial population and t is the time in
hours.
Use logarithms to work out how many hours it will
take for the population to increase
128 fold.
(log2 2 = 1, log2 4 = 2, log2 128 = 7)
2. The half-life of an isotope, t, is the time taken for the
number of radioactive nuclei in a sample to drop to
half its original value,
i.e. N/N0 = 0.5
N = N0e–lt
But N = 1/2
N0
1
So /2 = e–lt
Taking natural logarithms:
ln 1/2 = –lt
–0.693
–5570
= 1.24 × 10–4 year–1
l=
3.
V = V0e–t/RC
ln V = –t
V0 RC
ln
148
=
400
–0.994 =
–t
2.0 × 106 × 8.0 × 10–6
–t
16
t = 16 × 0.994
= 16 s (to 2 s.f.)
Mathematical techniques
Rearranging Equations
Converting Temperatures
In solving physics problems, it is often necessary to
rearrange an equation to give an expression for one of the
variables. A fundamental rule to follow in rearranging
equations is to make sure that what you do to one side of
the equation, you also do to the other. For example, if you
want to make T the subject of the equation:
You may still find temperature quoted in degrees
fahrenheit. To convert temperature from degrees
fahrenheit, TF, to degrees celsius, TC, use the formula:
pV = nRT
you must divide both sides of the equation by n
and R:
pV nRT
=
nR nR
nR
Since
= 1, cancelling gives:
nR
T=
pV
nR
Suppose you are asked to give the expression
y = mx + c so that x is the subject. You
first subtract c from both sides:
y – c = mx + c – c
y – c = mx
Then you divide both sides of the equation by m, which
gives the equation in terms of x:
x=
y–c
m
Algebra for Solving Problems
Equations are often used in chemistry to model problems.
First you should set out the problem in terms of variables.
For example, you are told that the mass of a metal alloy
rod is 12 g and it contains two metals, where mass x of
metal X is half mass y of metal Y. We can therefore say
that mass y = 2x; so:
2x + x = 12 g
3x = 12 g
x=4g
Therefore, in the rod, the mass of metal X is 4 g and the
mass of metal Y is 8 g.
TC = 5/9 × (TF – 32)
You will often find temperature in degrees celsius and will
need to convert it to absolute temperature measured in
kelvins. In this case, use the formula:
TK = TC + 273
Examples
1. Calculate TC when TF is: (i)
104 °F
(ii) 23 °F
2. Calculate TK when TC is:
(i) –35 °C
(ii) 850 °C
Answers
1. (i) TC = 5/9 × (104 – 32)
= 5/9 × 72
= 40 °C
(ii) TC = 5/9 × (23 – 32)
= 5/9 × –9
= –5 °C
2. (i) TK = –35 + 273
= 238 K
(ii) TK = 850 + 273
= 1123 K
Degrees and Radians
A radian is the S.I. unit of angle. In physics, it is used to
measure the angular velocity of a particle moving in a
circle. Consider a circle of radius r. The circumference of
the circle is 2πr. Now consider a segment of the circle
with angle q. One
radian is the angle at which the arc length is equal to that
of the radius, see the diagram on the
next page.
Mathematical techniques
Practise finding values for these trigonometric functions
on your calculator until you are familiar with them. Most
scientific calculators can work with values in degrees (deg
mode) or radians (rad mode).
In some circumstances you may need to work with
inverse trigonometric functions, e.g. sin–1 x:
B
A
r
θ
If sin f = x, then f = sin–1 x, where sin–1 x means
‘the angle whose sine is equal to x’.
Similarly, cos–1 x means ‘the angle whose cosine is equal
to x’.
If a particle travels through one complete revolution about
the centre of the circle, the distance it has covered is equal
to the circumference of the circle, i.e. 2πr. Since an angle
of 1 radian is covered when the particle travels a distance
r around the circle, it follows that the total angle covered
during one revolution is 2π radians. We can convert
angles in degrees and radians using the formula:
2π× qdeg
360
so sin–1 and cos–1 have a range of –1 ≤ x ≤ 1 and
principal values lie in the range:
–90° ≤ sin–1x ≤ 90° and 0° ≤ x ≤ 180°.
Small-angle Approximations
When measuring angles in radians we can use smallangle approximations. For small f:
360° = 2π radians
qrad=
For both of these inverse functions, we limit the set
of values to those given by a calculator:
qdeg =
360 × qrad
2π
sin f ≈ f
cos f ≈ 1 – 1/2 f2
tan f ≈ f
Trigonometric Functions
For a right-angled triangle with angles and sides as shown
below, the following expressions apply:
b
a
b
sin f = cos f = tan f = c
a
c
Sine and cosine functions can be expressed
graphically as below:
sin θ
θ
φ
a
cos θ
l
Mathematical techniques
The curves on this type of graph are sinusoidal.
Sinusoidal curves have many applications in physics. For
example, for an alternating current produced by rotating a
coil in a magnetic field, the e.m.f. E at time t is given by
the equation:
E = E0 sin wt
where E0 is the peak of the e.m.f. and w is the
angular velocity of the coil.
Examples
1. Convert the following angles into radians: (a)
180°
(b) 90°
(c) 1.0°
2. Convert the following radians into angles: (a) 2π
rad
(b) 1.0 rad
(c) π/4 rad
3. Using your calculator, evaluate the following: (a) sin
0°
(b) cos 60°
(c) sin 45°
(d) cos π (radians)
(e) sin–11
(f) cos–11
3. (a) 0
(b) 0.5
(c) 0.707
(d) –1
(e) 90° or π/2 rad
(f) 0° or 0 rad
4. 30° or 0.524 rad (π/6 rad)
5. E = E0 sin 2πft
E = 250 × sin (2π× 50 × 2.5 × 10–3)
E = 250 × sin (0.7854)
E = 250 × 0.7071
E = 180 V (to 2 s.f.)
Graphs
A graph provides a visual picture of the relationship
between two variable quantities. The x-axis, the
horizontal axis, represents a quantity
which changes (usually increases) in a regular way. The yaxis, the vertical axis, represents a quantity that varies with
the x-axis quantity, such as the
rate of a moving object varying with time.
1. Straight-line graph
The simplest type of graph is the straight-line graph,
as shown below.
y
gradient
4. Write down a value of x that satisfies the
expression sin x = 0.5
5. The e.m.f. E of an alternating current at time t
is found using the equation:
y-axis ntercept
E = E0 sin 2πft
where E0 = peak e.m.f, and f = frequency of the supply
in Hz (2πft is expressed in radians).
An a.c. electricity supply of 250 V peak value has a
frequency of 50 Hz. Calculate the e.m.f.
2.5 × 10–3 s after the start of the cycle.
Answers
1. (a) π rad
(b) π/2 rad (1.57 rad to 2 s.f.) (c)
0.017 rad
2. (a) 360°
(b) 57°
(c) 45°
The graph shown above can be represented by the
equation:
y = mx + c
where m is the gradient of the graph and c is a constant.
c is the point at which the line crosses the y-axis. The
gradient can be found using the formula:
m=
y2 – y1
x2 – x1
Mathematical techniques
2. Inverse proportion graphs
4. Inverse square law graphs
y
y
Inverse square graphs take the form of y = 1/x2 (above).
For example, Newton’s law of gravitation states that:
The type of graph above takes the form of
y = 1/x, and an example is the relationship between
pressure (x-axis) and volume (y-axis) for an ideal gas at
a constant temperature:
p × V = constant
So: p = constant × 1/V
where p = pressure and V = volume.
The constant can be found by plotting a
straight-line graph of p (x-axis) against 1/V (y-axis);
the constant is equal to the gradient of the graph.
F=
–Gm1m2
r2
where F is the force between two particles of masses m1
and m2, r is the distance between them and G is the
universal gravitational constant.
5. Exponential curves
Exponential growth curves take the form of
y = ex; y increases according to the curve below:
y
3. Square law graphs
Square law graphs take the form of y = x2 (see below). An
example of this is the relationship between the kinetic
energy EK of a flywheel and its angular velocity:
EK = 1/2Iw2
where w is the angular velocity and I the moment of
inertia of the flywheel.
y
A graph of y = e–x (shown below) shows an
exponential decay curve that can be applied to
radioactive decay experiments. N = N0e–lt,
where N is the number of undecayed atoms left at
time t, and N0 is the number of undecayed atoms present
when t = 0, and l is the decay constant.
y
Mathematical techniques
It is reasonable to assume that the activity A (counts
per second using a radioactivity detector) is directly
proportional to the number of undecayed atoms present.
The formula can be re- written A = A0e–lt. Plotting a
3. The table below shows the results of an experiment
investigating the relationship between temperature
and the resistance of a copper wire.
ln A = ln A0 – lt
Plotting ln A against t gives a straight line, with a
gradient of –l.
Examples
1. Sketch the graphs of the following equations where
m is a constant.
(a) y = mx + 2
(b) y = mx – 4
(c) y = k/x
(d) y = sin x
(e) y = e–kx
2. In an experiment to investigate Boyle’s law, the height
of an air column in a tube (of constant cross-sectional
area) was measured for different applied pressures. The
results are presented below:
Height of column/cm
41.9
35.8
30.2
26.25
23.3
21.2
18.9
17.5
16.2
15.0
Pressure/105 Pa
100
120
140
160
180
200
220
240
260
280
Temperature θ/°C
10
20
33
43
50
61
72
79
92
100
Resistance/Ω
23.5
24.5
25.5
26.4
27.0
28.0
29.0
29.4
31.0
31.6
graph of A against t
gives an exponential decay curve.
However, taking logs gives:
Draw a graph of resistance against temperature. Using
the formula Rq = R0(1 + aq) find Rq (the resistance of the
wire at 0 °C) and a (the temperature coefficient of
resistance).
Answers
1.
(a)
y
y
mx
2
y
mx
4
2
(b)
y
Use this data to plot a straight line graph. What does
the gradient of this graph give you?
4
Mathematical techniques
(c)
pV = k
p = pressure, V = volume, k = constant
Ix
y
The volume of the air in the tube is proportional to
the height of the air column since the crosssectional area A is constant
(V = Ah, V ∝ h).
pV = k but
V = Ah pAh
=k
k 1
h =
A p
(d)
y
y
Plotting h against 1/pressure gives a linear graph
with a gradient of k/A.
Plotting the graph of p against 1/height also
gives a graph with gradient of k/A.
in
3.
ResistanceIΩ
30
(e)
y
20
22.5
0
0
y
y-axis ntercept
20
80
40
60
TemperatureI°C
l00
–kx
Rq = R0(1 + aq)
From the graph, Rq is given by the intercept at the R
axis = 22.5 °C.
The gradient of the graph, m, is R0a
= (31.6 – 22.5)/100 = 0.091, so a = 0.091/22.5
= 0.0040 °C–1.
2.
Areas and volumes
50
Areas and volumes appear in many equations used in
physics. For example:
Height of columnIcm
40
density =
30
pressure =
20
0
0
and R =
20
40
p
60
Pa
80
04
l00
mass
volume
(kg m–3)
force
(N m–2 or Pa)
area
rl
A
where R is the resistance of a piece of uniform wire
(Ω), r is the resistivity (Ω m–1), l is its length, and
A is its cross-sectional area.
Mathematical techniques
1. Areas
The area of a rectangle is found by multiplying its height
by its length.
The area of a triangle, Atri, is given by the
expressions:
For a cylinder of length l and radius r, the total surface
area is:
Acyl = 2πrl + 2πr 2
The surface area of a sphere is:
Atri =
Atri =
1/ b × h
2
1/ ab sin
2
Asph = 4πr2
q
2. Volumes
a
h
θ
The volume of a cuboid can be calculated by
multiplying its height by its width and length.
The volume of a cylinder is calculated by
multiplying the cross-sectional area of the
cylinder by the cylinder’s length:
r
Vcyl = πr2l
A sphere has a volume related to its radius, r:
Vsph = 4/3πr3
The area of a circle, Acir, is given in terms of the
circle’s radius, r:
The circumference of a circle is 2πr and this
information is used in calculating the surface area of a
cylinder (e.g. a piece of metal cable). The surface area of
each end of the cylinder is that of a circle, πr2. The
remaining area can be calculated by picturing the curved
surface as being opened out into a rectangle (see below).
In some problems, you need to find the area under a curve
on a graph. For example, the area under
the curve on a velocity–time graph represents the distance
travelled in a certain time. If a body accelerates uniformly
from 0 to V m s–1 in time t1,
and then travels at this speed until time t2, its
velocity–time graph could be represented as shown
below:
t
t
VelocityIm s—l
Acir =
πr2
Areas under graphs
Vms
l
Y
X
r
2πr
l
2
TimeIs
The distance travelled is given by the area under
the graph. In this case, the total area is:
Atotal = area of triangle X + area of rectangle Y
= 1/2t1V + (t2 – t1)V
= (t2 – 1/2t1)V
Mathematical techniques
Examples
2. A wire of diameter 0.51 mm is suspended from a steel
beam. A load of 1.5 kg is attached to its lower end.
Calculate the applied stress on the wire using the
equation:
force
cross-sectional area
stress =
stress =
14.7
π× (2.55 × 10–4)2
= 7.2 × 107 N m–2 (or Pa)
3.
VelocityIm s—l
1. A solid cylinder of pure copper has a diameter of 5.00
cm and a length of 9.00 cm. Its mass is
1.58 kg. Calculate the density of copper in kg m–
3.
5
B
assume g is 9.8 N kg–1.
A
C
m s–1
3. A car accelerates uniformly from 0 to 15
in 4 seconds. It continues at 15 m s–1 for a further 20 s
before the brakes are applied. It decelerates uniformly
for 6 s before coming to a
stop. Calculate the distance travelled by the car.
Answers
1. V = πr2l
The diameter of the cylinder in metres
= 0.0500 m
so the radius r = 0.0250 m
The length of cylinder in metres = 0.0900 m
Therefore, the volume of the cylinder:
V = π× (0.0250)2 × 0.0900
= 1.767 × 10–4 m3
mass
volume
density =
1.58
1.767 × 10–4
= 8.94 × 103 kg m–3
=
2.
stress =
force
cross-sectional area
force = mg
= 1.5 × 9.8
= 14.7 N
area = πr2
wire diameter
r =
2
= 2.55 × 10–4 m
4
l0
20
24
30
TimeIs
distance travelled = area A + area B + area C
= 30 + 300 + 45
= 375 m
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