What is a formula, and how do you read one?

Imagine a trolley that travels two metres every second. Two metres in the first second, another two in the second, another two in the third. To find the total distance travelled, we can add three equal sections. Or we can write a short relationship that works for three seconds or five seconds. That is what a formula helps us do.

A physical formula uses symbols and mathematical operations to express a relationship between defined quantities. Reading it means finding out what the symbols represent, how the quantities relate and when the relationship applies. Only then do we substitute numbers together with their units. NIST: quantity equations, §7.11.

Put the relationship into words first

Let us make our imagined situation precise: the trolley moves straight at constant speed relative to the ground. It has finished speeding up and has not yet begun to brake. It travels equal distances in equal time intervals. We can therefore find the total distance by multiplying its speed by the elapsed time.

“Two metres per second” means that each second adds two metres to the distance travelled. Three seconds give us three such contributions: two plus two plus two — six metres. Multiplication is a shorter way of adding these equal parts.

We have specified a model, not measured a real trolley. For now the numbers help us see the relationship. Once we understand it in words, we can give the quantities short names.

Three letters replace three longer names

Let $d$ stand for distance travelled, $v$ for constant speed, and $t$ for the duration of the motion we are considering. Then:

d=vt.

Read it as “distance equals speed multiplied by time.” The two adjacent letters $vt$ mean the product $v\times t$ , not one long name “vt.” The equals sign says that the quantities on its two sides have the same value. It does not mean “you must now calculate something.”

A letter does not explain its own meaning. Another text might use $s$ or $l$ for distance, and give $t$ a different role. The definitions beside a formula are part of the explanation, not a footnote for specialists. Nor does a letter prescribe a unit: distance may be expressed in metres or kilometres. BIPM: quantity symbols and products, §§5.4.1, 5.4.6.

Read the formula in the figure, then find each symbol in the labels. The three equal strips show why multiplication matches our situation.

Three seconds — three equal sections of the path

d = v × t

d · Distance travelled: 6 m
v · Constant speed: 2 m/s
t · Elapsed time: 3 s

2 min 1 s

Total: 6 m

Teaching model: a trolley moves straight at constant speed relative to the ground. Each strip represents 2 m travelled in 1 s; equal strip lengths mean equal distances. These are specified conditions, not measurements. Acceleration and stopping occur outside this interval.

Substitute values with their units

This example uses metres and seconds: $v=2$ m/s and $t=3$ s. The unit symbol “m/s” means metres per second; the slash denotes division. Substituting both values gives:

\begin{aligned} d&=(2\ \text{m/s})\times(3\ \text{s})\\ &=6\ \text{m}. \end{aligned}

The seconds in the denominator of the speed unit cancel with the seconds of elapsed time. Metres remain — the unit of distance. This is a useful check: dividing speed by time instead of multiplying would give m/s², which is not a unit of length.

Units cannot be attached at random after doing the arithmetic. Express the same speed as 120 metres per minute, and three seconds must become 0.05 minutes. Multiplying 120 by 0.05 again gives 6 metres. Multiplying the bare numbers 120 and 3 would lose the distinction between seconds and minutes.

The right unit does not guarantee the right formula. “Twice the speed multiplied by time” would also have a unit of length, but would give the wrong distance for our trolley. Check both the units and the meaning.

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One equality can answer different questions

Suppose we know that the trolley travelled 6 m in 3 s. What was its constant speed? An equality works in both directions. Dividing both sides by a nonzero duration gives:

\begin{aligned} v&=\frac{d}{t}\\ &=\frac{6\ \text{m}}{3\ \text{s}}\\ &=2\ \text{m/s}. \end{aligned}

If we instead want the time needed to cover 6 m at a constant 2 m/s, divide distance by speed:

\begin{aligned} t&=\frac{d}{v}\\ &=\frac{6\ \text{m}}{2\ \text{m/s}}\\ &=3\ \text{s}. \end{aligned}

This division requires $v>0$ . At zero speed, a nonzero distance cannot be covered in any finite time in this model; we cannot divide by zero. Rearranging an equality adds no new physical law. It lets us ask a different question about the same relationship.

Check that this speed describes the whole interval

Imagine that an instrument showed 2 m/s only at the beginning, and the trolley then slowed down or stopped. Can we automatically multiply that reading by all three seconds? No. One reading describes the speed at a particular instant. Our initial calculation assumed that the same speed continued throughout the interval.

For motion with changing speed, dividing the total distance travelled by the total elapsed time gives the average speed. If that average is known for the interval we need, multiplying it by the duration also gives the total distance. An arbitrary instantaneous reading cannot replace it. OpenStax: average and instantaneous speed.

We also defined $d$ here as the length of the travelled path. If the trolley returns to the start, that distance does not disappear. Its displacement — the change from its initial position to its final position — becomes zero. Displacement is a different quantity and includes direction. Replacing distance with displacement changes the meaning of the formula. OpenStax: distance and displacement.

A formula can describe a relationship without explaining its cause. Writing $d=vt$ lets us find distance for a specified constant speed, but does not tell us what interactions could change the trolley’s motion. To explain acceleration, we need other information and a different model.

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A product, a subscript and a square mean different things

A small mark beside a letter can completely change how we read it:

$vt$ is the product of speed and time.
$v_1$ is speed with the subscript “one.” Let us define it as the speed in the first section. The subscript is a label, not multiplication by one.
$v^2$ is speed squared: $v\times v$ , not $2v$ .

For example, if $v=3$ m/s, then $2v=6$ m/s, whereas $v^2=9$ m²/s². Both the number and the unit change. Look up the author’s definition of a subscript; not every superscript in science denotes a power. NIST: symbols, subscripts and superscripts.

There is another distinction: $=$ means equality, while $\approx$ means approximate equality. In our specified example, 2 multiplied by 3 is exactly 6. In a real measurement, speed and time have uncertainties. Writing many digits would not make the prediction accurate to the last one.

Reading a formula includes checking its use

Try answering without another calculation: how will our trolley’s distance change if the same constant speed continues for twice as long? It doubles: each extra second contributes the same amount. Six seconds give 12 metres. At zero elapsed time, the distance in this model is zero.

A second check: the trolley travelled two metres out and two metres back to the start. Is its distance travelled zero? No: it travelled four metres, and its displacement is zero. This conclusion does not require pretending that speed stayed constant through the turnaround.

Before using an unfamiliar formula, take four steps: name the quantities in words, read the operations between them, make the units consistent, and check the model’s conditions. The numbers then have a physical meaning, and you can check the result.

We began with three equal sections of a path and obtained a short expression for a whole family of possible situations. A formula helps us reason about a relationship before choosing the numbers. For practice, change the duration to five seconds and explain why the result is ten metres. Later in the course, this way of reading will help us estimate space travel times, paying particular attention to when the constant-speed assumption is appropriate.