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A1.13 · Report calculations with suitable precision and significant figures

Learn to report calculations with suitable precision and significant figures through clear examples and targeted practice.

Ontario Grade 11 Physics

Scientific Investigation Skills and Career Exploration

Ontario Grade 11 Physics — A1.13

Physics calculations use measured values, and measurements have limits. A calculator can display many digits, but those digits are not automatically meaningful. Suitable precision means reporting a result in a way that matches the precision of the information used to calculate it. This lesson focuses on how to choose and present those digits. It does not change the physics relationship being used; it helps you report its result responsibly.

What you will learn

1. Start with what a measurement tells you

A measurement is a number paired with a unit. For example, a measured length might be recorded as 2.4 m2.4\ \mathrm{m}. The unit tells what kind of quantity was measured. The digits tell the recorded value and its precision.
Precision describes how finely a measurement is recorded. A value written as 2.4 m2.4\ \mathrm{m} is recorded to the nearest tenth of a metre. A value written as 2.40 m2.40\ \mathrm{m} is recorded to the nearest hundredth. The extra zero is meaningful: it communicates how the measurement was recorded. Precision does not by itself tell whether a measurement is close to the true value.
Significant figures are the meaningful digits in a recorded value. Non-zero digits count. Zeros between non-zero digits count. Zeros at the start do not count; they only locate the decimal point. A final zero after a decimal point counts. Scientific notation can make the intended number of significant figures clear: 3.0×102 m3.0\times10^2\ \mathrm{m} has two significant figures.
A scalar has magnitude only, such as time or distance. A vector has magnitude and direction, such as displacement or velocity. Significant figures affect the reported magnitude. For a vector, report its direction as well when the physics quantity requires one.

2. Choose the rounding rule from the operation

First calculate using the values given. Keep extra digits in the calculator while working, and round the final reported value only. Rounding too early can affect the last digit of the answer.
For multiplication or division, the result is reported with the same number of significant figures as the input value with the fewest significant figures. For example, if one measured value has two significant figures and another has three, the calculated product or quotient is normally reported with two significant figures.
For addition or subtraction, count decimal places instead of significant figures. The result is rounded to the same decimal place as the least precise value in the calculation. For instance, if one value is recorded to the nearest tenth and another to the nearest hundredth, their sum is reported to the nearest tenth.
These rules reflect the way the original values were recorded. Do not use the multiplication rule for addition, or the addition rule for division.
An exact number is not a measured value with limited precision. A defined conversion, such as 1 min=60 s1\ \mathrm{min}=60\ \mathrm{s}, is exact. A counted number of objects can also be exact. Exact numbers do not limit the significant figures of a calculated result. reported precision\text{reported precision} = precision allowed by the measured inputs

3. Report a complete physics result

Before calculating, identify the system, known values, unknown quantity, and direction convention if direction matters. The system is the object or objects being considered. A positive direction is the direction chosen to represent positive values; state it before using signed vector quantities.
Use the physics relationship that connects the known values to the unknown. Substitute values with their units. Check that the units fit the quantity being calculated, then round to suitable precision. A final report should include the value, its unit, and a direction when the quantity is a vector.
A sensible precision check asks whether the answer claims more detail than the measurements provide. A units check asks whether the result has the expected unit. A reasonableness check asks whether the size and direction fit the described situation. These checks do not replace the calculation; they help catch reporting and setup errors.
v=ΔdΔtv=\frac{\Delta d}{\Delta t}

Worked example

A speed from measured distance and time

A cart travels a measured distance of 3.42 m3.42\ \mathrm{m} in 1.6 s1.6\ \mathrm{s}. Find its average speed and report it suitably.
  1. Define the quantities
    The system is the cart. Distance and time are scalars, so no direction is needed for average speed. The known values are 3.42 m3.42\ \mathrm{m} and 1.6 s1.6\ \mathrm{s}; the unknown is average speed, vv.
  2. Choose the relationship
    Average speed is distance divided by elapsed time. Since this is division, the final result must have as many significant figures as the input with the fewest. The distance has three and the time has two.
    v=ΔdΔtv=\frac{\Delta d}{\Delta t}
  3. Substitute and calculate
    Substitute both measured values with their units. Keep the calculator digits until the final rounding step.
    v=3.42 m1.6 s=2.1375 m/sv=\frac{3.42\ \mathrm{m}}{1.6\ \mathrm{s}}=2.1375\ \mathrm{m/s}
  4. Round and check
    Two significant figures are allowed, so report 2.1 m/s2.1\ \mathrm{m/s}. The units reduce to metres per second, which is a speed unit. A speed near two metres per second is reasonable for a cart covering a few metres in a little over a second.
    2.1375 m/s≈2.1 m/s2.1375\ \mathrm{m/s}\approx2.1\ \mathrm{m/s}
Answer: The cart's average speed is 2.1 m/s2.1\ \mathrm{m/s}.
Check: The answer has two significant figures, matching the least precise input, and its unit is speed.

Worked example

A change in position with direction

A student records an initial position of 1.25 m1.25\ \mathrm{m} and a final position of 4.6 m4.6\ \mathrm{m} along a straight track. Find the displacement.
  1. Define the system and direction
    The system is the student. Choose the track direction from the initial position toward the final position as positive. Position and displacement are vectors along this track; a positive result points in the chosen positive direction. The unknown is displacement, Δd\Delta d.
  2. Use the position change
    Displacement is final position minus initial position. This is subtraction, so round the result to the least precise decimal place. The final position is recorded to the nearest tenth of a metre, while the initial position is recorded to the nearest hundredth.
    Δd=df−di\Delta d=d_f-d_i
  3. Substitute and calculate
    Substitute the positions with their units. The unrounded difference is retained until the reporting step.
    Δd=4.6 m−1.25 m=3.35 m\Delta d=4.6\ \mathrm{m}-1.25\ \mathrm{m}=3.35\ \mathrm{m}
  4. Round and check
    The least precise position is recorded to the nearest tenth, so report 3.4 m3.4\ \mathrm{m}. The positive sign means the displacement is toward the chosen positive direction. The unit is metres, and the student ends a few metres from the starting position, which fits the given positions.
    3.35 m≈+3.4 m3.35\ \mathrm{m}\approx+3.4\ \mathrm{m}
Answer: The student's displacement is +3.4 m+3.4\ \mathrm{m}, in the chosen positive direction.
Check: The subtraction result is rounded to one decimal place, and its sign agrees with the stated direction convention.

Worked example

Combining measured times

A signal takes 0.084 s0.084\ \mathrm{s} to travel through one section of a model and 0.12 s0.12\ \mathrm{s} through a second section. Find the total time.
  1. Define the system and unknown
    The system is the signal's travel through the two sections. Time is a scalar, so direction is not needed. The known times are 0.084 s0.084\ \mathrm{s} and 0.12 s0.12\ \mathrm{s}; the unknown is total time.
  2. Choose the operation and precision rule
    The total time is the sum of the section times. For addition, use decimal places: 0.084 s0.084\ \mathrm{s} has three decimal places, and 0.12 s0.12\ \mathrm{s} has two. The reported sum must be to two decimal places.
    ttotal=t1+t2t_{\mathrm{total}}=t_1+t_2
  3. Substitute and calculate
    Add the times while keeping seconds attached. The calculator result has three decimal places, but that does not mean all three can be reported.
    ttotal=0.084 s+0.12 s=0.204 st_{\mathrm{total}}=0.084\ \mathrm{s}+0.12\ \mathrm{s}=0.204\ \mathrm{s}
  4. Round and check
    Round to the hundredths place to get 0.20 s0.20\ \mathrm{s}. The zero after the decimal point is important: it shows the value is reported to the hundredths place. The unit remains seconds, and a total near two tenths of a second is reasonable.
    0.204 s≈0.20 s0.204\ \mathrm{s}\approx0.20\ \mathrm{s}
Answer: The total time is 0.20 s0.20\ \mathrm{s}.
Check: The sum is reported to two decimal places, as required by the less precise time.

Common mistakes and how to avoid them

Reporting every digit shown on the calculator.
Correction: Keep extra digits while calculating, then round the final answer using the precision of the measured inputs.
Using significant figures for addition and subtraction.
Correction: For addition and subtraction, compare decimal places. Use significant figures for multiplication and division.
Dropping a final zero that communicates precision.
Correction: Keep zeros that show how a value was recorded. For example, 0.20 s0.20\ \mathrm{s} communicates precision to the hundredths place.
Giving a vector's magnitude without its direction.
Correction: State the chosen positive direction and include the direction of the result, either with a sign tied to that convention or with words.
Leaving units off the final result.
Correction: Include the appropriate unit so the reader knows what physical quantity the number represents.

Lesson summary

Check your understanding

Question 1

Calculate 5.6 m/2.34 s5.6\ \mathrm{m}/2.34\ \mathrm{s}. Which reported value is suitable?
  1. 2.393 m/s2.393\ \mathrm{m/s}
  2. 2.4 m/s2.4\ \mathrm{m/s}
  3. 2.39 m/s2.39\ \mathrm{m/s}
  4. correctIndex": 1, "explanation": "Division uses the fewest significant figures. The numerator has two significant figures, so 5.6/2.34=2.393…5.6/2.34=2.393\ldots is reported as 2.4 m/s2.4\ \mathrm{m/s}, with two significant figures and the correct speed unit."
Show answer and explanation
2.4 m/s2.4\ \mathrm{m/s}
Division uses the fewest significant figures. The numerator has two significant figures, so 5.6/2.34=2.393…5.6/2.34=2.393\ldots is reported as 2.4 m/s2.4\ \mathrm{m/s}, with two significant figures and the correct speed unit.

Question 2

What is the suitable reported sum of 1.36 s+0.4 s1.36\ \mathrm{s}+0.4\ \mathrm{s}?
  1. 1.76 s1.76\ \mathrm{s}
  2. 1.8 s1.8\ \mathrm{s}
  3. 2 s2\ \mathrm{s}
  4. correctIndex": 1, "explanation": "The sum is 1.76 s1.76\ \mathrm{s}. Addition is rounded to the fewest decimal places: 0.4 s0.4\ \mathrm{s} is to the tenths place, so the result is 1.8 s1.8\ \mathrm{s}."
Show answer and explanation
1.8 s1.8\ \mathrm{s}
The sum is 1.76 s1.76\ \mathrm{s}. Addition is rounded to the fewest decimal places: 0.4 s0.4\ \mathrm{s} is to the tenths place, so the result is 1.8 s1.8\ \mathrm{s}.

Question 3

A position change is calculated as 2.3 m−3.1 m2.3\ \mathrm{m}-3.1\ \mathrm{m}. If right is positive, what is the suitable displacement?
  1. +0.8 m+0.8\ \mathrm{m} to the right
  2. −0.8 m-0.8\ \mathrm{m}, meaning left
  3. −0.80 m-0.80\ \mathrm{m}, meaning left
  4. correctIndex": 1, "explanation": "The difference is −0.8 m-0.8\ \mathrm{m}. The negative sign means left because right was defined as positive. Both values are recorded to the tenths place, so the result is also reported to the tenths place."
Show answer and explanation
−0.8 m-0.8\ \mathrm{m}, meaning left
The difference is −0.8 m-0.8\ \mathrm{m}. The negative sign means left because right was defined as positive. Both values are recorded to the tenths place, so the result is also reported to the tenths place.

Key terms

Precision
How finely a measurement is recorded.
Significant figures
The meaningful digits in a recorded value, including digits that communicate its precision.
Scalar
A quantity with magnitude only, such as time or distance.
Vector
A quantity with both magnitude and direction, such as displacement.
Positive direction
The chosen direction represented by positive values in a one-dimensional situation.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation A1.13. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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