How to Find the Uncertainty in a Graph Gradient

Your position-time graph gives a trolley velocity of \(0.80\ \mathrm{m\,s^{-1}}\). But each position measurement has an uncertainty. A slightly steeper or shallower straight line might fit the data just as well. So how certain is that velocity?

To find the uncertainty in a graph gradient, draw the steepest and shallowest straight lines that remain consistent with the measurement uncertainty intervals. Calculate both gradients, then take half their difference:

\[ \boxed{\Delta m=\frac{m_{\mathrm{high}}-m_{\mathrm{low}}}{2}} \]

Here, \(m_{\mathrm{high}}\) and \(m_{\mathrm{low}}\) are the numerically highest and lowest acceptable gradients, and \(\Delta m\) is the estimated absolute uncertainty in the gradient.

The best-fit gradient gives your central estimate. The maximum and minimum gradients tell you how much that estimate could change within the stated measurement uncertainties.

All experimental data and numerical examples in this resource are illustrative.

01Why does a graph gradient have uncertainty?

Imagine measuring the position of a moving trolley at five different times.

You plot the measurements and draw a straight line of best fit. The gradient represents the trolley’s velocity.

However, your position measurements aren’t perfectly precise. If one reading is \(1.10\ \mathrm{m}\), for example, its actual value might reasonably be represented by an interval from \(1.00\ \mathrm{m}\) to \(1.20\ \mathrm{m}\).

This means the plotted point doesn’t represent one perfectly known position. It represents a range of possible positions.

Several straight lines might pass through all the measurement intervals. Each line has a slightly different gradient, and therefore gives a slightly different estimate of velocity.

The important distinction is:

• The best-fit line gives your central estimate.
• The maximum gradient gives the highest acceptable gradient.
• The minimum gradient gives the lowest acceptable gradient.
• The difference between the extreme gradients estimates the uncertainty.

You aren’t calculating how far the experimental result is from the true answer. You’re estimating how much the gradient could vary given the uncertainty in your measurements.

02Start with the measurement interval

Before calculating anything, establish what each measurement’s uncertainty actually means.

Suppose a position measurement is:

\[ s=(1.10\pm0.10)\ \mathrm{m} \]

The measured position is \(1.10\ \mathrm{m}\), and its absolute uncertainty is \(0.10\ \mathrm{m}\).

The corresponding measurement interval is:

\[ 1.00\leq s\leq1.20\ \mathrm{m} \]

The error bar extends \(0.10\ \mathrm{m}\) above and below the measured value.

An error bar represents the measurement interval associated with your chosen uncertainty convention. It does not mean the measurement is definitely wrong by that amount, or that the true value is guaranteed to lie inside the interval.

Which uncertainty should you use?

The uncertainty convention depends on how the measurement was obtained.

For example, if a ruler has markings every \(1\ \mathrm{mm}\), a common convention is to assign an uncertainty of:

\[ \Delta L=\pm0.5\ \mathrm{mm} \]

For a digital instrument, don’t automatically assume the last displayed digit represents its entire measurement uncertainty. Its accuracy specification or the convention given in the question takes priority.

For repeated measurements, a common school convention is to use half the range:

\[ \Delta x=\frac{x_{\max}-x_{\min}}{2} \]

This is an estimate of absolute uncertainty. It is not automatically the same as standard deviation.

The key is to establish the uncertainty before drawing the error bars. If the question gives an uncertainty of \(\pm0.10\ \mathrm{m}\), don’t replace it with a different estimate simply because another method gives a smaller number.

03Uncertainty, error, and standard deviation are different

These terms are related, but they describe different things.

Measurement error is often unknown because the true value isn’t known.

Uncertainty can still be estimated even when the actual error is unknown.

Standard deviation describes the typical spread of repeated readings. It can be used as an uncertainty convention, but it isn’t automatically interchangeable with instrument uncertainty, half-range uncertainty, or uncertainty in the mean.

For example, a set of repeated measurements might have a standard deviation of \(0.03\ \mathrm{N}\), but a half-range uncertainty of \(0.04\ \mathrm{N}\).

If the question specifies half-range uncertainty, you must use \(0.04\ \mathrm{N}\) for the error bars. Don’t substitute the standard deviation.

This matters because the maximum and minimum gradients depend directly on the intervals you draw.

04How to calculate the uncertainty in a graph gradient

Once you’ve established the measurement intervals, follow this process.

Step 1: Identify the axes and their units

Start by checking what is plotted on each axis.

The gradient is:

\[ m=\frac{\Delta y}{\Delta x} \]

where:

• \(m\) is the gradient.
• \(\Delta y\) is the change in the vertical quantity.
• \(\Delta x\) is the change in the horizontal quantity.

The units of the gradient are the vertical units divided by the horizontal units.

For example, a position-time graph has gradient units of:

\[ \frac{\mathrm{m}}{\mathrm{s}}=\mathrm{m\,s^{-1}} \]

The gradient represents velocity.

If you need a refresher on what different HSC Physics graph gradients physically represent, see HSC Physics Graphs: How to Use Gradient and Area.

Step 2: Draw the uncertainty intervals

Plot each measured point and its corresponding error bars.

For example, if a position measurement is \(2.30\pm0.10\ \mathrm{m}\), the vertical error bar extends from \(2.20\ \mathrm{m}\) to \(2.40\ \mathrm{m}\).

Repeat this for every data point.

Step 3: Draw the best-fit line

Draw a straight line that represents the overall trend of the measurements.

It should balance the scatter of the points rather than simply joining them in order.

Calculate its gradient using two widely separated points on the line:

\[ m_{\mathrm{best}}=\frac{y_2-y_1}{x_2-x_1} \]

This is your central estimate.

Don’t automatically force the line through the origin. Only do this when the physical model and experimental conditions justify it.

Step 4: Draw the maximum gradient line

The maximum gradient is the numerically highest gradient that remains consistent with the measurement intervals.

For a positive relationship, this is generally the steepest acceptable line.

It will often pass through:

• The lower end of the first vertical error bar.
• The upper end of the last vertical error bar.

However, you must check that the line also passes through the intermediate error bars.

A line that misses an intermediate uncertainty interval is not an acceptable maximum-gradient line.

Step 5: Draw the minimum gradient line

The minimum gradient is the numerically lowest gradient that remains consistent with the measurement intervals.

For a positive relationship, this is generally the shallowest acceptable line.

It will often pass through:

• The upper end of the first vertical error bar.
• The lower end of the last vertical error bar.

Again, check every intermediate error bar.

Step 6: Calculate both extreme gradients

Use:

\[ m=\frac{y_2-y_1}{x_2-x_1} \]

Choose two points that are far apart on each extreme line. This reduces the effect of small reading errors when calculating the gradient.

You should now have:

\[ m_{\mathrm{high}} \]

and

\[ m_{\mathrm{low}} \]

Step 7: Calculate the uncertainty

Subtract the minimum gradient from the maximum gradient and divide by two:

\[ \boxed{\Delta m=\frac{m_{\mathrm{high}}-m_{\mathrm{low}}}{2}} \]

The result is the estimated absolute uncertainty in the gradient.

Its units are the same as the gradient’s units.

If you also need percentage uncertainty, use:

\[ \%\text{ uncertainty} = \frac{\Delta m}{|m_{\mathrm{best}}|} \times100\% \]

The absolute value ensures that the percentage uncertainty is positive.

A small qualification: the half-range formula gives a single symmetric estimate. If the best-fit gradient is not near the midpoint of the two extreme gradients, the more accurate way to report the result is with asymmetric uncertainties or the full interval. We’ll work through this shortly.

05Where the graph comes from: a trolley experiment

Consider an experiment investigating whether a trolley moves at constant velocity.

The experimental question is straightforward: does the trolley’s position change linearly with time?

A suitable apparatus consists of:

• A trolley moving along a straight track.
• An ultrasonic motion sensor positioned at one end of the track, facing the trolley.
• A data logger connected to the sensor.
• A computer displaying the recorded position-time data.

The sensor emits ultrasonic pulses towards the trolley and detects the returning echoes. It uses the travel time of each pulse to calculate the trolley’s position.

The data logger records position against time.

The prediction is that a trolley moving at constant velocity will produce a straight-line position-time graph.

Why?

Velocity is the rate of change of position:

\[ v=\frac{\Delta s}{\Delta t} \]

If velocity remains constant, equal time intervals produce equal changes in position. The gradient therefore remains constant.

For example, if the trolley moves from \(0.70\ \mathrm{m}\) to \(2.30\ \mathrm{m}\) in \(2.0\ \mathrm{s}\):

\[ v=\frac{2.30-0.70}{2.0} =0.80\ \mathrm{m\,s^{-1}} \]

The gradient gives the velocity.

Why the original measurement method matters

An initial design might use a metre ruler beside the track and a handheld stopwatch.

This introduces two avoidable problems:

• Human reaction time affects the start and stop of the stopwatch.
• Reading the ruler from an angle introduces parallax uncertainty.

A revised design uses the motion sensor and data logger. This removes much of the variation caused by human timing and manual position readings.

However, the sensor still has finite resolution and may have calibration uncertainty. The revised apparatus improves the measurements; it doesn’t make them perfectly exact.

When the results are plotted, the points should lie reasonably close to a straight line if the trolley moves at constant velocity. The error bars represent the remaining measurement uncertainty.

The maximum-minimum gradient method then estimates how much the calculated velocity could vary within those intervals.

06A second example: When the best-fit gradient is not centred

The first example had a best-fit gradient exactly halfway between the maximum and minimum gradients.

That won’t always happen.

Consider an electrical experiment in which a student investigates the resistance of a resistor.

The student plots potential difference \(V\) vertically against current \(I\) horizontally.

From Ohm’s law:

\[ V=IR \]

The gradient therefore represents resistance:

\[ m=R \]

Assume current uncertainty is negligible and the voltage uncertainties are given below.

The following data are illustrative:

The best-fit line has a gradient of \(4.84\ \Omega\).

Calculate the maximum and minimum gradients, then determine the uncertainty.

Step 1: Calculate the maximum gradient.

Use the lower end of the first error bar and the upper end of the final error bar:

\[ \begin{aligned} m_{\mathrm{high}} &=\frac{2.54-0.50}{0.50-0.10}\\ &=\frac{2.04}{0.40}\\ &=5.10\ \Omega \end{aligned} \]

Check that this line passes through the intermediate error bars.

For example, at \(I=0.30\ \mathrm{A}\), the line gives \(V=1.52\ \mathrm{V}\), which is inside the interval \(1.37\)-\(1.53\ \mathrm{V}\).

Step 2: Calculate the minimum gradient.

Use the upper end of the first error bar and the lower end of the final error bar:

\[ \begin{aligned} m_{\mathrm{low}} &=\frac{2.42-0.54}{0.50-0.10}\\ &=\frac{1.88}{0.40}\\ &=4.70\ \Omega \end{aligned} \]

This line also passes through all the intermediate uncertainty intervals.

Step 3: Calculate the half-range uncertainty.

\[ \begin{aligned} \Delta m &=\frac{5.10-4.70}{2}\\ &=0.20\ \Omega \end{aligned} \]

The graphical half-range is \(0.20\ \Omega\).

Step 4: Check whether a symmetric uncertainty is appropriate.

The best-fit gradient is \(4.84\ \Omega\).

However, the midpoint of the extreme gradients is:

\[ \frac{5.10+4.70}{2}=4.90\ \Omega \]

The best-fit gradient is not exactly at this midpoint.

The full interval is:

\[ 4.70\leq R\leq5.10\ \Omega \]

Relative to the best-fit value of \(4.84\ \Omega\), the uncertainty is:

• \(0.14\ \Omega\) below the best-fit value.
• \(0.26\ \Omega\) above the best-fit value.

A more faithful report is therefore:

\[ R=4.84^{+0.26}_{-0.14}\ \Omega \]

The conventional half-range estimate is still \(0.20\ \Omega\). However, writing \(4.84\pm0.20\ \Omega\) would give an interval of \(4.64\)-\(5.04\ \Omega\), which does not include the maximum gradient.

This is why you should check whether the best-fit gradient is close to the midpoint before reporting a symmetric uncertainty.

Step 5: Calculate the percentage uncertainty using the half-range convention.

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.20}{4.84}\times100\\ &\approx4.13\% \end{aligned} \]

The half-range percentage uncertainty is approximately \(4.1\%\).

The important lesson is that the half-range formula estimates the width of the plausible gradient range. It doesn’t guarantee that the best-fit value is exactly at its centre.

07What if the horizontal axis also has uncertainty?

The examples above assume that uncertainty in the horizontal measurements is negligible.

That isn’t always true.

If both horizontal and vertical measurements have significant uncertainty, each data point represents a rectangular region rather than just a vertical error bar.

A candidate line must pass through the uncertainty rectangle to be considered consistent with that measurement.

For a positive gradient:

• To increase the gradient, you generally want a larger vertical change and a smaller horizontal separation.
• To decrease the gradient, you generally want a smaller vertical change and a larger horizontal separation.

You must still check that the lines pass through the intermediate uncertainty rectangles.

Ignoring significant horizontal uncertainty can underestimate the uncertainty in the gradient.

If a question explicitly tells you to ignore horizontal uncertainty, use vertical error bars only. Otherwise, account for both axes.

For a more detailed treatment of drawing error bars and selecting extreme gradients, see Error Bars and Maximum-Minimum Gradients: HSC Physics Guide.

08Common mistakes when calculating gradient uncertainty

One more important point: if no measurement uncertainty or uncertainty convention is provided, you cannot calculate a defensible maximum-minimum gradient uncertainty. You need a basis for drawing the error bars first.

09Questions and solutions

Question 1

A student plots a position-time graph for a moving object.

The best-fit gradient is \(2.4\ \mathrm{m\,s^{-1}}\). The maximum acceptable gradient is \(2.6\ \mathrm{m\,s^{-1}}\), and the minimum acceptable gradient is \(2.2\ \mathrm{m\,s^{-1}}\).

Calculate:

a) The absolute uncertainty in the gradient.

b) The percentage uncertainty.

c) The final velocity, including its uncertainty.

Solution 1

The absolute uncertainty is \(0.20\ \mathrm{m\,s^{-1}}\), and the percentage uncertainty is approximately \(8.3\%\).

Step 1: Calculate the absolute uncertainty.

\[ \begin{aligned} \Delta m &=\frac{m_{\mathrm{high}}-m_{\mathrm{low}}}{2}\\ &=\frac{2.6-2.2}{2}\\ &=0.20\ \mathrm{m\,s^{-1}} \end{aligned} \]

Step 2: Calculate the percentage uncertainty.

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.20}{2.4}\times100\\ &=8.33\% \end{aligned} \]

Step 3: Report the final result.

\[ v=(2.4\pm0.2)\ \mathrm{m\,s^{-1}} \]

The velocity is estimated to be \(2.4\ \mathrm{m\,s^{-1}}\), with an absolute uncertainty of \(0.2\ \mathrm{m\,s^{-1}}\).

The half-range method works because the maximum and minimum gradients define the plausible range of slopes.

Question 2

A student investigates the acceleration of a trolley by plotting velocity against time.

The graph has vertical uncertainty bars, and uncertainty in time is negligible.

The following gradients are obtained:

• Best-fit gradient: \(0.80\ \mathrm{m\,s^{-2}}\).
• Candidate maximum line A: \(0.92\ \mathrm{m\,s^{-2}}\).
• Candidate maximum line B: \(0.89\ \mathrm{m\,s^{-2}}\).
• Minimum acceptable line: \(0.73\ \mathrm{m\,s^{-2}}\).

Line A passes through the first and last uncertainty bars but misses the third. Line B passes through all five uncertainty bars.

Calculate the uncertainty in acceleration and explain which maximum gradient should be used.

Solution 2

The correct maximum gradient is \(0.89\ \mathrm{m\,s^{-2}}\), because line B passes through all the uncertainty intervals. The absolute uncertainty is \(0.08\ \mathrm{m\,s^{-2}}\).

Step 1: Select the correct maximum gradient.

Line A has a larger numerical gradient, but it misses the third uncertainty bar.

It is therefore not consistent with all five measurements.

Line B is the steepest acceptable line and must be used.

\[ m_{\mathrm{high}}=0.89\ \mathrm{m\,s^{-2}} \]

Step 2: Calculate the uncertainty.

\[ \begin{aligned} \Delta m &=\frac{0.89-0.73}{2}\\ &=\frac{0.16}{2}\\ &=0.08\ \mathrm{m\,s^{-2}} \end{aligned} \]

Step 3: Report the acceleration.

\[ a=(0.80\pm0.08)\ \mathrm{m\,s^{-2}} \]

The percentage uncertainty is:

\[ \frac{0.08}{0.80}\times100=10\% \]

The tempting mistake is to choose the line with the largest gradient automatically. However, the line must also be consistent with the measurement intervals. A steep line that misses an intermediate error bar cannot be accepted.

Question 3

A student takes five repeated measurements of the force acting on an object at a fixed position.

The results are:

The mean force is \(1.04\ \mathrm{N}\).

The half-range uncertainty is \(0.04\ \mathrm{N}\), while the sample standard deviation is approximately \(0.032\ \mathrm{N}\).

The experimental instructions explicitly specify that half-range uncertainty must be used to draw the graph’s error bars.

a) Calculate the measurement interval that should be represented by the error bar.

b) Explain why the standard deviation should not be substituted.

c) What interval would be used if the instructions instead specified error bars of plus or minus one standard deviation?

Solution 3

The correct interval is \(1.00\)-\(1.08\ \mathrm{N}\). The half-range convention must be used because it is explicitly specified in the question.

Step 1: Write the measurement and uncertainty.

\[ F=(1.04\pm0.04)\ \mathrm{N} \]

Step 2: Calculate the lower and upper limits.

\[ F_{\mathrm{low}}=1.04-0.04=1.00\ \mathrm{N} \]

\[ F_{\mathrm{high}}=1.04+0.04=1.08\ \mathrm{N} \]

Therefore:

\[ 1.00\leq F\leq1.08\ \mathrm{N} \]

Step 3: Explain the difference between half-range and standard deviation.

The half-range is based on the difference between the largest and smallest measurements.

The standard deviation describes the typical spread of the individual readings around their mean.

They measure different aspects of the data. The standard deviation is not automatically the uncertainty required for the graph.

If the question had specified plus or minus one standard deviation, the interval would instead be:

\[ 1.04\pm0.032\ \mathrm{N} \]

Giving:

\[ 1.008\leq F\leq1.072\ \mathrm{N} \]

The error bars would be narrower because the standard deviation is smaller than the half-range.

The choice of convention changes the intervals, which can change the maximum and minimum gradients. Always follow the convention specified in the question.

Question 4

A student investigates the cooling of hot water by plotting temperature against time.

The best-fit gradient is:

\[ m_{\mathrm{best}}=-0.42\ ^\circ\mathrm{C\,min^{-1}} \]

The two extreme acceptable gradients are:

\[ -0.50\ ^\circ\mathrm{C\,min^{-1}} \]

and

\[ -0.36\ ^\circ\mathrm{C\,min^{-1}} \]

Calculate:

a) The maximum and minimum numerical gradients.

b) The absolute uncertainty in the gradient.

c) The percentage uncertainty.

d) The final gradient, including its uncertainty.

Solution 4

The absolute uncertainty is \(0.07\ ^\circ\mathrm{C\,min^{-1}}\), and the percentage uncertainty is approximately \(16.7\%\).

Step 1: Identify the maximum and minimum numerical gradients.

Remember that maximum and minimum refer to numerical values.

Since:

\[ -0.36>-0.50 \]

the maximum numerical gradient is:

\[ m_{\mathrm{high}}=-0.36\ ^\circ\mathrm{C\,min^{-1}} \]

The minimum numerical gradient is:

\[ m_{\mathrm{low}}=-0.50\ ^\circ\mathrm{C\,min^{-1}} \]

Notice that the minimum gradient is actually the steepest falling line.

Step 2: Calculate the absolute uncertainty.

\[ \begin{aligned} \Delta m &=\frac{m_{\mathrm{high}}-m_{\mathrm{low}}}{2}\\ &=\frac{-0.36-(-0.50)}{2}\\ &=\frac{0.14}{2}\\ &=0.07\ ^\circ\mathrm{C\,min^{-1}} \end{aligned} \]

Step 3: Calculate the percentage uncertainty.

Use the absolute value of the best-fit gradient:

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.07}{|-0.42|}\times100\\ &=16.7\% \end{aligned} \]

Step 4: Report the result.

\[ m=(-0.42\pm0.07)\ ^\circ\mathrm{C\,min^{-1}} \]

The negative sign must remain because temperature decreases as time increases.

If the question asks for the rate of cooling as a positive magnitude, report \(0.42\pm0.07\ ^\circ\mathrm{C\,min^{-1}}\).

The trap is assuming that the steepest line must be the maximum gradient. For negative gradients, the steepest line has the smallest numerical value.

Question 5

A student investigates a trolley’s motion using a position-time graph.

The measured best-fit gradient is:

\[ m_{\mathrm{best}}=0.800\ \mathrm{m\,s^{-1}} \]

The maximum and minimum gradients are:

\[ m_{\mathrm{high}}=0.830\ \mathrm{m\,s^{-1}} \]

\[ m_{\mathrm{low}}=0.770\ \mathrm{m\,s^{-1}} \]

The student then discovers two calibration issues:

• Every position reading is \(0.20\ \mathrm{m}\) too large because the position sensor has a fixed zero offset. The same offset applies to every reading.
• The data logger records time intervals that are \(2.0\%\) too long. The calibration factor is known exactly, so \(t_{\mathrm{measured}}=1.02t_{\mathrm{true}}\).

The student suggests adding \(0.20\ \mathrm{m}\) to every vertical error bar and adding \(2.0\%\) to the percentage uncertainty in the gradient.

Determine:

a) Whether the position zero offset affects the gradient.

b) Whether the time calibration issue requires a correction or an additional uncertainty.

c) The corrected best-fit velocity.

d) The corrected absolute uncertainty in the velocity.

e) Whether repeating the experiment removes either calibration issue.

Assume the original maximum and minimum gradients already account for all independent measurement uncertainties, and that the calibration factor is exact.

Solution 5

The corrected velocity is \(0.816\ \mathrm{m\,s^{-1}}\), with an absolute uncertainty of \(0.0306\ \mathrm{m\,s^{-1}}\). The student’s proposed treatment is incorrect because a common position offset does not change the gradient, while the known time-scale error requires a correction rather than an additional uncertainty.

Step 1: Consider the position zero offset.

Suppose every measured position is \(0.20\ \mathrm{m}\) too large.

This means:

\[ s_{\mathrm{measured}}=s_{\mathrm{true}}+0.20 \]

Consider two measurements:

\[ \Delta s_{\mathrm{measured}} = (s_{2,\mathrm{true}}+0.20) - (s_{1,\mathrm{true}}+0.20) \]

The common offset cancels:

\[ \Delta s_{\mathrm{measured}} = s_{2,\mathrm{true}}-s_{1,\mathrm{true}} \]

Therefore, the change in position is unaffected.

Since gradient depends on the change in position divided by the change in time, the position zero offset does not affect the gradient.

It shifts the entire graph vertically and changes its intercept, but leaves its slope unchanged.

Adding \(0.20\ \mathrm{m}\) to every error bar would be incorrect. The error bars represent uncertainty in the measurements, not the common offset itself.

Step 2: Consider the time calibration issue.

The data logger records time intervals that are \(2.0\%\) too long.

Therefore:

\[ t_{\mathrm{measured}}=1.02t_{\mathrm{true}} \]

The measured gradient is:

\[ m_{\mathrm{measured}} = \frac{\Delta s_{\mathrm{true}}} {1.02\Delta t_{\mathrm{true}}} \]

The true velocity is consequently:

\[ v_{\mathrm{true}}=1.02m_{\mathrm{measured}} \]

The measured gradient is too small because the recorded time interval is too large.

The calibration factor is known exactly, so this is a correction, not an additional uncertainty.

Step 3: Correct the best-fit gradient.

\[ \begin{aligned} v_{\mathrm{true}} &=1.02\times0.800\\ &=0.816\ \mathrm{m\,s^{-1}} \end{aligned} \]

The corrected best-fit velocity is:

\[ 0.816\ \mathrm{m\,s^{-1}} \]

Step 4: Correct the maximum and minimum gradients.

The same correction factor applies to both extreme gradients.

Maximum:

\[ m_{\mathrm{high,true}} =1.02\times0.830 =0.8466\ \mathrm{m\,s^{-1}} \]

Minimum:

\[ m_{\mathrm{low,true}} =1.02\times0.770 =0.7854\ \mathrm{m\,s^{-1}} \]

Step 5: Calculate the corrected absolute uncertainty.

\[ \begin{aligned} \Delta v &=\frac{0.8466-0.7854}{2}\\ &=\frac{0.0612}{2}\\ &=0.0306\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore, the corrected result is:

\[ v=(0.816\pm0.031)\ \mathrm{m\,s^{-1}} \]

Step 6: Consider what happens when the experiment is repeated.

Repeating measurements can reduce random variation and improve the precision of the experimental result.

However, repeating measurements does not automatically remove systematic calibration errors.

The common position offset still shifts every reading by the same amount, although it does not affect the gradient.

The time-scale error still makes every recorded time interval too large. More measurements won’t correct that bias.

The time calibration must be corrected independently.

This is the important distinction: an additive offset and a multiplicative scale error can affect a graph differently. One can change the intercept without changing the gradient, while the other changes the gradient itself.

10What to do with your result

Once you’ve calculated the uncertainty in a graph gradient, you have two pieces of information:

• The best-fit gradient, which gives your central experimental estimate.
• The uncertainty, which describes the range of gradients supported by your measurement intervals.

Your next step is to interpret that gradient physically and compare it with the prediction of the relevant model.

For example, a position-time graph gives velocity, a voltage-current graph can give resistance, and a force-extension graph can give a spring constant.

Don’t stop at reporting the numerical gradient. Check its units, its physical meaning, and whether its uncertainty makes the experimental result consistent with the theoretical prediction.