Describing Motion Around Us

A flitting butterfly, a slithering snake, a galloping horse, dust dancing in a sunbeam, tides rising and falling, clouds gathering before a storm — everything in nature is in motion, and the variety is staggering. Faced with such richness, how does science even begin? The way it always begins with a complicated phenomenon: study it first in its simplest, idealised form. This chapter builds that simple toolkit for the two cleanest kinds of motion you can picture — motion in a straight line and uniform circular motion — and gives you a precise vocabulary for something you already do informally every day: describing how something moves. You already know distance, time and speed from earlier grades. This chapter sharpens that toolkit with four new physical quantities — displacement, average velocity, average acceleration, and the graphs that picture all of them — and shows that distance and speed alone are not enough once direction starts to matter. A runner who ends up back where she started has covered real distance but zero displacement; a car cruising at a rock-steady speed around a bend is, surprisingly, accelerating. Untangling exactly why these things are true, and building the equations and graphs that let you calculate with them, is the work of this chapter. By the end you will read a position-time or velocity-time graph the way you read a sentence — seeing at a glance whether an object is at rest, moving steadily, speeding up, or slowing down — and use three compact kinematic equations to predict where a moving object will be, and how fast, at any future instant, as long as its acceleration stays constant. The chapter closes by lifting motion off the straight line and onto a circle, the second idealised shape nature loves, from a child on a merry-go-round to a satellite orbiting the Earth.

Key takeaways

Position needs a reference point. Distance and direction from a chosen origin fix an object's position; if that position changes with time, it is in motion.
Distance and displacement differ. Distance travelled is a scalar; displacement is the net change in position — a vector — and its magnitude can never exceed distance travelled.
Average speed and average velocity can disagree sharply. Speed uses total distance; velocity uses displacement, so it can be zero even when the object never stopped, as for a swimmer who returns to the start.
Average acceleration is the rate of change of velocity, $a = (v - u)/(t_2 - t_1)$ , needing a direction — with velocity if speed rises, against it if speed falls.
A straight-line position-time graph means constant velocity; a curved one means changing velocity. Its slope gives velocity.
On a velocity-time graph, slope gives acceleration and the enclosed area gives displacement — one graph, two quantities.
Three kinematic equations connect five quantities ( $s$ , $t$ , $u$ , $v$ , $a$ ) whenever acceleration is constant: $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , and $v^2 = u^2 + 2as$ .
Uniform circular motion is constant speed around a circle, and it is always accelerated, because the direction of velocity keeps changing even while its magnitude stays fixed.
A tangent touches a circle at exactly one point; the velocity of an object in circular motion always points along that tangent, in the direction of travel.
Key idea: motion is described completely once position is tracked from a reference point over time using the vector pair displacement and velocity, whose rates of change define acceleration, all readable as slope and area on position-time and velocity-time graphs, reducing to three constant-acceleration kinematic equations in a straight line and to an always-accelerated picture for uniform circular motion.

The core insight

Describing motion precisely means tracking an object's position from a fixed reference point over time using vector quantities — displacement, average velocity, average acceleration — whose values and rates of change appear directly as the slope and enclosed area of position-time and velocity-time graphs, are captured algebraically by three kinematic equations for constant acceleration in a straight line, and extend naturally to uniform circular motion, whose constantly changing direction of velocity makes it accelerated motion even though its speed never changes.

What this chapter covers

The chapter has four connected parts. It opens with motion in a straight line: fixing position with a reference point and sign convention, then building distance versus displacement and average speed versus average velocity, then average acceleration. It turns to graphs: a position-time graph's slope is velocity, and a velocity-time graph's slope is acceleration while its enclosed area is displacement. Third come the three kinematic equations, finding any one of displacement, time, initial velocity, final velocity or acceleration from the other four, whenever acceleration is constant. It ends with motion in a plane, on uniform circular motion and the surprising fact that constant speed on a curved path still means acceleration.

The core ideas

Position, reference point, and what "in motion" means

To describe where something is, you first need a reference point (or origin) — a fixed point you choose to measure from. The position of an object at any instant is given by its distance and direction from that reference point. An object is in motion if its position relative to the reference point changes with time, and at rest if it does not. For motion on a straight line, direction has only two possibilities, so it is shown with a plus or minus sign: positions on one side of the origin are positive, on the other, negative — exactly as on a number line. Keep two related ideas separate: an instant of time is a single clock reading, while a time interval is the duration between two instants. Almost every quantity in this chapter — displacement, average velocity, average acceleration — is defined over a time interval, not at a single instant.

Distance travelled and displacement

Once an object moves, two different questions can be asked about "how far": how much ground did it actually cover, and how far — and in which direction — did it end up from where it started?

Total distance travelled is the actual length of the path covered, regardless of direction — a scalar, needing only a numerical value (with units). Displacement is the net change in position between two instants — the straight-line distance from the starting position to the final position, plus the direction from the first toward the second. A quantity needing both a numerical value (its magnitude) and a direction is a vector; one needing only a numerical value, like distance, is a scalar. Both distance and displacement share the SI unit of length, the metre (m). Picture an athlete starting at reference point $O$ on a straight track, running to point $A$ , $100\ \text{m}$ away, and then jogging back to point $B$ , $60\ \text{m}$ from $A$ (so $B$ is $40\ \text{m}$ from $O$ ). The total distance travelled is $OA + AB = 100\ \text{m} + 60\ \text{m} = 160\ \text{m}$ . But her displacement — the net change in position from $O$ to her final position at $B$ — is only $40\ \text{m}$ , in the direction from $O$ toward $A$ . The two numbers, $160\ \text{m}$ and $40\ \text{m}$ , are simply not the same quantity, and confusing them is the single most common slip in this whole chapter.

Figure

A number line showing an athlete starting at reference point O, running out to point A one hundred metres away, then back to point B forty metres from O, with the outward and return legs of the actual path marked above the line summing to a total distance of one hundred sixty metres, and a single straight arrow below the line from O to B marking the net displacement of forty metres

One special case is worth remembering: for straight-line motion that never turns back, distance travelled and the magnitude of displacement are equal. As soon as the object reverses direction even once, the two pull apart, and displacement's magnitude becomes strictly less than distance travelled.

Average speed and average velocity

You already met average speed: the total distance travelled divided by the time interval taken.

\text{average speed} = \frac{\text{total distance travelled}}{\text{time interval}}. \qquad (1)

where: distance travelled has no direction, so average speed also has no direction — a scalar, with SI unit metre per second (m/s), also commonly given in kilometre per hour (km/h). If an object covers equal distances in every equal interval of time, it is in uniform motion; if it covers unequal distances, it is in non-uniform motion — speeding up, slowing down, or both. Speed alone says nothing about direction. Average velocity supplies it: displacement divided by the time interval in which it occurs.

v_{av} = \frac{\text{displacement}}{\text{time interval}} = \frac{s}{\Delta t}. \qquad (2)

where: $v_{av}$ is average velocity, $s$ is displacement, and $\Delta t$ is the time interval. Because displacement is a vector, average velocity is one too, sharing its direction and shown, in straight-line motion, with a plus or minus sign. It is equivalently the average rate of change of position with time. Here the split becomes sharp. A swimmer who takes $50\ \text{s}$ to swim a $25\ \text{m}$ pool's length and straight back travels $25+25=50\ \text{m}$ , giving average speed $50/50 = 1\ \text{m/s}$ . But his displacement is zero, since he ends where he began — so his average velocity is $0\ \text{m/s}$ , even though he swam hard the whole time. Average speed and the magnitude of average velocity are equal only when motion never reverses direction.

Average acceleration

Velocity, like speed, can stay constant or change with time. The quantity that measures how fast velocity itself is changing is average acceleration: the change in velocity of an object over a time interval, divided by that interval.

a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time interval}} = \frac{v - u}{t_2 - t_1}. \qquad (3)

where: $u$ is the velocity at the earlier instant $t_1$ , $v$ at the later instant $t_2$ , and $a$ is the average acceleration over that interval, in m/s $^2$ . Acceleration needs a direction too: if velocity's magnitude is increasing, acceleration acts along it; if decreasing, opposite to it. It can arise from a change in magnitude, a change in direction, or both — the fact that matters most in uniform circular motion below. Acceleration measures how quickly velocity changes, not how large it is. A car cruising a highway at a rock-steady $100\ \text{km/h}$ has zero acceleration, however fast that feels. One everyday constant-acceleration example worth remembering: an object falling freely near Earth's surface (ignoring air resistance) speeds up by very nearly $9.8\ \text{m/s}$ every second it falls — the acceleration due to gravity, $g \approx 9.8\ \text{m/s}^2$ , directed downward, the same for every falling object regardless of mass.

Reading motion from a position-time graph

A position-time graph plots position (vertical axis) against time (horizontal axis). A straight-line graph means constant velocity — equal displacements in equal time intervals. A curved graph means changing velocity — non-uniform, accelerated motion. A line parallel to the time axis (zero slope) means position is not changing — the object is at rest. The graph also gives velocity directly, through its slope. For two points at times $t_1, t_2$ with positions $s_1, s_2$ ,

v = \frac{s_2 - s_1}{t_2 - t_1}, \qquad (4)

where: the numerator is the change in position between the two instants and the denominator is the time interval — Equation (2) again, read as "rise over run." A steeper line means larger velocity; comparing two graphs on the same axes, the steeper line belongs to the faster object.

Reading motion from a velocity-time graph, and reading area as displacement

A velocity-time graph plots velocity against time, yielding two quantities from two different features of one picture. The slope gives the acceleration:

a = \frac{v - u}{t_2 - t_1}, \qquad (5)

where: $u$ and $v$ are the velocities read off the graph at the earlier and later instants. A horizontal line (zero slope) means zero acceleration. A line sloping upward means velocity is increasing with constant acceleration, along the velocity's direction; sloping downward means velocity is decreasing, opposite to it. The second feature: the area enclosed between the graph and the time axis, over any interval, equals the displacement in that interval. For constant velocity, this area is a rectangle, velocity times time — Equation (2) rearranged. For constant non-zero acceleration, the shape is a trapezium (rectangle plus triangle), and adding the two areas gives displacement even while velocity keeps changing.

Figure

A velocity-time graph for a bus that accelerates from ten to fifteen metres per second over ten seconds, then cruises at fifteen metres per second for fifteen seconds, then brakes to a stop in five seconds, with the slope of each straight segment labelled as the acceleration in that phase and the shaded area under each segment labelled as the displacement in that phase

The three kinematic equations for constant acceleration

For the special but very common case of motion in a straight line with constant acceleration, the five quantities displacement ( $s$ ), time interval ( $t$ ), initial velocity ( $u$ ), final velocity ( $v$ ) and acceleration ( $a$ ) are tied together by three equations, called the kinematic equations:

v = u + at, \qquad s = ut + \frac{1}{2}at^2, \qquad v^2 = u^2 + 2as. \qquad (6)

where: $u$ is the velocity at the start of the interval ( $t=0$ ), $v$ at the end (time $t$ later), and $a$ is the constant acceleration throughout. These equations find any one of the five quantities once the other three are known, without needing a graph at every step. They hold only while acceleration truly stays constant — built from the velocity-time graph in the first derivation below.

Motion in a plane and uniform circular motion

Motion is not always confined to a straight line. A vehicle overtaking another, a kicked ball's curving path, or a satellite looping the Earth are all motion in a plane. The chapter's one detailed case is circular motion: motion along a circular path. Picture a child on a merry-go-round moving from point $A$ to $B$ to $C$ along the rim. The distance travelled is the length of arc $ABC$ ; the displacement is the straight chord from $A$ to $C$ — two different numbers. Over one complete revolution around a circle of radius $R$ , distance travelled equals the circumference, $2\pi R$ , while displacement is exactly zero, since the child returns to the start. If one revolution takes time $T$ , the average speed over that revolution is

v_{av} = \frac{2\pi R}{T}, \qquad (7)

where: $R$ is the radius and $T$ is the time for one revolution (the period). When speed stays exactly constant, this is called uniform circular motion. Because speed is constant everywhere on the circle here, Equation (7) also gives the speed at any point, not merely the lap average. Here is the surprising part: even though speed never changes in uniform circular motion, the object is still accelerating, because velocity is a vector whose direction keeps changing continuously — always along the tangent to the circle (a straight line touching it at exactly one point) in the direction of travel. Since a vector changes if either its magnitude or direction changes, a continuously turning velocity means continuously non-zero acceleration, even at fixed speed — worked through with numbers in the second derivation below. Uniform circular motion is an idealised model, but a foundational one, underlying everything from a vehicle taking a curve to a planet orbiting the Sun.

Derivation ADeriving the three kinematic equations of motion for constant acceleration

The goal is to build all three kinematic equations, $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , and $v^2 = u^2 + 2as$ , from nothing but the definition of average acceleration and the area-under-a-velocity-time-graph rule already established — no calculus, only algebra and the geometry of a trapezium. Because the acceleration is constant, its value at every instant equals its average over any interval, which is what allows one straight line on the velocity-time graph to represent the whole motion.

Step 1. Start from the definition of acceleration. With initial velocity $u$ at time $0$ and final velocity $v$ at time $t$ , Equation (3) gives $a = (v-u)/t$ . Rearranging, $at = v - u$ , so $v = u + at$ — the first kinematic equation, giving velocity at any later time from the initial velocity and the constant acceleration.

Step 2. Draw the velocity-time graph: a straight line from height $u$ at $t=0$ to height $v$ at time $t$ . Displacement is the area between this line and the time axis — a trapezium, splitting into a rectangle of height $u$ , width $t$ , plus a triangle on top with base $t$ , height $(v-u)$ .

Step 3. Add the areas: rectangle $u \times t$ , triangle $\tfrac{1}{2} \times t \times (v-u)$ , so $s = ut + \tfrac{1}{2}t(v-u)$ .

Step 4. Substitute $(v-u) = at$ from Step 1: $s = ut + \tfrac{1}{2}t(at) = ut + \tfrac{1}{2}at^2$ — the second kinematic equation.

Step 5. Eliminate $t$ to get a third equation needing no time. From Step 1, $t=(v-u)/a$ ; substituting into $s = ut+\tfrac12at^2$ and simplifying over the common denominator $2a$ gives $s = \dfrac{2uv-2u^2+v^2-2uv+u^2}{2a} = \dfrac{v^2-u^2}{2a}$ , which rearranges to $v^2 = u^2 + 2as$ . Verified: taking $u = 15\ \text{m/s}$ , $a = -4\ \text{m/s}^2$ , $v = 0$ gives $s = (v^2-u^2)/2a = (0 - 225)/(-8) = 28.1\ \text{m}$ from the third equation; substituting the same numbers into $v = u+at$ first gives $t = 15/4 = 3.75\ \text{s}$ , and then $s = ut + \tfrac12 at^2 = 15(3.75) - 2(3.75)^2 = 56.25 - 28.125 = 28.1\ \text{m}$ from the second equation — the two independent routes agree.

Go deeper · Derivation B · 2 minWhy uniform circular motion is always accelerated motion

Derivation BWhy uniform circular motion is always accelerated motion

The goal is to show, using only the definitions already built and simple vector subtraction, that an object moving at perfectly constant speed around a circle is nonetheless accelerating at every instant. The reasoning follows the chapter's own polygon-to-circle picture and checks it with a numerical example.

Step 1. Recall what acceleration measures: the rate of change of velocity, a vector carrying both magnitude (speed) and direction. A vector changes — producing a non-zero acceleration — if either its magnitude changes, or its direction changes, or both. Everyday intuition wrongly assumes only the first case counts.

Step 2. Picture an athlete running at constant speed around a polygon track — first a square, then a hexagon. On each straight side, velocity is genuinely constant. But at every corner, direction changes abruptly while speed does not, so the velocity vector changes at each corner — a burst of acceleration exactly at each turn.

Step 3. Let the number of sides increase without limit. Each side shrinks and each turn gentles, but turns happen more often. In the limit, the polygon becomes a perfect circle, and the once-occasional turns merge into a turn happening continuously, at every instant, along the whole path.

Step 4. Conclude for the circle: direction of velocity changes continuously, even though its magnitude (speed) is fixed by "uniform." Since the velocity vector differs from one instant to the next — purely from its rotating direction — the average acceleration between any two nearby instants is non-zero. Uniform circular motion is accelerated motion throughout, despite the speedometer never budging.

Step 5. Check with numbers. An object moves at constant speed $10\ \text{m/s}$ around a circle; at two instants a quarter-turn apart, velocity rotates from due east to due north. As (east, north) components, $\vec{v}_1 = (10,0)$ and $\vec{v}_2 = (0,10)\ \text{m/s}$ — both magnitude $10\ \text{m/s}$ , so speed truly hasn't changed. But $\vec{v}_2 - \vec{v}_1 = (-10,10)\ \text{m/s}$ , magnitude $\sqrt{200}\approx14.1\ \text{m/s}$ — not zero. Dividing by the quarter-turn's time gives a non-zero average acceleration, exactly as predicted. Verified: $|\vec{v}_1| = |\vec{v}_2| = 10\ \text{m/s}$ (speed constant) while $|\vec{v}_2 - \vec{v}_1| = \sqrt{200} \approx 14.1\ \text{m/s} \ne 0$ (velocity, and hence acceleration, is not constant) — the two checks together confirm constant speed with changing velocity.

Worked examples

Board level - Two postmen walking toward each other (adapted from the Ganitakaumudi, 14th century CE)

Two postmen walk toward each other from 210 yojanas apart, one at 9 yojanas/day and the other at 5 yojanas/day. In how many days do they meet, and how far will each have walked?

Step 1. Combined rate of closing the gap: $9 + 5 = 14$ yojanas per day.

Step 2. Time to close the 210-yojana gap: $t = 210/14 = 15$ days.

Step 3. Distance each walked: $9\times15 = 135$ yojanas, and $5\times15 = 75$ yojanas.

Step 4. Check: $135+75 = 210$ yojanas, the original gap. They meet after 15 days, with the first postman having walked 135 yojanas and the second 75 yojanas.

Board level - A bus that speeds up, then brakes (extending the chapter's own bus example)

A bus at 36 km/h speeds up to 54 km/h over 10 s, cruises there for 8 s, then brakes to a stop in 5 s. Find the average acceleration in the speeding-up and braking phases, and the total distance covered across all three phases.

Step 1. Convert to SI: $36\ \text{km/h} = 10\ \text{m/s}$ ; $54\ \text{km/h} = 15\ \text{m/s}$ .

Step 2. Speeding up: $a_1 = (15-10)/10 = 0.5\ \text{m/s}^2$ . Distance: $s_1 = ut+\tfrac12at^2 = 10(10)+\tfrac12(0.5)(10)^2 = 125\ \text{m}$ .

Step 3. Cruising: constant $15\ \text{m/s}$ for $8\ \text{s}$ gives zero acceleration and $s_2 = 15\times8 = 120\ \text{m}$ .

Step 4. Braking: $a_2 = (0-15)/5 = -3\ \text{m/s}^2$ . $s_3 = 15(5)+\tfrac12(-3)(5)^2 = 37.5\ \text{m}$ .

Step 5. Total: $125+120+37.5 = 282.5\ \text{m}$ . Speeding-up acceleration = 0.5 m/s², braking acceleration = -3 m/s², total distance = 282.5 m.

Higher-order thinking (HOTS / olympiad-style) - Will the bus stop in time?

A bus at 36 km/h sees an obstacle 30 m ahead. The driver takes 0.5 s to react, then brakes with constant acceleration of magnitude 2.5 m/s². Does the bus stop before the obstacle, and with how much room to spare?

Step 1. Identify the two phases: a reaction phase, moving at unchanged velocity since the brake is not yet applied, and a braking phase, decelerating. Each needs its own distance, then the two are added.

Step 2. Convert velocity to SI: $36\ \text{km/h} = 10\ \text{m/s}$ .

Step 3. Reaction-phase distance: for $0.5\ \text{s}$ at constant $10\ \text{m/s}$ (no acceleration yet), $s_{\text{reaction}} = 10\times0.5 = 5\ \text{m}$ .

Step 4. Braking-phase distance: $u=10\ \text{m/s}$ , $v=0$ , $a=-2.5\ \text{m/s}^2$ . Using $v^2=u^2+2as$ : $0=100+2(-2.5)s_{\text{brake}}$ , so $s_{\text{brake}} = 100/5 = 20\ \text{m}$ .

Step 5. Total: $5+20 = 25\ \text{m}$ , less than the $30\ \text{m}$ available. The bus stops in time, with 5 m to spare before the obstacle. Timing note: budget five to six minutes. The trap is applying a kinematic equation to the whole event at once, forgetting that during the 0.5 s reaction time the bus travels at constant velocity — simple distance = speed × time, not $v^2=u^2+2as$ . Always split a reaction-time-plus-braking problem into its two phases first.

Common traps

Treating distance and displacement as the same number. They agree only in one unreversed direction; any round trip makes displacement smaller, sometimes zero.
Treating average speed and average velocity's magnitude as always equal. True only when direction never changes.
Dropping the sign on displacement, velocity or acceleration. These are vectors; an inconsistent sign invalidates the answer even with a correct magnitude.
Assuming a curved position-time graph means a curved path. It does not — the axes are position and time; a curve means changing velocity, on whatever path is actually followed.
Reading a velocity-time graph's slope as displacement, or its area as acceleration. The other way round: slope is acceleration, area is displacement.
Forgetting to convert km/h to m/s before using a kinematic equation.
Assuming the kinematic equations apply to any motion. They hold only while acceleration is constant.
Believing constant speed always means zero acceleration. True on a straight line, false the instant the path curves.

Common trapConstant speed does not mean zero acceleration

The most persistent misconception in this chapter is assuming that if speed never changes, an object cannot be accelerating. That is true only for straight-line motion. The instant a path curves — even at a perfectly steady speed, as in uniform circular motion — the direction of velocity changes continuously, and since velocity is a vector, a change in direction alone produces a real, non-zero acceleration, exactly as shown in the second derivation above. A car taking a bend at a constant speedometer reading, a child on a merry-go-round, and a satellite in orbit are all accelerating the entire time, even though a speedometer never reveals it. Whenever a question mentions circular motion or changing direction, check direction as carefully as magnitude before deciding acceleration is zero.

You should be able to…

Use a reference point and a sign convention to describe an object's position and decide whether it is in motion or at rest.
Distinguish distance travelled from displacement, and explain why displacement's magnitude can never exceed distance.
Calculate average speed and average velocity from given data, and explain why the two can differ, including cases where velocity is exactly zero.
Calculate average acceleration from initial and final velocity and a time interval, including its sign.
Read a position-time graph to decide whether an object is at rest, moving at constant velocity, or accelerating, and find velocity from its slope.
Read a velocity-time graph to find acceleration from its slope and displacement from the enclosed area.
State and apply the three kinematic equations for constant acceleration, choosing the one matching the given quantities.
Explain why uniform circular motion is accelerated motion despite constant speed, in terms of changing velocity direction.
Solve multi-phase motion problems, such as reaction-time-plus-braking situations, by treating each phase with its own correct equation.

Ask the tutor: if a problem mixes phases — like a reaction time followed by braking, or speeding up followed by cruising and braking — ask the tutor to walk through how to split it into separate constant-acceleration stages before picking equations for each.

Can displacement ever be negative, or is it always positive like distance?

Displacement can absolutely be negative — that is exactly why direction matters for it and not for distance. Once you fix a reference point and a positive direction, an object ending up on the negative side of the origin has a negative displacement, even though distance travelled (a scalar) is always reported as positive. For example, walking 5 m forward from home and then 12 m back past home gives a total distance of 17 m, but a displacement of -7 m — 7 m backward from where you started. The sign is not a quirk; it is direction information distance can never carry.

If a car's speedometer shows a constant reading, does that mean it definitely is not accelerating?

Not necessarily — it depends on whether the car moves in a straight line or a curve. A speedometer only shows speed, the magnitude of velocity, never direction. On a straight road with a steady reading, both speed and velocity are constant, so acceleration is truly zero. But the same steady reading around a bend or roundabout means direction is changing every instant even though magnitude is not — enough to make the car accelerate. This is exactly the reasoning behind uniform circular motion, and the single biggest trap in this chapter.

My average velocity for a round trip came out to zero, but I was clearly moving the whole time. Did I do something wrong?

No, that is exactly correct, and it is a useful check. Average velocity uses displacement, not distance, and displacement measures only the net change in position between start and end points. If you end up back where you started — a lap, a swim there and back, a round trip to the shop — your displacement is zero regardless of how far you actually travelled, so average velocity over that interval is genuinely zero. Average speed, from the actual distance covered, would be a large positive number instead. Both answers are correct; they answer different questions.

Why does everything fall with the same acceleration due to gravity, whether it is heavy or light?

Near Earth's surface, ignoring air resistance, every falling object gains about 9.8 m/s of downward velocity each second it falls, regardless of mass — this acceleration is denoted g. You can check the "constant" part from data: if speed goes 0, 9.8, 19.6, 29.4 m/s at the ends of successive seconds, the change each second is always 9.8 m/s², confirming a genuinely constant acceleration. Why mass does not affect this value is a deeper question this chapter does not derive; it belongs to the study of force and gravity later, so for now treat g = 9.8 m/s² as a standing fact about free fall.