Graphing and fitting a straight line

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 4-1Core

   The distance $s$ travelled by a trolley was measured as a function of time $t$, and the line fitted to the data points passes through the points $(1.0\,\text{s};\ 2.3\,\text{m})$ and $(7.0\,\text{s};\ 10.1\,\text{m})$.

   What is the slope of the line? Give the answer in m/s.

   m/sCheck
   Answer

   $1.3\,\text{m/s}$

   Show the solution

   Given

   two points on the line $(t_1$; $s_1) = (1.0\,\text{s}$; $2.3\,\text{m})$ and $(t_2$; $s_2) = (7.0\,\text{s}$; $10.1\,\text{m})$

   Find

   the slope $k$ of the line in m/s

   Principle

   On a straight line, $s$ increases by the same amount per unit of time everywhere, so the slope is the same whichever two points of the line it is calculated from. The points are on the line and far apart, so the reading error has only a small effect on the result.

   Formula

   $k$$= \dfrac{\Delta s}{\Delta t}$

   $= \dfrac{s_2 - s_1}{t_2 - t_1}$

   Conversions

   None needed: the distances are in metres and the times in seconds, so the result comes directly in m/s.

   Substitution

   $k$$= \dfrac{10.1\,\text{m} - 2.3\,\text{m}}{7.0\,\text{s} - 1.0\,\text{s}}$

   $= \dfrac{7.8\,\text{m}}{6.0\,\text{s}}$

   $= 1.3\,\text{m/s}$

   Answer

   $1.3\,\text{m/s}$

   Two significant figures, because the differences $7.8\,\text{m}$ and $6.0\,\text{s}$ are known to two figures.

   Reasonableness

   The slope is positive, because the distance increases with time. Between the points the trolley travels about $8\,\text{m}$ in about 6 seconds, that is, about 1.3 metres per second, which is the speed of a leisurely walk.

2. Exercise 4-2Core

   Water is let into a swimming pool. According to the measurements, the height $h$ of the water as a function of time $t$ follows the line $h = 0.50\,\tfrac{\text{cm}}{\text{min}} \cdot t + 12\,\text{cm}$.

   How high is the water surface when the tap has been open for 30 minutes? Give the answer in centimetres.

   cmCheck
   Answer

   $27\,\text{cm}$

   Show the solution

   Given

   the line $h = 0.50\,\tfrac{\text{cm}}{\text{min}} \cdot t + 12\,\text{cm}$, that is, $k = 0.50\,\tfrac{\text{cm}}{\text{min}}$ and $b = 12\,\text{cm}$; time $t = 30\,\text{min}$

   Find

   the height $h$ of the water at $t = 30\,\text{min}$ in centimetres

   Principle

   The line fitted to the measurements is a linear model of the filling of the pool: the slope tells you the rise per minute and the constant term the height at the start. Substituting the time into the equation of the line gives the height.

   Formula

   $h = k t + b$

   Conversions

   None needed: the time is in minutes as in the unit of the slope, and the height comes out in centimetres.

   Substitution

   $h$$= 0.50\,\tfrac{\text{cm}}{\text{min}} \cdot 30\,\text{min} + 12\,\text{cm}$

   $= 15\,\text{cm} + 12\,\text{cm}$

   $= 27\,\text{cm}$

   Answer

   $27\,\text{cm}$

   Two significant figures, like the slope $0.50$ and the constant term $12\,\text{cm}$.

   Reasonableness

   In 30 minutes the surface rises about $0.5 \cdot 30 = 15\,\text{cm}$ from the starting height of $12\,\text{cm}$, so the height is about $27\,\text{cm}$, which is greater than at the start, as it should be when the pool is filling.

3. Exercise 4-3

   The height $h$ of a lawn was followed as a function of days $t$, and the fitted line is $h = 0.60\,\tfrac{\text{cm}}{\text{day}} \cdot t + 4.0\,\text{cm}$.

   On which day from the start of the measurements is the lawn $10\,\text{cm}$ tall? Give the answer in days.

   daysCheck
   Answer

   $10\,\text{days}$

   Show the solution

   Given

   the line $h = 0.60\,\tfrac{\text{cm}}{\text{day}} \cdot t + 4.0\,\text{cm}$, that is, $k = 0.60\,\tfrac{\text{cm}}{\text{day}}$ and $b = 4.0\,\text{cm}$; length $h = 10\,\text{cm}$

   Find

   the time $t$ in days

   Principle

   The line gives the length as a function of time. Here the length is known and the time is asked for, so the equation of the line is solved for the time.

   Formula

   $h = k t + b$, from which $t = \dfrac{h - b}{k}$

   Conversions

   None needed: the lengths are in centimetres and the slope gives centimetres per day.

   Substitution

   $t$$= \dfrac{10\,\text{cm} - 4.0\,\text{cm}}{0.60\,\tfrac{\text{cm}}{\text{day}}}$

   $= \dfrac{6.0\,\text{cm}}{0.60\,\tfrac{\text{cm}}{\text{day}}}$

   $= 10\,\text{days}$

   Answer

   $10\,\text{days}$

   Two significant figures, because $6.0\,\text{cm}$ and $0.60\,\tfrac{\text{cm}}{\text{day}}$ are known to two figures.

   Reasonableness

   The lawn grows six centimetres from the starting $4.0\,\text{cm}$, and that takes about $6 : 0.6 = 10$ days. Check: $0.60 \cdot 10 + 4.0 = 10\,\text{cm}$.

4. Exercise 4-4

   A candle burns steadily. The line fitted to the data points passes through the points $(0.0\,\text{h};\ 20.0\,\text{cm})$ and $(4.0\,\text{h};\ 14.0\,\text{cm})$, with the length of the candle on the $y$-axis.

   What is the slope of the line? Give the answer with its sign in cm/h.

   cm/hCheck
   Answer

   $-1.5\,\text{cm/h}$

   Show the solution

   Given

   points on the line $(t_1$; $y_1) = (0.0\,\text{h}$; $20.0\,\text{cm})$ and $(t_2$; $y_2) = (4.0\,\text{h}$; $14.0\,\text{cm})$, where $y$ is the length of the candle

   Find

   the slope $k$ of the line in cm/h

   Principle

   The candle gets shorter steadily, so the length is a linear function of time. The slope is the change in length divided by the change in time, and it is negative when the quantity decreases.

   Formula

   $k$$= \dfrac{\Delta y}{\Delta t}$

   $= \dfrac{y_2 - y_1}{t_2 - t_1}$

   Conversions

   None needed: the lengths are in centimetres and the times in hours.

   Substitution

   $k$$= \dfrac{14.0\,\text{cm} - 20.0\,\text{cm}}{4.0\,\text{h} - 0.0\,\text{h}}$

   $= \dfrac{-6.0\,\text{cm}}{4.0\,\text{h}}$

   $= -1.5\,\tfrac{\text{cm}}{\text{h}}$

   Answer

   $-1.5\,\text{cm/h}$

   Two significant figures, like the differences $-6.0\,\text{cm}$ and $4.0\,\text{h}$.

   Reasonableness

   The minus sign belongs in the result, because the length of the candle decreases. A candle $20\,\text{cm}$ long would burn out in about $20 : 1.5 \approx 13$ hours, which suits an ordinary candle.

5. Exercise 4-5

   The mass $m$ of a stack of copy paper is directly proportional to the number of sheets $N$, so the mass is represented by a line through the origin. When $N = 250$, the mass is $1.2\,\text{kg}$.

   What is the mass of one sheet? Give the answer in grams.

   gCheck
   Answer

   $4.8\,\text{g}$

   Show the solution

   Given

   the stack has $N = 250$ sheets and a mass $m = 1.2\,\text{kg}$; the mass is represented by a line through the origin

   Find

   the mass of one sheet, that is, the slope $k$ in grams per sheet

   Principle

   The mass is directly proportional to the number of sheets, so the line passes through the origin and its equation is $m = kN$. The slope tells you the mass of one sheet.

   Formula

   $m = kN$, from which $k = \dfrac{m}{N}$

   Conversions

   $1.2\,\text{kg} = 1200\,\text{g}$, because $1\,\text{kg} = 1000\,\text{g}$.

   Substitution

   $k$$= \dfrac{1200\,\text{g}}{250}$

   $= 4.8\,\text{g}$

   Answer

   $4.8\,\text{g}$

   Two significant figures, like the mass $1.2\,\text{kg}$ (the number of sheets is an exact count).

   Reasonableness

   The mass of an ordinary A4 sheet is about $5\,\text{g}$, so the result is reasonable. Check:

   $250 \cdot 4.8\,\text{g}$$= 1200\,\text{g}$

   $= 1.2\,\text{kg}$
6. Exercise 4-6

   An experiment gave six data points that lie roughly on a line but scatter around it because of measurement uncertainty.

   Which of the following correctly describes how a line is fitted to the points?

   Choose an option.

   • The points are joined into a broken line, because every measurement is equally important.
   • The line is drawn through the first and the last point, because they mark the limits of the measured range.
   • The line is drawn through as many points as possible and the remaining points are ignored.
   • The line passes through the middle of the set of points, with points left on both sides of it.

   Answer
   I have done my solution.
   Answer

   The line passes through the middle of the set of points, with points left on both sides of it.

   Show the solution

   Find

   how a line is fitted to scattered data points.

   Principle

   Data points do not usually lie exactly on one line because of measurement uncertainty. Every point tells you about the position of the line, but every point also has random scatter, so the line is chosen to be close to all the points, with roughly as many points above it as below it. A fitted line is a model, not a combination of the points.

   Model answer

   The line passes through the middle of the set of points, with points left on both sides of it. It does not have to pass through any single data point. The line is not tied to two end points and the points are not joined into a broken line, because then the scatter would be read as a property of the phenomenon. A point may be ignored only if a gross error has been identified in it and the reason is recorded.

7. Exercise 4-7

   The mass $m$ of objects of different sizes made of the same metal was measured in grams as a function of the volume $V$ in cubic centimetres. The line fitted to the data points passes through the origin, and its slope is $7.9$.

   What does the slope tell you?

   Choose an option.

   • The slope is $7.9\,\text{g}$ and tells you the mass of the object when the volume is zero.
   • The slope is $7.9\,\text{g/cm}^3$ and tells you how many grams of mass there are per cubic centimetre.
   • The slope is just a number without a unit, because it only tells you how steep the line is.
   • The slope is $7.9\,\text{cm}^3/\text{g}$ and tells you how large a volume one gram fills.

   Answer
   I have done my solution.
   Answer

   The slope is $7.9\,\text{g/cm}^3$ and tells you how many grams of mass there are per cubic centimetre.

   Show the solution

   Find

   what the slope $7.9$ of the line in the mass–volume graph of objects made of the same metal tells you.

   Principle

   The slope is $k = \Delta y : \Delta x$, so its unit is the unit of $y$ divided by the unit of $x$. Here $y$ is the mass in grams and $x$ the volume in cubic centimetres, so the unit is $\text{g/cm}^3$. The line passes through the origin, so the constant term is $b = 0$ and the slope tells you the steepness of the whole line.

   Model answer

   The slope is $7.9\,\text{g/cm}^3$. For every cubic centimetre added, the mass increases by $7.9$ grams, so the slope is the density of this metal. The slope is the density only because all the objects are made of the same material; one line would not be fitted to objects made of different materials. It is not the mass at zero volume (that would be the constant term), nor is it a number without a unit or the inverse $\text{cm}^3/\text{g}$.

8. Exercise 4-8

   The length $l$ of a spring was measured as a function of the mass $m$ of the load. The line fitted to the data points passes through the point $(3.0\,\text{kg};\ 7.0\,\text{cm})$, and its slope is $1.5\,\tfrac{\text{cm}}{\text{kg}}$.

   How long is the spring without a load? Give the answer in centimetres.

   cmCheck
   Answer

   $2.5\,\text{cm}$

   Show the solution

   Given

   a point on the line $(m_1$; $l_1) = (3.0\,\text{kg}$; $7.0\,\text{cm})$ and the slope $k = 1.5\,\tfrac{\text{cm}}{\text{kg}}$

   Find

   the length of the spring without a load, that is, the constant term $b$ in centimetres

   Principle

   The length of the spring is a linear function of the load, $l = km + b$. Without a load $m = 0$, so the length of the spring is where the line crosses the vertical axis, that is, the constant term $b$. It is calculated from one point of the line and the slope.

   Formula

   $l_1 = km_1 + b$, from which $b = l_1 - km_1$

   Conversions

   None needed: the lengths are in centimetres and the masses in kilograms, as in the unit of the slope.

   Substitution

   $b$$= 7.0\,\text{cm} - 1.5\,\tfrac{\text{cm}}{\text{kg}} \cdot 3.0\,\text{kg}$

   $= 7.0\,\text{cm} - 4.5\,\text{cm}$

   $= 2.5\,\text{cm}$

   Answer

   $2.5\,\text{cm}$

   Two significant figures, like the given values $7.0\,\text{cm}$ and $1.5\,\tfrac{\text{cm}}{\text{kg}}$.

   Reasonableness

   The load $3.0\,\text{kg}$ stretches the spring by about $4.5\,\text{cm}$, so the spring without a load is this much shorter than $7.0\,\text{cm}$. The result is positive and smaller than the loaded length, as it should be.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 4-9Core

   The distance $s$ travelled by a cyclist was measured in kilometres and the time $t$ in minutes. The line fitted to the data points passes through the points $(2.0\,\text{min};\ 0.30\,\text{km})$ and $(10.0\,\text{min};\ 1.50\,\text{km})$.

   What is the slope of the line in m/s?

   m/sCheck
   Answer

   $2.5\,\text{m/s}$

   Show the solution

   Given

   points on the line $(t_1$; $s_1) = (2.0\,\text{min}$; $0.30\,\text{km})$ and $(t_2$; $s_2) = (10.0\,\text{min}$; $1.50\,\text{km})$

   Find

   the slope $k$ in m/s

   Principle

   The slope of the line in a distance–time graph is the change in distance divided by the change in time. It is the same for any two points on the line, and the units must be converted to SI units before the result is given.

   Formula

   $k$$= \dfrac{\Delta s}{\Delta t}$

   $= \dfrac{s_2 - s_1}{t_2 - t_1}$

   Conversions

   $\Delta s$$= 1.50\,\text{km} - 0.30\,\text{km}$

   $= 1.20\,\text{km}$

   $= 1200\,\text{m}$

   $\Delta t$$= 10.0\,\text{min} - 2.0\,\text{min}$

   $= 8.0\,\text{min}$

   $= 480\,\text{s}$

   Substitution

   $k$$= \dfrac{1200\,\text{m}}{480\,\text{s}}$

   $= 2.5\,\text{m/s}$

   Answer

   $2.5\,\text{m/s}$

   Two significant figures, because the time interval $8.0\,\text{min}$ is given to two figures.

   Reasonableness

   The result is about $2.5 \cdot 3.6 = 9\,\text{km/h}$. Check with kilometres and minutes:

   $1.20\,\text{km} : 8.0\,\text{min}$

   $= 0.15\,\text{km/min}$

   $= 9\,\text{km/h}$

   which is the speed of leisurely cycling.

2. Exercise 4-10Core

   The graph shows the distance $s$ travelled by a trolley moving steadily as a function of time $t$ in six measurements. Fit a straight line to the points by eye (plot the points on squared paper or lay a ruler over the figure) and use it to estimate how far the trolley has travelled at the moment $t = 8.0\,\text{s}$, when the motion continues in the same way. Give the answer in metres.

   mCheck
   Answer

   $s \approx 3.9\,\text{m}$

   Show the solution

   Given

   the measured points $(t$; $s)$ of the graph: $(1\,\text{s}$; $0.6\,\text{m})$, $(2\,\text{s}$; $0.8\,\text{m})$, $(3\,\text{s}$; $1.6\,\text{m})$, $(4\,\text{s}$; $2.1\,\text{m})$, $(5\,\text{s}$; $2.4\,\text{m})$ and $(6\,\text{s}$; $2.9\,\text{m})$; the time asked about is $t = 8.0\,\text{s}$

   Find

   the distance $s$ travelled by the trolley at $8.0\,\text{s}$ in metres

   Principle

   The trolley moves steadily, so the distance increases linearly with time and a line $s = kt + b$ is fitted to the points. The points deviate randomly from the line, so the line is drawn through the middle of the points, with points on both sides of it, and not through any two points. The prediction assumes that the motion continues in the same way after the measured range too.

   Reading the graph

   The line drawn through the middle of the points passes almost through the origin, and at $6.0\,\text{s}$ it is at $2.9\,\text{m}$. Read these two points from the line, far apart, $(0\,\text{s}$; $0.0\,\text{m})$ and $(6.0\,\text{s}$; $2.9\,\text{m})$, and extend the line to the time $8.0\,\text{s}$.

   Formula

   $k = \dfrac{s_2 - s_1}{t_2 - t_1}$ and $s = s_1 + k\,(t - t_1)$

   Conversions

   None needed: the times are in seconds and the distances in metres.

   Substitution

   $k$$= \dfrac{2.9\,\text{m} - 0.0\,\text{m}}{6.0\,\text{s} - 0\,\text{s}}$

   $= 0.4833\,\tfrac{\text{m}}{\text{s}}$

   so

   $s$$= 0.0\,\text{m} + 0.4833\,\tfrac{\text{m}}{\text{s}} \cdot 8.0\,\text{s}$

   $= 3.87\,\text{m}$

   Answer

   $s \approx 3.9\,\text{m}$

   Two significant figures, because the values read from the line are known to two figures. A fit by eye varies, so about $3.6\ldots4.2\,\text{m}$ is a good answer.

   Reasonableness

   The trolley travels about half a metre per second, so from about $2.9\,\text{m}$ at $6\,\text{s}$ it moves about a metre further in two seconds: $2.9\,\text{m} + 1.0\,\text{m} \approx 3.9\,\text{m}$. A prediction outside the measured range is more uncertain than a value read within the measured range.

3. Exercise 4-11

   The combined mass $m$ of a water canister and the water in it was measured as a function of the volume $V$ of the water. The fitted line passes through the points $(3.0\,\textsf{l};\ 3.9\,\text{kg})$ and $(9.0\,\textsf{l};\ 9.9\,\text{kg})$.

   What is the mass of the empty canister? Give the answer in kilograms.

   kgCheck
   Answer

   $0.9\,\text{kg}$

   Show the solution

   Given

   points on the line $(V_1$; $m_1) = (3.0\,\textsf{l}$; $3.9\,\text{kg})$ and $(V_2$; $m_2) = (9.0\,\textsf{l}$; $9.9\,\text{kg})$, where $m$ is the combined mass of the canister and the water

   Find

   the mass of the empty canister, that is, the constant term $b$ in kilograms

   Principle

   The combined mass is $m = kV + b$: the water adds mass by the amount of the slope per litre, and when $V = 0$ only the mass of the empty canister remains. This is the point where the line crosses the $m$-axis, the constant term $b$.

   Formula

   $k = \dfrac{m_2 - m_1}{V_2 - V_1}$, and since $m_1 = kV_1 + b$, we get $b = m_1 - kV_1$

   Conversions

   None needed: the masses are in kilograms and the volumes in litres, so $k$ comes out in kg/l.

   Substitution

   $k$$= \dfrac{9.9\,\text{kg} - 3.9\,\text{kg}}{9.0\,\textsf{l} - 3.0\,\textsf{l}}$

   $= 1.0\,\tfrac{\text{kg}}{\textsf{l}}$

   $b$$= 3.9\,\text{kg} - 1.0\,\tfrac{\text{kg}}{\textsf{l}} \cdot 3.0\,\textsf{l}$

   $= 0.9\,\text{kg}$

   Answer

   $0.9\,\text{kg}$

   One decimal place like the given masses $3.9\,\text{kg}$ and $9.9\,\text{kg}$.

   Reasonableness

   The slope $1.0\,\text{kg/l}$ is the density of water, so the liquid is correctly identified as water. An empty plastic canister has a mass of about $1\,\text{kg}$, so $0.9\,\text{kg}$ is reasonable.

4. Exercise 4-12

   The price of a taxi ride is a starting charge of $4.0\,\text{€}$ plus $2.5\,\text{€}$ for every kilometre driven. The price is a function of the distance, $y = kx + b$.

   How many times as expensive is a $10\,\text{km}$ ride as a $5.0\,\text{km}$ ride? Give the answer as a number.

   timesCheck
   Answer

   $1.8$

   Show the solution

   Given

   starting charge $b = 4.0\,\text{€}$ and price per kilometre $k = 2.5\,\tfrac{\text{€}}{\text{km}}$; distances $x_1 = 5.0\,\text{km}$ and $x_2 = 10\,\text{km}$

   Find

   the ratio of the prices $r = \dfrac{y(x_2)}{y(x_1)}$

   Principle

   The price is a linear function of distance, $y = kx + b$. Since the constant term $b$ is not zero, the price is not directly proportional to the distance: twice the distance does not cost twice the price.

   Formula

   $y = kx + b$, so $r = \dfrac{kx_2 + b}{kx_1 + b}$

   Conversions

   None needed: the distances are in kilometres and the prices in euros.

   Substitution

   $r$$= \dfrac{2.5\,\tfrac{\text{€}}{\text{km}} \cdot 10\,\text{km} + 4.0\,\text{€}}{2.5\,\tfrac{\text{€}}{\text{km}} \cdot 5.0\,\text{km} + 4.0\,\text{€}}$

   $= \dfrac{29.0\,\text{€}}{16.5\,\text{€}}$

   $= 1.76$

   Answer

   $1.8$

   Times, two significant figures, like the price per kilometre $2.5\,\tfrac{\text{€}}{\text{km}}$.

   Reasonableness

   The ratio is less than 2, because the starting charge $4.0\,\text{€}$ is paid even for a short trip. If the price were directly proportional to the distance ($b = 0$), the ratio would be exactly 2.

5. Exercise 4-13

   Different masses were hung from a steel spring and the extension $x$ of the spring was measured as a function of the mass $m$ of the load. Plot the data points, fit a straight line to them and determine its slope.

   How many millimetres does the spring stretch per additional gram? Give the answer in mm/g.

   Extension of a spring as a function of the load

   $m\,(\text{g})$	$x\,(\text{mm})$
   50.0	4.1
   100.0	8.3
   150.0	11.8
   200.0	16.5
   250.0	20.2
   300.0	23.9

   mm/gCheck
   Answer

   $0.080\,\text{mm/g}$

   Show the solution

   Given

   the data points of the table $(m$; $x)$: $(50.0\,\text{g}$; $4.1\,\text{mm})$, $(100.0\,\text{g}$; $8.3\,\text{mm})$, $(150.0\,\text{g}$; $11.8\,\text{mm})$, $(200.0\,\text{g}$; $16.5\,\text{mm})$, $(250.0\,\text{g}$; $20.2\,\text{mm})$ and $(300.0\,\text{g}$; $23.9\,\text{mm})$

   Find

   the slope $k$ of the fitted line in mm/g

   Principle

   The extension of a spring increases steadily with the load as long as the spring is not stretched too far, so a linear model $x = km + b$ fits the points. The slope tells you the increase in extension per additional gram. The line is drawn through the middle of the set of points, and the slope is calculated from points on the line, not from data points.

   Reading the graph

   The points lie on a rising line with points on both sides of it. The fitted line passes approximately through the points $(0\,\text{g}$; $0.2\,\text{mm})$ and $(300\,\text{g}$; $24.1\,\text{mm})$. They are read from the line and chosen far apart.

   Formula

   $k$$= \dfrac{\Delta x}{\Delta m}$

   $= \dfrac{x_2 - x_1}{m_2 - m_1}$

   Conversions

   None needed: the extension is in millimetres and the mass in grams, as in the requested unit.

   Substitution

   $k$$= \dfrac{24.1\,\text{mm} - 0.2\,\text{mm}}{300\,\text{g} - 0\,\text{g}}$

   $= \dfrac{23.9\,\text{mm}}{300\,\text{g}}$

   $= 0.0797\,\tfrac{\text{mm}}{\text{g}}$

   Answer

   $0.080\,\text{mm/g}$

   Two significant figures, because the line is fitted by eye and the slope is therefore obtained only to two figures. Fitting by eye varies, so about $0.076\text{–}0.084\,\text{mm/g}$ is a good answer.

   Reasonableness

   A load of $100\,\text{g}$ stretches the spring by about $0.080 \cdot 100 = 8\,\text{mm}$, which matches the second row of the table. The line passes almost through the origin, as it should, because an unloaded spring does not stretch.

6. Exercise 4-14

   A trolley moved steadily along a track, and its position $x$ from a mark on the track was measured as a function of time $t$. Plot the data points and fit a straight line to them.

   What is the constant term $b$ of the line in metres? Explain why the line does not pass through the origin and what the constant term means in this experiment.

   Position of a trolley as a function of time

   $t\,(\text{s})$	$x\,(\text{m})$
   0.0	0.82
   1.0	1.31
   2.0	1.68
   3.0	2.22
   4.0	2.57
   5.0	3.09

   mCheck
   Answer

   $0.83\,\text{m}$

   Show the solution

   Given

   the data points of the table $(t$; $x)$: $(0.0\,\text{s}$; $0.82\,\text{m})$, $(1.0\,\text{s}$; $1.31\,\text{m})$, $(2.0\,\text{s}$; $1.68\,\text{m})$, $(3.0\,\text{s}$; $2.22\,\text{m})$, $(4.0\,\text{s}$; $2.57\,\text{m})$ and $(5.0\,\text{s}$; $3.09\,\text{m})$

   Find

   the constant term $b$ of the fitted line in metres, and an explanation of why the line does not pass through the origin

   Principle

   The trolley moves steadily, so the position is a linear function of time, $x = vt + b$. The constant term is the position at the moment $t = 0$, that is, where the line crosses the position axis. It is read from the fitted line and not from a data point, because the points have scatter.

   Reading the graph

   The points lie on a rising line that starts on the vertical axis clearly above zero. The fitted line passes approximately through the points $(0.0\,\text{s}$; $0.83\,\text{m})$ and $(5.0\,\text{s}$; $3.07\,\text{m})$, so the intercept is $0.83\,\text{m}$. The points $(0.0\,\text{s}$; $0.83\,\text{m})$ and $(5.0\,\text{s}$; $3.07\,\text{m})$ also give the slope, that is, the speed of the trolley.

   Formula

   $b = x_1 - vt_1$, where $v = \dfrac{x_2 - x_1}{t_2 - t_1}$; when $t_1 = 0$, $b = x_1$

   Conversions

   None needed: the times are in seconds and the positions in metres.

   Substitution

   $v$$= \dfrac{3.07\,\text{m} - 0.83\,\text{m}}{5.0\,\text{s} - 0.0\,\text{s}}$

   $= 0.448\,\tfrac{\text{m}}{\text{s}}$

   $b$$= 0.83\,\text{m} - 0.448\,\tfrac{\text{m}}{\text{s}} \cdot 0.0\,\text{s}$

   $= 0.83\,\text{m}$

   Answer

   $0.83\,\text{m}$

   Two significant figures, like the positions. Fitting by eye varies, so about $0.75\text{–}0.91\,\text{m}$ is a good answer. The line does not pass through the origin, because when the clock was started the trolley was already $0.83\,\text{m}$ from the mark on the track from which the position is measured. The constant term is therefore the starting position of the trolley, and a line through the origin would not fit this measurement.

   Reasonableness

   A speed of $0.45\,\text{m/s}$ takes the trolley about $2.2\,\text{m}$ further in five seconds: $0.83\,\text{m} + 2.2\,\text{m} \approx 3.1\,\text{m}$, which matches the last measurement. The constant term is positive, because the trolley was already ahead of the mark.

7. Exercise 4-15

   From each of two uniform metals, A and B, objects of different volumes made of the same material were produced, and the mass of each was weighed.

   Plot the data points of both materials in the same coordinate system and fit a separate line to each. How many times greater is the slope of the line of B than that of A? What does the slope mean in this graph, and which material is denser?

   Mass of the objects as a function of volume

   $V\,(\text{cm}^3)$	$m_\text{A}\,(\text{g})$	$m_\text{B}\,(\text{g})$
   10.0	27.4	78.6
   20.0	53.8	156.8
   30.0	82.1	237.1
   40.0	108.9	313.5
   50.0	135.2	392.0

   timesCheck
   Answer

   $2.9$

   Show the solution

   Given

   the volumes of the table $V = 10.0\text{–}50.0\,\text{cm}^3$ and the masses $m_\text{A}$: $27.4$, $53.8$, $82.1$, $108.9$ and $135.2\,\text{g}$ and $m_\text{B}$: $78.6$, $156.8$, $237.1$, $313.5$ and $392.0\,\text{g}$

   Find

   the ratio $\dfrac{k_\text{B}}{k_\text{A}}$, the meaning of the slope in this graph and which material is denser

   Principle

   Each metal is uniform and the objects are made of the same material, so the mass is a linear function of the volume, $m = kV$. The slope $k = \Delta m : \Delta V$ is the mass per unit of volume, that is, in this graph the density of the material. A separate line is fitted for each material in the same coordinate system, and the steeper line belongs to the denser material.

   Reading the graph

   Both sets of points rise along lines close to the origin, but the points of B are much higher than those of A at every volume. The fitted lines pass approximately through the points $(0\,\text{cm}^3$; $0.3\,\text{g})$ and $(50\,\text{cm}^3$; $135.6\,\text{g})$ (A) and $(0\,\text{cm}^3$; $0.5\,\text{g})$ and $(50\,\text{cm}^3$; $392.3\,\text{g})$ (B).

   Formula

   $k = \dfrac{\Delta m}{\Delta V}$ and $\dfrac{k_\text{B}}{k_\text{A}}$

   Conversions

   None needed: the ratio has no unit, and the slope of each line comes out in $\text{g/cm}^3$.

   Substitution

   $k_\text{A}$$= \dfrac{135.6\,\text{g} - 0.3\,\text{g}}{50\,\text{cm}^3}$

   $= 2.71\,\tfrac{\text{g}}{\text{cm}^3}$

   $k_\text{B}$$= \dfrac{392.3\,\text{g} - 0.5\,\text{g}}{50\,\text{cm}^3}$

   $= 7.84\,\tfrac{\text{g}}{\text{cm}^3}$

   so

   $\dfrac{k_\text{B}}{k_\text{A}}$$= \dfrac{7.84}{2.71}$

   $= 2.89$

   Answer

   $2.9$

   Times, two significant figures, because the lines are fitted by eye and the slopes $2.7\,\text{g/cm}^3$ and $7.8\,\text{g/cm}^3$ are reliable only to two figures, although the masses are given to three. Here the slope means the density of the material (mass per cubic centimetre), and B is denser, because its slope is greater.

   Reasonableness

   The slopes $2.7\,\text{g/cm}^3$ and $7.8\,\text{g/cm}^3$ are close to the densities of aluminium and iron, so the materials could be aluminium and iron. At the same volume of $50\,\text{cm}^3$, B has roughly three times the mass of A, which matches the ratio.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 4-16

   A student measured the position $s$ of a trolley as a function of time $t$, but started the stopwatch only when the trolley had already travelled some distance from the start of the track.

   The line fitted to the data points is $s = 0.85\,\tfrac{\text{m}}{\text{s}} \cdot t + 0.40\,\text{m}$, where $s$ is measured from the start of the track.

   What time on the stopwatch corresponds to the moment when the trolley was at the start of the track ($s = 0$)? Give the answer in seconds with its sign and justify the sign.

   sCheck
   Answer

   $-0.47\,\text{s}$

   Show the solution

   Given

   the line $s = 0.85\,\tfrac{\text{m}}{\text{s}} \cdot t + 0.40\,\text{m}$, that is, $k = 0.85\,\tfrac{\text{m}}{\text{s}}$ and $b = 0.40\,\text{m}$; the trolley is at the start of the track when $s = 0$

   Find

   the stopwatch reading $t$ when $s = 0$

   Principle

   The constant term $b = 0.40\,\text{m}$ is the position at $t = 0$, that is, when the stopwatch was started: the trolley was already $0.40\,\text{m}$ from the start of the track. So the trolley was at the start of the track before the watch was started, and extending the line backwards gives a negative stopwatch reading.

   Formula

   $s = kt + b$, from which $t = \dfrac{s - b}{k}$

   Conversions

   None needed: the position is in metres and the slope gives metres per second.

   Substitution

   $t$$= \dfrac{0 - 0.40\,\text{m}}{0.85\,\tfrac{\text{m}}{\text{s}}}$

   $= -0.4706\,\text{s}$

   Answer

   $-0.47\,\text{s}$

   Two significant figures like the slope $0.85\,\tfrac{\text{m}}{\text{s}}$ and the constant term $0.40\,\text{m}$. The minus sign means that the trolley was at the start of the track $0.47\,\text{s}$ before the stopwatch was started. The result assumes that the trolley was already moving at the same speed before the watch was started, because the line is extended outside the measured range ($t < 0$).

   Reasonableness

   Check: in 0.47 seconds at $0.85\,\text{m/s}$ the trolley travels about $0.85 \cdot 0.47 = 0.40\,\text{m}$, which is exactly the constant term. The result is the same as if the watch had been started $0.47\,\text{s}$ earlier and the constant term were zero.

2. Exercise 4-17

   The temperature $T$ of coffee was measured as a function of time $t$: $0\,\text{min}$: $90\,{}^\circ\text{C}$, $10\,\text{min}$: $72\,{}^\circ\text{C}$, $20\,\text{min}$: $60\,{}^\circ\text{C}$ and $30\,\text{min}$: $51\,{}^\circ\text{C}$.

   How many times greater is the change in temperature per minute (the absolute value of the slope) in the first ten-minute interval than in the last? Justify why one line does not describe all the measurements.

   timesCheck
   Answer

   $2.0$

   Show the solution

   Given

   the temperature $T$ of the coffee: $90\,{}^\circ\text{C}$ ($0\,\text{min}$), $72\,{}^\circ\text{C}$ ($10\,\text{min}$), $60\,{}^\circ\text{C}$ ($20\,\text{min}$) and $51\,{}^\circ\text{C}$ ($30\,\text{min}$)

   Find

   the ratio of the absolute values of the slopes $\dfrac{|k_1|}{|k_3|}$ in the first and last ten-minute intervals, and a justification of why one line does not fit

   Principle

   If the data points followed a straight line, the slope would be the same in every time interval. Calculate the slope for the first and last intervals and compare them.

   Formula

   $k = \dfrac{\Delta T}{\Delta t}$ and $\dfrac{|k_1|}{|k_3|}$

   Conversions

   None needed: a temperature difference is the same in degrees Celsius and in kelvins, and the time is in minutes.

   Substitution

   $k_1$$= \dfrac{72\,{}^\circ\text{C} - 90\,{}^\circ\text{C}}{10\,\text{min}}$

   $= -1.8\,\tfrac{{}^\circ\text{C}}{\text{min}}$

   $k_3$$= \dfrac{51\,{}^\circ\text{C} - 60\,{}^\circ\text{C}}{10\,\text{min}}$

   $= -0.90\,\tfrac{{}^\circ\text{C}}{\text{min}}$

   so

   $\dfrac{|k_1|}{|k_3|}$$= \dfrac{1.8}{0.90}$

   $= 2.0$

   Answer

   $2.0$

   Times, two significant figures like the temperature differences. One line does not fit, because the slope is not constant: the coffee first cools twice as fast as at the end, and the interval 10–$20\,\text{min}$ gives a third value, $-1.2\,\tfrac{{}^\circ\text{C}}{\text{min}}$. The points do not lie on a straight line.

   Reasonableness

   The slopes are negative, because the coffee cools. The cooling slows down because the coffee approaches room temperature and the temperature difference to the surroundings gets smaller, which fits what you know about a hot drink cooling.

3. Exercise 4-18

   Two points are read from the line $y = 2.0\,x$, and the reading error of each $y$ value is at most $0.1$ units. The slope is calculated either from the points $x = 4.0$ and $x = 4.5$ or from the points $x = 1.0$ and $x = 9.0$.

   How many times greater is the relative uncertainty of the slope when it is calculated from the close points? Use the worst case, in which the reading errors add up in the difference $\Delta y$. Justify why points far apart are the better choice.

   timesCheck
   Answer

   $16$

   Show the solution

   Given

   line $y = 2.0\,x$; the uncertainty of a $y$ reading is about $\Delta y_\text{read} = 0.1$, so the uncertainty of the difference $\Delta y$ of two readings is, estimated cautiously, about $2\,\Delta y_\text{read} = 2 \cdot 0.1 = 0.2$; close together $\Delta x_1 = 4.5 - 4.0 = 0.5$, far apart $\Delta x_2 = 9.0 - 1.0 = 8.0$

   Find

   how many times greater the relative uncertainty of the slope is when it is calculated from the close points

   Principle

   The slope is $k = \Delta y : \Delta x$. When the $x$ values are known accurately, the relative uncertainty of the slope is equal to the relative uncertainty of the difference $\Delta y$. On a straight line $\Delta y = k\,\Delta x$, so the difference grows with the interval, but the reading error stays the same.

   Formula

   $\dfrac{\Delta k}{k}$$= \dfrac{2\,\Delta y_\text{read}}{\Delta y}$

   $= \dfrac{2\,\Delta y_\text{read}}{k\,\Delta x}$

   so

   $\dfrac{(\Delta k : k)_\text{near}}{(\Delta k : k)_\text{far}}$$= \dfrac{\Delta x_2}{\Delta x_1}$

   Conversions

   None needed: the ratio has no unit.

   Substitution

   Near:

   $\Delta y$$= 2.0 \cdot 0.5$

   $= 1.0$

   $\dfrac{\Delta k}{k}$$= \dfrac{0.2}{1.0}$

   $= 0.20$

   Far:

   $\Delta y$$= 2.0 \cdot 8.0$

   $= 16$

   $\dfrac{\Delta k}{k}$$= \dfrac{0.2}{16}$

   $= 0.0125$

   The ratio is

   $\dfrac{0.20}{0.0125}$$= \dfrac{8.0}{0.5}$

   $= 16$

   Answer

   $16$

   Times. The reading error cancels out of the ratio, so the number is determined only by the ratio of the $\Delta x$ values, $8.0 : 0.5 = 16$, which is exact; two significant figures are enough. Points far apart are the better choice, because the difference $\Delta y$ is large compared with the reading error.

   Reasonableness

   An error of $20\,\%$ for the close points is too large, while about $1.3\,\%$ for the far points is good. The ratio 16 is the same as the ratio of the $\Delta x$ values, $8.0 : 0.5$, as it must be according to the formula.

4. Exercise 4-19

   The extension $x$ of a spring was measured as a function of the mass $m$ of the load with greater loads than usual. Plot the data points.

   Over which interval of loads does the linear model hold? Fit a line to that interval and give its slope in mm/g. Justify why the model does not hold over the whole measured range.

   Extension of a spring with large loads

   $m\,(\text{g})$	$x\,(\text{mm})$
   100.0	8.1
   200.0	16.3
   300.0	24.2
   400.0	32.5
   500.0	40.1
   600.0	51.8
   700.0	68.9

   mm/gCheck
   Answer

   $0.080\,\text{mm/g}$

   Show the solution

   Given

   the data points of the table $(m$; $x)$ over the range $100\text{–}700\,\text{g}$; extensions $8.1$, $16.3$, $24.2$, $32.5$, $40.1$, $51.8$ and $68.9\,\text{mm}$

   Find

   the interval of loads over which the linear model holds, the slope $k$ of its line in mm/g, and a justification of why the model does not hold over the whole range

   Principle

   The linear model holds only over the measured range where the points lie on a line. Fit a line to the points that are on the line, and check where the other points no longer fit. Among the measured points, over the range $100\text{–}500\,\text{g}$ the extension increases steadily, but with greater loads the spring stretches more than the line predicts.

   Reading the graph

   The first five points lie on a line with points on both sides of it. The fitted line passes approximately through the points $(0\,\text{g}$; $0.2\,\text{mm})$ and $(500\,\text{g}$; $40.3\,\text{mm})$. The points $600\,\text{g}$ and $700\,\text{g}$ are above the line, and the difference grows: the line would give $48.3\,\text{mm}$ and $56.3\,\text{mm}$, but the measured values are $51.8\,\text{mm}$ and $68.9\,\text{mm}$. The deviations are $3.5\,\text{mm}$ and $12.6\,\text{mm}$, while inside the range the deviation is at most $0.3\,\text{mm}$.

   Formula

   $k$$= \dfrac{\Delta x}{\Delta m}$

   $= \dfrac{x_2 - x_1}{m_2 - m_1}$

   Conversions

   None needed: the extension is in millimetres and the mass in grams.

   Substitution

   $k$$= \dfrac{40.3\,\text{mm} - 0.2\,\text{mm}}{500\,\text{g} - 0\,\text{g}}$

   $= \dfrac{40.1\,\text{mm}}{500\,\text{g}}$

   $= 0.0802\,\tfrac{\text{mm}}{\text{g}}$

   Answer

   $0.080\,\text{mm/g}$

   Two significant figures, like the extensions. Among the measured points, the linear model holds over the range $100\text{–}500\,\text{g}$, and by extrapolation also for smaller loads all the way to an unloaded spring, because the line passes approximately through the origin and an unloaded spring does not stretch. The loads $600\,\text{g}$ and $700\,\text{g}$ deviate from the line clearly more than the scatter of the measurements explains: the spring is overloaded, that is, stretched beyond its elastic range, so the extension is no longer directly proportional to the load and the spring may be left permanently stretched. The line must not be extended to this range of loads.

   Reasonableness

   The slope of the first part, $0.080\,\text{mm/g}$, corresponds to a spring under a small load (compare $8\,\text{mm}$ with a load of $100\,\text{g}$). If a line were fitted to all seven points, the slope would increase and the range $100\text{–}500\,\text{g}$ would be described worse.

5. Exercise 4-20

   The distance $s$ travelled by a trolley was measured as a function of time $t$. Plot the data points. The steepest line that still fits among the points passes through the points $(1.0\,\text{s};\ 0.75\,\text{m})$ and $(6.0\,\text{s};\ 3.50\,\text{m})$, and the gentlest line through the points $(2.0\,\text{s};\ 1.40\,\text{m})$ and $(5.0\,\text{s};\ 2.60\,\text{m})$.

   A student estimated the slope as the mean of the slopes of these lines and its uncertainty as their difference: "The steepest line gives $0.55\,\text{m/s}$ and the gentlest $0.40\,\text{m/s}$. The mean is $0.475\,\text{m/s}$ and the uncertainty is $0.55\,\text{m/s} - 0.40\,\text{m/s} = 0.15\,\text{m/s}$, so the uncertainty is $32\,\%$ of the slope."

   Find the student's error and calculate the correct uncertainty as a percentage of the slope. Explain what this estimate tells you and what it does not guarantee.

   Distance of a trolley as a function of time

   $t\,(\text{s})$	$s\,(\text{m})$
   1.0	0.75
   2.0	1.40
   3.0	1.80
   4.0	2.30
   5.0	2.60
   6.0	3.50

   %Check
   Answer

   $16\,\%$

   Show the solution

   Given

   the data points of the table $(t$; $s)$: $(1.0\,\text{s}$; $0.75\,\text{m})$, $(2.0\,\text{s}$; $1.40\,\text{m})$, $(3.0\,\text{s}$; $1.80\,\text{m})$, $(4.0\,\text{s}$; $2.30\,\text{m})$, $(5.0\,\text{s}$; $2.60\,\text{m})$ and $(6.0\,\text{s}$; $3.50\,\text{m})$; the steepest line through the points $(1.0\,\text{s}$; $0.75\,\text{m})$ and $(6.0\,\text{s}$; $3.50\,\text{m})$, the gentlest line through the points $(2.0\,\text{s}$; $1.40\,\text{m})$ and $(5.0\,\text{s}$; $2.60\,\text{m})$

   Find

   the student's error and the uncertainty of the slope as a percentage of the slope

   Principle

   The steepest and the gentlest line that still fit among the points bound an estimated range for the slope. The value of the slope is the middle of the range and the uncertainty is half the width of the range, because the value is in the middle and the range extends by the uncertainty to both sides. The estimate is simple and does not guarantee that the true value is within the range.

   Reading the graph

   The plotted points rise steadily, and the steepest and the gentlest line both pass through the set of points so that points are left on both sides. Between the points $1.0\,\text{s}$ and $6.0\,\text{s}$ the steepest line rises $3.50\,\text{m} - 0.75\,\text{m} = 2.75\,\text{m}$, and between the points $2.0\,\text{s}$ and $5.0\,\text{s}$ the gentlest rises $2.60\,\text{m} - 1.40\,\text{m} = 1.20\,\text{m}$. The student's mean and slopes are correct; the error is in the calculation of the uncertainty.

   Formula

   $k$$= \dfrac{\Delta s}{\Delta t}$

   $\;k$$= \dfrac{k_\text{max} + k_\text{min}}{2}$

   $\;\Delta k$$= \dfrac{k_\text{max} - k_\text{min}}{2}$

   and the relative uncertainty $\dfrac{\Delta k}{k} \cdot 100\,\%$

   Conversions

   None needed: the slopes come out in m/s, and the percentage is a ratio without a unit multiplied by $100$.

   Substitution

   $k_\text{max}$$= \dfrac{2.75\,\text{m}}{5.0\,\text{s}}$

   $= 0.55\,\tfrac{\text{m}}{\text{s}}$

   $k_\text{min}$$= \dfrac{1.20\,\text{m}}{3.0\,\text{s}}$

   $= 0.40\,\tfrac{\text{m}}{\text{s}}$

   The mean is

   $k$$= \dfrac{0.55 + 0.40}{2}\,\tfrac{\text{m}}{\text{s}}$

   $= 0.475\,\tfrac{\text{m}}{\text{s}}$

   $\Delta k$$= \dfrac{0.55 - 0.40}{2}\,\tfrac{\text{m}}{\text{s}}$

   $= 0.075\,\tfrac{\text{m}}{\text{s}}$

   so

   $\dfrac{\Delta k}{k} \cdot 100\,\%$$= \dfrac{0.075}{0.475} \cdot 100\,\%$

   $= 15.8\,\%$

   Answer

   The student used the whole difference of the slopes, $0.15\,\tfrac{\text{m}}{\text{s}}$, as the uncertainty instead of half of it; the correct uncertainty is $16\,\%$, two significant figures, like the coordinates of the points (for example $0.75\,\text{m}$). The slope is therefore about $(0.48 \pm 0.08)\,\tfrac{\text{m}}{\text{s}}$. The estimate tells you how much the slope can vary because of the scatter, but it does not guarantee that the true speed is within this range: the range is an estimated bound and not a certain limit.

   Reasonableness

   The correct range $0.475 \pm 0.075\,\tfrac{\text{m}}{\text{s}}$ is $0.40\text{–}0.55\,\tfrac{\text{m}}{\text{s}}$, that is, exactly the gentlest and the steepest line. The student's range $0.475 \pm 0.15$ would be $0.33\text{–}0.63\,\tfrac{\text{m}}{\text{s}}$, which extends clearly beyond both lines, so $32\,\%$ is double. The average speed obtained from the end points of the whole measuring interval,

   $\tfrac{3.50 - 0.75}{5.0}\,\tfrac{\text{m}}{\text{s}}$$= 0.55\,\tfrac{\text{m}}{\text{s}}$

   is inside the range.