Why Units Matter in Motion
In a description of motion, a number alone is not enough. A velocity of \(12\) could mean \(12\) metres per second, \(12\) kilometres per hour, or something else. Those rates are not interchangeable. The units tell us what the number measures and how to interpret calculations involving position, velocity, and acceleration.
Units also help reveal errors. If position is measured in kilometres and time in minutes, then velocity must be measured in kilometres per minute, while acceleration must be measured in kilometres per minute squared. Converting these quantities requires attention to every distance and time factor in the unit.
Units of Position, Velocity, and Acceleration
Position describes a location relative to a chosen origin, so it has units of length, such as metres (\(\mathrm{m}\)), feet (\(\mathrm{ft}\)), or kilometres (\(\mathrm{km}\)). If position is represented by \(x(t)\), with \(t\) denoting time, then velocity is the derivative of position with respect to time. Its units are position units divided by time units.
Acceleration is the derivative of velocity with respect to time. Its units therefore have one more factor of time in the denominator. For example, if position is measured in metres and time in seconds, then velocity is measured in metres per second, and acceleration in metres per second squared.
The same pattern applies to other choices of units. With position in miles and time in hours, velocity is in miles per hour and acceleration is in miles per hour squared. The word “squared” applies to the time unit: \(\mathrm{mi}/\mathrm{h}^2\) means miles divided by hours squared.
Why Differentiation Produces These Units
The units of a derivative come from the difference quotient used to define it. For a change in time \(\Delta t\), the change in position \(x(t+\Delta t)-x(t)\) has the same units as position. Dividing this change by \(\Delta t\) gives position units divided by time units. Taking a limit as \(\Delta t\) approaches zero changes the numerical value into the derivative; it does not change the units of the quotient.
Applying the same reasoning to velocity, its change has velocity units, and dividing by a time change gives velocity per time. Since velocity already has distance per time units, acceleration has distance per time squared units. This explains why acceleration uses a squared time unit in the denominator.
Coefficients Must Carry Units Too
A motion formula must be consistent term by term: every term added to another term must have the same units. This includes the coefficients multiplying powers of time. For example, if \(t\) is measured in seconds and \(x(t)\) in metres, the coefficient of \(t^2\) in a position model must have units \(\mathrm{m}/\mathrm{s}^2\). Multiplying it by \(t^2\) then produces metres.
Writing units alongside coefficients may seem detailed, but it makes the meaning of a formula clear and helps prevent incorrect conversions. Consider the model in the first example.
Worked Example: Differentiate a Position Model with Units
A particle’s position is given by \(x(t)=(2\ \mathrm{m}/\mathrm{s}^2)t^2+(3\ \mathrm{m}/\mathrm{s})t+4\ \mathrm{m}\), where \(t\) is measured in seconds. Find position, velocity, and acceleration at \(t=2\ \mathrm{s}\).
First check the terms in the position formula. The quadratic term has units \((\mathrm{m}/\mathrm{s}^2)(\mathrm{s}^2)=\mathrm{m}\). The linear term has units \((\mathrm{m}/\mathrm{s})(\mathrm{s})=\mathrm{m}\), and the constant term is in metres. Thus every term represents position.
At \(t=2\ \mathrm{s}\), the position is \(x(2)=(2)(2^2)+(3)(2)+4=8+6+4=18\ \mathrm{m}\).
Differentiate the position model with respect to \(t\): \(x'(t)=2(2\ \mathrm{m}/\mathrm{s}^2)t+3\ \mathrm{m}/\mathrm{s}=(4\ \mathrm{m}/\mathrm{s}^2)t+3\ \mathrm{m}/\mathrm{s}\). Each term has units of velocity. At \(t=2\ \mathrm{s}\), \(x'(2)=(4)(2)+3=11\ \mathrm{m}/\mathrm{s}\).
Differentiate again: \(x''(t)=4\ \mathrm{m}/\mathrm{s}^2\). So the acceleration at \(t=2\ \mathrm{s}\) is \(4\ \mathrm{m}/\mathrm{s}^2\).
Check: Differentiating the stated position model gives the velocity above, and differentiating that velocity gives the stated constant acceleration. The units also agree: \((\mathrm{m}/\mathrm{s}^2)(\mathrm{s})=\mathrm{m}/\mathrm{s}\) in the velocity’s first term.
Converting Units of Velocity and Acceleration
To convert a rate, convert every unit in it. A velocity has one distance factor and one time factor. An acceleration has one distance factor and two time factors. Consequently, converting acceleration between different time units requires applying the time conversion twice.
For instance, \(1\ \mathrm{min}=60\ \mathrm{s}\), so \(1\ \mathrm{min}^2=(60\ \mathrm{s})^2=3600\ \mathrm{s}^2\). It is not \(60\ \mathrm{s}^2\). The distance conversion is applied once; the time conversion is applied as many times as the time unit appears in the denominator.
For example, \(\mathrm{km}/\mathrm{min}^2\) means kilometres divided by minutes squared.
Use equivalent quantities, such as \(1\ \mathrm{km}=1000\ \mathrm{m}\) and \(1\ \mathrm{min}=60\ \mathrm{s}\).
For acceleration, square the time conversion factor because time is squared in the denominator.
Keep the units visible through the arithmetic so that cancellations can be checked.
Worked Example: Convert a Model from Kilometres per Minute
A position model is \(x(t)=(0.02\ \mathrm{km}/\mathrm{min}^2)t^2+(0.30\ \mathrm{km}/\mathrm{min})t+1.2\ \mathrm{km}\), with \(t\) in minutes. Find position, velocity, and acceleration at \(t=10\ \mathrm{min}\), then express velocity in kilometres per hour and acceleration in metres per second squared.
Each term in the position formula has units of kilometres. At \(t=10\ \mathrm{min}\), \(x(10)=(0.02)(10^2)+(0.30)(10)+1.2=2+3+1.2=6.2\ \mathrm{km}\).
Differentiating gives \(x'(t)=(0.04\ \mathrm{km}/\mathrm{min}^2)t+0.30\ \mathrm{km}/\mathrm{min}\). At \(t=10\ \mathrm{min}\), this is \(x'(10)=(0.04)(10)+0.30=0.70\ \mathrm{km}/\mathrm{min}\).
To convert velocity to kilometres per hour, multiply by \(60\ \mathrm{min}/\mathrm{h}\): \(0.70\ \mathrm{km}/\mathrm{min}\times60\ \mathrm{min}/\mathrm{h}=42\ \mathrm{km}/\mathrm{h}\). The minute units cancel, leaving kilometres per hour.
Differentiating velocity gives \(x''(t)=0.04\ \mathrm{km}/\mathrm{min}^2\). Convert this to metres per second squared: \(0.04\ \mathrm{km}/\mathrm{min}^2\times1000\ \mathrm{m}/\mathrm{km}\times(1\ \mathrm{min}/60\ \mathrm{s})^2\). The calculation is \(40/3600=1/90\), so the acceleration is \(1/90\ \mathrm{m}/\mathrm{s}^2\).
Check: Reversing the velocity conversion gives \(42\ \mathrm{km}/\mathrm{h}\div60=0.70\ \mathrm{km}/\mathrm{min}\). Reversing the acceleration conversion gives \((1/90)\times3600/1000=0.04\ \mathrm{km}/\mathrm{min}^2\). Both recover the original rates.
A Conversion Involving Miles per Hour
Conversions can involve different distance units as well as different time units. The conversion factors must still be applied to the units in the rate, not just to its numerical value. For example, \(1\ \mathrm{mi}=5280\ \mathrm{ft}\) and \(1\ \mathrm{h}=3600\ \mathrm{s}\), so one mile per hour equals \(5280/3600=22/15\) feet per second.
Worked Example: Convert Velocity and Acceleration to Feet and Seconds
Suppose velocity is \(v(t)=30\ \mathrm{mi}/\mathrm{h}+(3\ \mathrm{mi}/\mathrm{h}/\mathrm{s})t\), where \(t\) is in seconds. Find the velocity at \(t=5\ \mathrm{s}\) in feet per second and the acceleration in feet per second squared.
Substitute \(t=5\ \mathrm{s}\) into the velocity model: \(v(5)=30+3(5)=45\ \mathrm{mi}/\mathrm{h}\). The product of the acceleration coefficient and time has units \((\mathrm{mi}/\mathrm{h}/\mathrm{s})(\mathrm{s})=\mathrm{mi}/\mathrm{h}\), as required.
Convert the velocity: \(45\ \mathrm{mi}/\mathrm{h}\times5280\ \mathrm{ft}/\mathrm{mi}\times1\ \mathrm{h}/3600\ \mathrm{s}\). The numerical calculation is \(45\times5280/3600=66\), so the velocity is \(66\ \mathrm{ft}/\mathrm{s}\).
Differentiate the velocity model with respect to time. Its derivative is the constant acceleration \(3\ \mathrm{mi}/\mathrm{h}/\mathrm{s}\). Since the hour is part of the velocity unit and the rate is per second, convert miles per hour to feet per second, retaining the additional “per second”: \(3\ (\mathrm{mi}/\mathrm{h})/\mathrm{s}\times5280\ \mathrm{ft}/\mathrm{mi}\times1\ \mathrm{h}/3600\ \mathrm{s}=4.4\ \mathrm{ft}/\mathrm{s}^2\).
Check: Converting \(66\ \mathrm{ft}/\mathrm{s}\) back gives \(66\times3600/5280=45\ \mathrm{mi}/\mathrm{h}\). Converting \(4.4\ \mathrm{ft}/\mathrm{s}^2\) back using \(3600/5280\) gives \(3\ (\mathrm{mi}/\mathrm{h})/\mathrm{s}\). Also, differentiating \(v(t)\) with respect to seconds gives its stated acceleration coefficient.
A Common Unit Error
A frequent mistake is to convert the distance unit correctly but apply the time conversion only once when converting acceleration. For example, someone converting \(2\ \mathrm{km}/\mathrm{min}^2\) to metres per second squared might calculate \(2\times1000/60\) and report \(33.3\ \mathrm{m}/\mathrm{s}^2\). This is incorrect because the original unit contains minutes squared.
The correction is to square the minute-to-second conversion factor. The incorrect calculation treated the denominator as though it contained just one minute; it therefore failed to produce the correct conversion for acceleration.
Why Consistent Units Are Useful
Consistent units make motion descriptions interpretable and calculations checkable. They help distinguish a position value from a rate, and a velocity rate from an acceleration rate. They also show how a numerical value changes when the units change: a velocity expressed in metres per second will generally have a different number from the same velocity expressed in kilometres per hour.
Before using a motion formula, check that each added term has the same units. Before converting a rate, write the conversion factors so units cancel visibly. For acceleration, count two time factors in the denominator. These checks do not replace the mathematical calculation, but they can catch mistakes before they affect the interpretation.
Check Your Understanding
Use the units as well as the numerical values when answering these questions.
- If position is measured in feet and time in seconds, what are the units of velocity and acceleration?
- A position model uses time in minutes and position in kilometres. What units must the coefficient of \(t^2\) have?
- Convert \(1.5\ \mathrm{km}/\mathrm{min}\) to kilometres per hour.
- Convert \(0.5\ \mathrm{km}/\mathrm{min}^2\) to metres per second squared. Which conversion factor must be squared?
- Why must all terms being added in a position formula have the same units?
