2.1 Introduction

Chapter 2 — Motion in a Straight Line

On the afternoon of 30 April 1006, sky-watchers across Egypt, China, Iraq, Japan and Europe looked up and saw something impossible. A “new star” had appeared in the constellation Lupus — so bright that, according to observers of the time, one could read by its light at night. It hung in the sky for weeks, dimmed slowly, and finally faded from view months later. Today we know what they saw: SN 1006, one of the brightest supernovae ever recorded, the death-flash of a star roughly 7,200 light-years away.

But here is the strange part. That star did not explode in 1006. It exploded around 6,200 BCE — long before any recorded human history. The light from that explosion had been travelling toward Earth for over seven thousand years, moving in a nearly straight line through empty space at a fixed speed, before finally arriving above medieval skies.

Everything in this universe is in motion — the light from that supernova, the Earth beneath your feet, the blood in your veins, the electrons in your phone. Some of these motions are fast, some slow; some in straight lines, some in curves. Before we can understand why things move (that comes in Chapter 4), we first need a precise language to describe motion. That is the business of kinematics, and we begin with the simplest case of all — motion along a straight line.

Figure to come

Fig. 2.0 – Split illustration: (left) a medieval observer pointing at a bright new star in the night sky; (right) a straight arrow showing light travelling from the distant exploding star toward Earth across a vast distance, labelled with time and speed.

Curiosity Corner

Q1. If a car’s speedometer reads 60 km/h, does that tell you how fast it is going right now, or how fast it has been going on average? Are these the same thing?

Q2. A ball thrown straight up momentarily stops at its highest point. At that instant, is its acceleration also zero?

Q3. When you drop a stone and a feather in air, the stone falls faster. But in a vacuum, both fall together. What single quantity describes how anything falls freely near the Earth’s surface?

Q4. A car’s brakes bring it to a halt over 10 metres when travelling at 30 km/h. Roughly how far will the same car take to stop if it is travelling at 60 km/h — twice, or four times as far?

Q5. If two trains move in opposite directions on parallel tracks, why does each appear to rush past the other far faster than either is actually moving?

By the end of this chapter, you will have the tools — position, velocity, acceleration, and the kinematic equations — to answer every one of these questions precisely.

Look around for a moment. A pen sits on your desk, perhaps rolling slowly. A fan turns overhead. Outside, a bike goes past, leaves flutter on a tree, a bird cuts across the sky. Even if you sit perfectly still, air moves in and out of your lungs, blood flows through your arteries and veins, and your heart contracts and relaxes many times each minute.

Motion is not the exception in our universe — it is the rule. The Earth spins on its axis once every 24 hours and travels around the Sun once every year. The Sun itself carries the Earth along as it orbits the centre of our galaxy, the Milky Way, and the Milky Way in turn drifts through its own local group of galaxies. From atoms vibrating in a solid to entire galaxy clusters flying apart, something is always moving.

Figure to come

Fig. 2.1 – A collage of motions at very different scales: a child running, a car on a road, water flowing from a dam, the Earth orbiting the Sun, and a galaxy — with a small arrow on each indicating the direction of motion.

So what, exactly, do we mean by motion? At its simplest, motion is a change in the position of an object with time. If nothing about where a body sits in space changes as the clock ticks, we say it is at rest. If its position shifts — even slightly — as time passes, we say it is in motion.

Definition

Motion is the change in position of an object with time.

This chapter takes on the question: how do we describe motion in a precise, mathematical way? To do so, we will build two central quantities — velocity, which tells us how fast an object’s position is changing, and acceleration, which tells us how fast the velocity itself is changing. Together, these two quantities let us capture the complete story of any moving object.

To keep our first study manageable, we restrict ourselves to motion along a single straight line — a car on a straight stretch of highway, a ball thrown vertically up, a bead sliding down a straight wire. This kind of motion is called rectilinear motion.

Definition

Rectilinear motion is motion of an object along a straight line.

Within rectilinear motion, one special case turns out to be remarkably useful: motion in which the acceleration stays constant throughout. For such uniformly accelerated motion, we will derive a small set of simple equations — the kinematic equations — that connect position, velocity, acceleration, and time. A ball dropped from a building, or a car braking uniformly to a stop, are examples of such motion that we will meet later in the chapter.

Finally, once we can describe motion along a line, we take one more step: what happens when two objects move, and we ask how fast one appears to move as seen from the other? This is the idea of relative velocity, which explains everyday puzzles like why an oncoming train seems to rush past you far faster than either train is actually moving on its own.

There is one more simplification we will use throughout. We shall treat every moving object as a point object — a body whose size can be ignored in comparison with the distance it travels. A cricket ball is roughly 7 cm across; if it flies 30 m from bowler to batter, its size is tiny compared to its journey, and treating it as a point costs us nothing. But a bus that only creeps forward by half a metre in a traffic jam cannot be called a point — its length and its motion are comparable, and its size matters.

Definition

A point object is an object whose size is much smaller than the distance it moves in the time interval of interest, so that its size can be neglected.

Quick Question

Can the same object be treated as a point object in one situation and not in another?

Yes. The Earth can be treated as a point object when we study its year-long orbit around the Sun — it travels hundreds of millions of kilometres, while its own diameter is only about 13,000 km. But when we study the rotation of the Earth on its axis, or the motion of a ship sailing across an ocean on Earth’s surface, we cannot ignore the Earth’s size. Whether a body is a “point” depends on the motion being studied, not on the body itself.

One last piece of framing before we begin. The branch of mechanics that describes how things move — without asking why — is called kinematics. In this chapter and the next, we stay strictly within kinematics: positions, velocities, accelerations, times. The question of what causes motion — forces, mass, and Newton’s laws — is the subject of Chapter 4.

Definition

Kinematics is the branch of mechanics that describes the motion of objects without considering the causes (forces) that produce the motion.

With these ideas in place — motion as change of position, the straight-line simplification, the point-object approximation, and the kinematic viewpoint — we are ready to start measuring motion. We begin, in the next section, by asking the most natural question of all: how fast is something moving right now?

2.2 Instantaneous Velocity and Speed

Imagine you drive from home to school, a distance of 10 km, and it takes you half an hour. Your average velocity for the trip is simply the total displacement divided by the total time — in this case, 20 km/h. Recall from earlier study that for motion along a straight line, if an object is at position \(x_1\) at time \(t_1\) and position \(x_2\) at time \(t_2\), the average velocity over that interval is

\[\bar{v} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}\]

But this single number of 20 km/h hides a lot. You may have crawled at 5 km/h through traffic near your home, opened up to 60 km/h on a clear stretch, and stopped altogether at a red light. The average tells you nothing about how fast you were moving at any particular moment.

That is why we need a new quantity — one that answers the question, “how fast is the object moving right now, at this very instant?” We call it the instantaneous velocity, or simply the velocity, \(v\), at an instant \(t\).

Curiosity Corner

Q. If a car’s speedometer reads 60 km/h, does that tell you how fast it is going right now, or how fast it has been going on average? Are these the same thing? A. It tells you how fast the car is going right now: the speedometer shows the instantaneous speed, responding to how fast the wheels are turning at that very moment, which is why the reading changes the instant the accelerator or the brake is pressed. The two are not the same thing in general. An average velocity is a total displacement divided by a total time, and a single average of 20 km/h can hide a crawl through traffic at 5 km/h, a clear stretch at 60 km/h, and a complete stop at a red light. Only in uniform motion, when the velocity never changes, does the instantaneous value coincide with the average.

From average to instantaneous — the limiting process

How do we go from average velocity, which needs a time interval, to velocity at a single instant? The trick is to shrink the interval.

Take the average velocity over a small time interval \(\Delta t\) centred at the instant we care about. Then make \(\Delta t\) smaller. Then smaller still. As \(\Delta t\) becomes vanishingly small — closer and closer to zero — the average velocity settles down to a definite value. That limiting value is defined as the instantaneous velocity at that instant.

In symbols,

\[v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} \tag{2.1a}\]

\[= \frac{dx}{dt} \tag{2.1b}\]

The symbol \(\lim_{\Delta t \to 0}\) means “the value that the quantity on the right approaches as \(\Delta t\) is made infinitesimally small.” In the language of calculus, the right-hand side of Eq. (2.1a) is called the differential coefficient, or derivative, of \(x\) with respect to \(t\), written \(\dfrac{dx}{dt}\). Physically, it is the rate of change of position with respect to time, at that instant.

Definition

The instantaneous velocity of an object at an instant \(t\) is the limit of its average velocity as the time interval \(\Delta t\) around \(t\) becomes infinitesimally small: \(v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}\).

Two ways to find instantaneous velocity

Equation (2.1a) can be used to find velocity at an instant in two ways — graphically or numerically.

Graphical method. Suppose we want to find the velocity of a moving car at time \(t = 4\,\text{s}\) (point P on its position–time curve), as shown in Fig. 2.1. Start with a time interval \(\Delta t = 2\,\text{s}\) centred at \(t = 4\,\text{s}\) — that is, from \(t = 3\,\text{s}\) to \(t = 5\,\text{s}\). The slope of the line \(P_1 P_2\) joining the positions at these two instants gives the average velocity over that interval.

Figure to come

Fig. 2.1 – Position–time graph (curve x vs t) with point P at t = 4 s; secant lines P₁P₂ (from t = 3 s to 5 s) and Q₁Q₂ (from t = 3.5 s to 4.5 s) drawn through P, along with the tangent at P labelled as giving the instantaneous velocity.

Now shrink the interval — say, to \(\Delta t = 1\,\text{s}\), from \(t = 3.5\,\text{s}\) to \(t = 4.5\,\text{s}\). The chord \(P_1P_2\) becomes a shorter chord \(Q_1Q_2\), and its slope gives the average velocity over this smaller interval. As we keep decreasing \(\Delta t\), the chord rotates closer and closer to the tangent to the curve at point P. In the limit \(\Delta t \to 0\), the chord becomes the tangent, and its slope gives the velocity at \(t = 4\,\text{s}\).

Principle / Law

The instantaneous velocity at any instant is equal to the slope of the tangent drawn to the position–time graph at that instant.

Graphically, taking a limit is hard to show cleanly — the chord almost coincides with the tangent, and the eye cannot separate them.

Numerical method. The idea becomes much clearer if we work with numbers. Suppose the car’s position varies as \(x = 0.08 \, t^3\) (with \(x\) in metres, \(t\) in seconds). Table 2.1 lists \(\Delta x / \Delta t\) for successively smaller intervals \(\Delta t\), each centred at \(t = 4.0\,\text{s}\).

The second and third columns give \(t_1 = t - \dfrac{\Delta t}{2}\) and \(t_2 = t + \dfrac{\Delta t}{2}\). The fourth and fifth columns give \(x(t_1) = 0.08\, t_1^3\) and \(x(t_2) = 0.08\, t_2^3\). The sixth column lists \(\Delta x = x(t_2) - x(t_1)\), and the last column gives \(\Delta x / \Delta t\), which is the average velocity over that interval.

\(\Delta t\) (s)	\(t_1\) (s)	\(t_2\) (s)	\(x(t_1)\) (m)	\(x(t_2)\) (m)	\(\Delta x\) (m)	\(\Delta x/\Delta t\) (m s⁻¹)
2.0	3.0	5.0	2.16	10.0	7.84	3.92
1.0	3.5	4.5	3.43	7.29	3.86	3.86
0.5	3.75	4.25	4.21875	6.14125	1.9225	3.845
0.1	3.95	4.05	4.93039	5.31441	0.38402	3.8402
0.01	3.995	4.005	5.100824	5.139224	0.0384	3.8400

Table 2.1 – Limiting value of \(\Delta x / \Delta t\) at \(t = 4\) s.

Read the last column downward. As \(\Delta t\) shrinks from 2.0 s to 0.01 s, the average velocity settles from 3.92 m s⁻¹ toward a clear limiting value of 3.84 m s⁻¹. That is the instantaneous velocity at \(t = 4.0\,\text{s}\), i.e., the value of \(\dfrac{dx}{dt}\) at \(t = 4.0\,\text{s}\). In this way, we can find velocity at every instant of the car’s motion.

Using calculus directly

The graphical method needs a carefully drawn curve, and the numerical method needs a table of positions. Both work, but neither is convenient if we already know the position as a mathematical function of time. In that case, we can just differentiate the expression using calculus and read off \(\dfrac{dx}{dt}\) at any instant.

Solved Example 2.1

The position of an object moving along the x-axis is given by \(x = a + b t^2\), where \(a = 8.5\,\text{m}\), \(b = 2.5\,\text{m s}^{-2}\), and \(t\) is in seconds. Find its velocity at \(t = 0\,\text{s}\) and \(t = 2.0\,\text{s}\). Also find the average velocity between \(t = 2.0\,\text{s}\) and \(t = 4.0\,\text{s}\).

Answer

In the language of calculus, the instantaneous velocity is obtained by differentiating \(x\) with respect to \(t\):

\[v = \frac{dx}{dt} = \frac{d}{dt}\left(a + b t^2\right) = 2 b t = 5.0\, t \ \text{m s}^{-1}\]

At \(t = 0\,\text{s}\), \(v = 0\,\text{m s}^{-1}\). At \(t = 2.0\,\text{s}\), \(v = 10\,\text{m s}^{-1}\).

For the average velocity between \(t = 2.0\,\text{s}\) and \(t = 4.0\,\text{s}\), we use the ratio of displacement to time interval:

\[\bar{v} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{(a + 16b) - (a + 4b)}{2.0} = \frac{12b}{2.0} = 6.0 \times b\]

\[= 6.0 \times 2.5 = 15\,\text{m s}^{-1}\]

Notice that the two instantaneous velocities (0 and 10 m/s) and the average velocity over the interval (15 m/s) are all different — a reminder that instantaneous and average velocities are not the same thing in general.

A special case — uniform motion. If an object moves with a constant velocity (no acceleration), then its velocity at every instant is the same, and its average velocity over any interval is also the same number. So for uniform motion, the instantaneous velocity equals the average velocity at every instant. It is only when velocity changes with time that the distinction between instantaneous and average matters.

Numerical 2.1

The position of a particle moving along a straight line is given by \(x = 3 t^2 - 2 t + 1\) (with \(x\) in metres and \(t\) in seconds). Find (a) its instantaneous velocity at \(t = 2\,\text{s}\), and (b) its average velocity between \(t = 1\,\text{s}\) and \(t = 3\,\text{s}\). Compare the two values.

Instantaneous speed

Instantaneous speed, or simply speed, is the magnitude of the instantaneous velocity — it drops the direction and keeps only “how fast.” A velocity of \(+24\,\text{m s}^{-1}\) and a velocity of \(-24\,\text{m s}^{-1}\) both have the same speed of \(24\,\text{m s}^{-1}\); they differ only in direction.

Definition

Instantaneous speed is the magnitude of the instantaneous velocity at that instant.

An important subtlety: for average quantities over a finite time interval, average speed is generally greater than or equal to the magnitude of average velocity, because average speed uses total path length while average velocity uses net displacement. If a runner runs 100 m north and then 100 m back south in 40 s, her displacement is zero (so average velocity is zero) but she has still covered 200 m, so her average speed is 5 m/s.

At an instant, however, this distinction disappears — instantaneous speed is always exactly equal to the magnitude of the instantaneous velocity.

Quick Question

Why do the two definitions collide only at an instant?

Over a finite interval, path length and displacement can differ, because the object may reverse direction and add extra distance without adding to net displacement. At a single instant, there is no interval to reverse direction over — the object is moving in one direction (or momentarily not moving at all). So over an infinitesimal interval \(dt\), path length and \(|dx|\) are the same, and their ratio with \(dt\) is the same.

2.3 Acceleration

So far we have learned how to describe how fast an object is moving at any instant — that is what velocity tells us. But in most real motions, velocity itself does not stay the same. A car speeds up when its driver presses the accelerator and slows down when the brakes are applied. A ball thrown straight up gets slower as it rises, momentarily stops, and then speeds up as it falls. In all these cases, the velocity is changing — and we need a new quantity to describe how quickly it changes.

How should we measure “change in velocity”?

Suppose an object’s velocity is changing. A natural question arises: change with respect to what? With respect to distance travelled, or with respect to time elapsed?

Both sound reasonable, and this was a genuine puzzle in the seventeenth century. It was Galileo Galilei who, through his careful studies of freely falling objects and balls rolling down inclined planes, settled the question. He discovered that for objects in free fall, the rate of change of velocity with time is constant, but the rate of change of velocity with distance is not.

Real Incident / Discovery

Galileo could not directly time a falling stone with the crude clocks of his day, so he “diluted” gravity by letting bronze balls roll down long, gently inclined wooden grooves and used a water clock — measuring how much water flowed out during each descent. By repeating the experiment with different inclinations, he found that in equal time intervals the ball gained equal amounts of velocity, no matter how steep the slope. This constant-per-unit-time gain is what convinced him that time — not distance — is the right variable to track velocity change against.

This led to the modern concept of acceleration: the rate at which velocity changes with time.

Average acceleration

If an object has velocity \(v_1\) at time \(t_1\) and velocity \(v_2\) at a later time \(t_2\), its average acceleration \(\bar{a}\) over that interval is defined as the change in velocity divided by the time interval:

\[\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t} \tag{2.2}\]

Here \(\Delta v = v_2 - v_1\) is the change in velocity (in m s⁻¹) and \(\Delta t = t_2 - t_1\) is the time interval over which the change occurs (in s). The SI unit of acceleration is therefore metre per second squared, written \(\text{m s}^{-2}\). Physically, an acceleration of, say, \(3\,\text{m s}^{-2}\) means the object’s velocity is changing by 3 m s⁻¹ every second.

Definition

The average acceleration of an object over a time interval is the change in its velocity divided by that time interval: \(\bar{a} = \dfrac{\Delta v}{\Delta t}\).

On a plot of velocity versus time, the average acceleration between two instants is simply the slope of the straight line joining the points \((t_1, v_1)\) and \((t_2, v_2)\).

Instantaneous acceleration

Just as with velocity, the average value over an interval does not tell us what is happening at any particular moment. To get the acceleration at a single instant, we shrink the interval \(\Delta t\) to zero, exactly the same way we did for instantaneous velocity:

\[a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \tag{2.3}\]

Definition

The instantaneous acceleration is the limit of the average acceleration as the time interval \(\Delta t\) becomes infinitesimally small: \(a = \dfrac{dv}{dt}\).

Graphically, the instantaneous acceleration at any instant equals the slope of the tangent drawn to the velocity–time (\(v\)–\(t\)) curve at that instant.

Principle / Law

The instantaneous acceleration of an object equals the slope of the tangent to its velocity–time graph at that instant.

What kinds of changes count as acceleration?

Since velocity is a vector — it has both magnitude (speed) and direction — a change in velocity can arise in three distinct ways:

A change in speed (magnitude), with direction unchanged. Example: a car speeding up on a straight road.
A change in direction, with speed unchanged. Example: a stone whirled at the end of a string in a circle at constant speed.
A change in both speed and direction together.

Any one of these changes produces an acceleration. In this chapter, since we are dealing with rectilinear (straight-line) motion, the direction reduces to just a sign (positive or negative along our chosen axis). But keep the vector idea in mind — it will matter in later chapters on motion in a plane.

Acceleration, like velocity, can be positive, negative, or zero. The sign is not just a formality; it tells you which way the velocity is changing relative to your chosen positive direction. We will unpack this in more detail shortly.

What acceleration looks like on graphs

Fig. 2.2 shows position–time graphs for three types of motion — with positive, negative, and zero acceleration.

Figure to come

Fig. 2.2 – Three position–time (x–t) graphs side by side: (a) curve bending upward, labelled “Positive a”; (b) curve bending downward, labelled “Negative a”; (c) straight inclined line, labelled “a = 0”.

Notice a clear pattern:

For positive acceleration, the \(x\)-\(t\) graph curves upward (concave up) — the object is gaining velocity, so it covers more distance per unit time as time progresses.
For negative acceleration, the graph curves downward (concave down) — the object is losing velocity.
For zero acceleration, the graph is a straight line — the velocity is constant, so the object covers equal distances in equal times.

Restriction to constant acceleration

Although acceleration can, in general, vary from one instant to the next (a driver may press the accelerator harder and harder), in this chapter we restrict ourselves to the simplest and most common case: motion with constant acceleration. When acceleration is constant, the average acceleration over any interval is equal to that same constant value, and the instantaneous acceleration equals it too.

For an object whose velocity is \(v_0\) at \(t = 0\) and \(v\) at some later time \(t\), taking the interval from \(0\) to \(t\) in the definition of average acceleration gives

\[\bar{a} = \frac{v - v_0}{t - 0} = a \quad \text{or} \quad v = v_0 + a\, t \tag{2.4}\]

This is our very first kinematic equation. It says: after time \(t\), the velocity is the initial velocity plus the extra velocity gained (\(a \times t\)) due to acceleration.

Quick Question

At the highest point of a ball thrown straight up, its velocity is momentarily zero. Is its acceleration also zero at that instant?

No. The ball is still under gravity, so its acceleration is \(-g\) (downward) at every instant of its flight — including the highest point. Zero velocity does not mean zero acceleration; the ball’s velocity is passing through zero, changing from positive (upward) to negative (downward), and that change itself requires acceleration.

Curiosity Corner

Q. A ball thrown straight up momentarily stops at its highest point. At that instant, is its acceleration also zero? A. No. The ball remains under gravity for every instant of its flight, so its acceleration there is still \(-g\), directed downward. Zero velocity does not imply zero acceleration: at the highest point the velocity is passing through zero on its way from positive (upward) to negative (downward), and producing that very change is what the acceleration is doing.

Velocity–time graphs for constant acceleration

Fig. 2.3 shows velocity–time graphs for four cases of constant acceleration.

Figure to come

Fig. 2.3 – Four v–t graphs: (a) v rising from v₀ > 0 with positive slope (positive direction, positive a); (b) v falling from v₀ > 0 to lower positive values (positive direction, negative a); (c) v starting near zero and dropping to more negative values (negative direction, negative a); (d) v starting positive, crossing zero at t₁, becoming negative, showing an object that reverses direction between times t₁ and t₂.

Reading each graph:

An object moves in the positive direction with positive acceleration — its velocity keeps growing.
An object moves in the positive direction but with negative acceleration — its velocity is positive but shrinking.
An object moves in the negative direction with negative acceleration — its velocity is negative and becoming more negative (i.e., its speed is increasing in the negative direction).
An object starts moving in the positive direction, decelerates to zero velocity at time \(t_1\), and then reverses direction, moving in the negative direction between \(t_1\) and \(t_2\). Throughout, the acceleration is a single constant negative value; only the velocity reverses sign.

Quick Question

In case (c), the acceleration is negative but the object’s speed is increasing. Doesn’t a negative acceleration mean slowing down?

No — that is a common misreading. The sign of acceleration only tells you which way (positive or negative axis direction) the velocity is being pushed. If velocity is already negative and acceleration is also negative, the two agree in direction, so velocity becomes more negative, i.e., speed increases. An object slows down only when acceleration and velocity have opposite signs.

Area under a v–t curve equals displacement

One of the most useful features of the velocity–time graph is this: the area under the \(v\)–\(t\) curve, between two instants, equals the displacement of the object over that time interval.

A full proof needs integral calculus, but we can see why it works for the simplest case — an object moving with constant velocity \(u\). Its velocity-time graph is a horizontal straight line at height \(u\), as shown in Fig. 2.4.

Figure to come

Fig. 2.4 – A v–t graph showing a horizontal line at height u from t = 0 to t = T, forming a rectangle with the time axis of height u and base T; the rectangle is shaded to indicate area equals displacement uT.

Between \(t = 0\) and \(t = T\), the area under the curve is just the area of the rectangle: height \(u\) times base \(T\), which is \(u T\). And \(u T\) is exactly the displacement of an object moving at constant velocity \(u\) for time \(T\). So the area does equal the displacement.

A subtle caution about sharp corners

You will notice that many \(x\)–\(t\), \(v\)–\(t\), and \(a\)–\(t\) graphs in this chapter have sharp kinks — for instance, a v–t graph might suddenly change slope at some instant. Mathematically, at a sharp kink the function is not differentiable — the slope on the left does not match the slope on the right, so no single tangent (and hence no single acceleration) can be defined at that instant.

In any real situation, however, the change would take place over a very small but non-zero time interval, so all these graphs would be smooth curves without truly sharp corners. What this means physically is that velocity and acceleration cannot change abruptly at a single instant — real changes are always continuous. The kinks in our diagrams are just idealisations for clarity.

2.4 Kinematic Equations for Uniformly Accelerated Motion

When the acceleration of an object stays constant throughout its motion, a beautiful simplification takes place: just three simple equations are enough to tell us everything about the motion at any instant. These are the kinematic equations of motion, and they connect five quantities:

displacement, \(x\)
time taken, \(t\)
initial velocity, \(v_0\)
final velocity, \(v\)
(constant) acceleration, \(a\)

If you know any three of these, the two equations you pick will give you the other two. That is why these equations show up in almost every problem on straight-line motion — thrown balls, braking cars, falling stones, sliding blocks — and why they are worth deriving carefully once and remembering forever.

Equation 1 — from the definition of acceleration

The first equation is nothing new. We already got it in Section 2.3 by applying the definition of average acceleration between \(t = 0\) and time \(t\):

\[v = v_0 + a t \tag{2.4}\]

This tells us that after time \(t\), the velocity is the starting velocity plus whatever extra velocity the constant acceleration has added.

Equation 2 — from the area under the v–t graph

To get the second equation, we use the useful fact from the previous section: the area under a velocity–time curve equals the displacement. Fig. 2.5 shows the velocity–time graph for an object with constant acceleration — a straight line starting from \(v_0\) at \(t = 0\) and rising to \(v\) at time \(t\).

Figure to come

Fig. 2.5 – A v–t graph with a straight line from point A (at height v₀) to point B (at height v) over the interval from O (t = 0) to D (time t). The region under the line is split into a rectangle OACD (of height v₀, base t) and a triangle ABC (of height v − v₀, base t), together shaded to represent the displacement.

The area under the line between \(t = 0\) and time \(t\) can be split into two simple shapes:

a rectangle OACD of width \(t\) and height \(v_0\), with area \(v_0 \, t\)
a triangle ABC on top of it, with base \(t\) and height \((v - v_0)\), with area \(\tfrac{1}{2}(v - v_0)\, t\)

Adding the two,

\[x = \frac{1}{2}(v - v_0)\, t + v_0\, t \tag{2.5}\]

Since \(v - v_0 = a t\) (from Eq. 2.4), we can replace \(v - v_0\) with \(a t\) inside the bracket:

\[x = \frac{1}{2} a t^2 + v_0 t\]

or, more commonly written,

\[x = v_0 t + \frac{1}{2} a t^2 \tag{2.6}\]

This is the second kinematic equation. It gives displacement directly in terms of time.

Eq. (2.5) can also be re-arranged into another useful form:

\[x = \frac{v + v_0}{2}\, t = \bar{v}\, t \tag{2.7a}\]

where

\[\bar{v} = \frac{v + v_0}{2} \quad \text{(constant acceleration only)} \tag{2.7b}\]

Equations (2.7a) and (2.7b) carry a nice physical meaning: for constant acceleration, the average velocity over any interval is simply the arithmetic mean of the initial and final velocities. This is only true for constant acceleration — do not use it otherwise.

Equation 3 — eliminating time

The third equation removes \(t\) altogether. From Eq. (2.4), \(t = (v - v_0)/a\). Substituting this into Eq. (2.7a),

\[x = \bar{v}\, t = \left(\frac{v + v_0}{2}\right)\left(\frac{v - v_0}{a}\right) = \frac{v^2 - v_0^2}{2 a}\]

Rearranging,

\[v^2 = v_0^2 + 2 a x \tag{2.8}\]

You can also obtain this equation by substituting \(t\) from Eq. (2.4) into Eq. (2.6) — try it as a small exercise.

The three equations together

Collecting our results, we have the three kinematic equations of motion for constant acceleration:

\[v = v_0 + a t\]

\[x = v_0 t + \frac{1}{2} a t^2\]

\[v^2 = v_0^2 + 2 a x \tag{2.9a}\]

Principle / Law

For an object in rectilinear motion with constant acceleration \(a\), initial velocity \(v_0\), final velocity \(v\), displacement \(x\) and time \(t\): \(v = v_0 + a t\), \(x = v_0 t + \tfrac{1}{2} a t^2\), and \(v^2 = v_0^2 + 2 a x\). These equations connect any four of the five quantities.

These equations were derived assuming that at \(t = 0\), the object is at position \(x = 0\). What if it starts at some non-zero initial position \(x_0\)? Then \(x\) in every equation is replaced by the change in position, \(x - x_0\):

\[v = v_0 + a t\]

\[x = x_0 + v_0 t + \frac{1}{2} a t^2 \tag{2.9b}\]

\[v^2 = v_0^2 + 2 a (x - x_0) \tag{2.9c}\]

That is the fully general form.

Deriving the equations using calculus

Everything above can also be obtained cleanly using calculus. The calculus approach is more powerful because it extends to non-uniform acceleration as well.

Solved Example 2.2

Obtain the equations of motion for constant acceleration using the method of calculus.

Answer

By the definition of instantaneous acceleration,

\[a = \frac{dv}{dt} \quad \Rightarrow \quad dv = a\, dt\]

Integrating both sides from time 0 (velocity \(v_0\)) to time \(t\) (velocity \(v\)), and using the fact that \(a\) is constant so it can come out of the integral,

\[\int_{v_0}^{v} dv = \int_0^t a\, dt = a \int_0^t dt\]

\[v - v_0 = a t \quad \Rightarrow \quad v = v_0 + a t\]

Next, from the definition of instantaneous velocity,

\[v = \frac{dx}{dt} \quad \Rightarrow \quad dx = v\, dt\]

Integrating from position \(x_0\) at time 0 to position \(x\) at time \(t\), and substituting the expression for \(v\),

\[\int_{x_0}^{x} dx = \int_0^t v\, dt = \int_0^t (v_0 + a t)\, dt\]

\[x - x_0 = v_0 t + \frac{1}{2} a t^2 \quad \Rightarrow \quad x = x_0 + v_0 t + \frac{1}{2} a t^2\]

For the third equation, we use a chain-rule trick:

\[a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\, \frac{dv}{dx}\]

so that \(v\, dv = a\, dx\). Integrating both sides,

\[\int_{v_0}^{v} v\, dv = \int_{x_0}^{x} a\, dx\]

\[\frac{v^2 - v_0^2}{2} = a\, (x - x_0) \quad \Rightarrow \quad v^2 = v_0^2 + 2 a\, (x - x_0)\]

The great advantage of the calculus method is that it also works when acceleration is not constant — you would then have to keep \(a\) inside the integral rather than pulling it out.

Applying the equations — vertical motion under gravity

The most familiar case of straight-line motion with constant acceleration is motion of a body near the Earth’s surface, where gravity produces a constant downward acceleration. Let us use the equations we just derived to analyse such a motion.

Solved Example 2.3

A ball is thrown vertically upwards with a velocity of \(20\,\text{m s}^{-1}\) from the top of a multistorey building. The height of the launch point above the ground is \(25.0\,\text{m}\). Taking \(g = 10\,\text{m s}^{-2}\): (a) how high will the ball rise, and (b) how long will it be before the ball hits the ground?

Answer

Choose the y-axis pointing vertically upward, with \(y = 0\) at the ground level, as shown in Fig. 2.6.

[Diagram: Fig. 2.6 – A tall building of height 25 m; a ball is shown at point A on the roof (y₀ = 25 m), rising to a highest point B (of height y − y₀ above A), then falling all the way to the ground at C. Arrows mark the upward direction as positive y, and the constant acceleration a = −10 m s⁻² acting downward.]

At launch, \(v_0 = +20\,\text{m s}^{-1}\) (upward). Gravity acts downward, so \(a = -g = -10\,\text{m s}^{-2}\). At the highest point, the ball is momentarily at rest, so \(v = 0\). Using the third kinematic equation (which does not need time),

\[v^2 = v_0^2 + 2 a\, (y - y_0)\]

\[0 = (20)^2 + 2(-10)(y - y_0)\]

Solving, \((y - y_0) = 20\,\text{m}\). The ball rises 20 m above its launch point, reaching a maximum height of \(25 + 20 = 45\,\text{m}\) above the ground.

The time-of-flight until the ball hits the ground can be found in two ways. Note both — the comparison is instructive.

First method — split the motion into up and down. Going up (A to B): using \(v = v_0 + a t\) with final velocity \(v = 0\),

\[0 = 20 - 10\, t_1 \quad \Rightarrow \quad t_1 = 2\,\text{s}\]

Coming down (B to C): now the ball starts from rest at height 45 m above the ground, so \(y_0 = 45\,\text{m}\), \(v_0 = 0\), \(a = -10\,\text{m s}^{-2}\), and final position \(y = 0\). Using \(y = y_0 + v_0 t + \tfrac{1}{2} a t^2\),

\[0 = 45 + 0 + \tfrac{1}{2}(-10)\, t_2^2 \quad \Rightarrow \quad t_2 = 3\,\text{s}\]

Total time = \(t_1 + t_2 = 2\,\text{s} + 3\,\text{s} = 5\,\text{s}\).

Second method — treat the full flight as one motion. Take the launch point as the reference position: \(y_0 = 25\,\text{m}\), \(y = 0\) at the ground, \(v_0 = 20\,\text{m s}^{-1}\), \(a = -10\,\text{m s}^{-2}\). Apply \(y = y_0 + v_0 t + \tfrac{1}{2} a t^2\) directly:

\[0 = 25 + 20\, t + \tfrac{1}{2}(-10)\, t^2\]

\[5 t^2 - 20 t - 25 = 0\]

Solving the quadratic, \(t = 5\,\text{s}\) (rejecting the negative root).

The second method is more elegant. Since the motion is under constant acceleration throughout, one equation covers the whole trajectory — we do not need to know at what instant the ball turned around or split the journey artificially. This is a general principle: whenever acceleration is constant, treat the motion as one continuous piece.

Free fall — a special case of uniform acceleration

The motion of the ball in Example 2.3 is an example of free fall — motion under gravity alone, with air resistance neglected.

Solved Example 2.4

Free fall — discuss the motion of an object under free fall. Neglect air resistance.

Answer

An object released near the surface of the Earth is accelerated downward under the influence of gravity. The magnitude of this acceleration is called the acceleration due to gravity, denoted \(g\). If air resistance is negligible, the object is said to be in free fall. So long as the fall is over a height very small compared to the Earth’s radius, \(g\) can be treated as a constant equal to about \(9.8\,\text{m s}^{-2}\). Free fall is thus a case of motion with uniform acceleration.

[Definition:] Free fall is motion of an object under the influence of gravity alone, with air resistance neglected. The magnitude of the acceleration produced is \(g \approx 9.8\,\text{m s}^{-2}\), directed vertically downward.

Curiosity Corner

Q. When you drop a stone and a feather in air, the stone falls faster. But in a vacuum, both fall together. What single quantity describes how anything falls freely near the Earth’s surface? A. The acceleration due to gravity, \(g\). When air resistance is neglected, an object near the Earth’s surface is in free fall, and the magnitude of its downward acceleration is about \(9.8\,\text{m s}^{-2}\), provided the fall is over a height very small compared with the Earth’s radius. Because \(g\) appears in the kinematic equations without any reference to the falling body itself, free fall is simply motion with this one uniform acceleration.

Take the y-axis pointing upward as positive. Since gravity pulls downward, its acceleration is negative:

\[a = -g = -9.8\,\text{m s}^{-2}\]

For an object released from rest at \(y = 0\), we have \(v_0 = 0\), and the three kinematic equations become

\[v = 0 - g t = -9.8\, t \ \text{m s}^{-1}\]

\[y = 0 - \tfrac{1}{2} g t^2 = -4.9\, t^2 \ \text{m}\]

\[v^2 = 0 - 2 g y = -19.6\, y \ \text{m}^2 \text{s}^{-2}\]

These three equations give velocity as a function of time, position as a function of time, and velocity as a function of position. Fig. 2.7 plots how acceleration, velocity, and distance vary with time during free fall.

Figure to come

Fig. 2.7 – Three graphs stacked: (a) acceleration vs time — horizontal line at a = −9.8 m s⁻²; (b) velocity vs time — straight line through the origin with negative slope, becoming more negative with time; (c) distance vs time — downward-opening parabola showing y becoming more negative (deeper below the launch point) as time progresses.

Notice the shapes: acceleration is a flat line (constant), velocity is a straight sloping line (linear in \(t\)), and displacement is a parabola (quadratic in \(t\)). These are the visual signatures of uniform acceleration.

Solved Example 2.5

Galileo’s law of odd numbers: “The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1 : 3 : 5 : 7 …].” Prove it.

Answer

Divide the total time of the fall into equal intervals \(\tau\), and find the object’s position at the end of each interval. Since the initial velocity is zero, position at any time \(t\) is

\[y = -\tfrac{1}{2} g t^2\]

Using this, we compute the position after times \(0, \tau, 2\tau, 3\tau, \ldots\), and then the distance covered during each successive interval (i.e., the difference between two consecutive positions).

If we let \(y_0 = -(1/2) g \tau^2\) — the (negative) position after the first interval — the results fit neatly into Table 2.2:

\(t\)	\(y\)	\(y\) in units of \(y_0\)	Distance in successive \(\tau\)	Ratio
0	0	0	—	—
\(\tau\)	\(-(1/2)g\tau^2\)	\(y_0\)	\(y_0\)	1
\(2\tau\)	\(-4(1/2)g\tau^2\)	\(4\, y_0\)	\(3\, y_0\)	3
\(3\tau\)	\(-9(1/2)g\tau^2\)	\(9\, y_0\)	\(5\, y_0\)	5
\(4\tau\)	\(-16(1/2)g\tau^2\)	\(16\, y_0\)	\(7\, y_0\)	7
\(5\tau\)	\(-25(1/2)g\tau^2\)	\(25\, y_0\)	\(9\, y_0\)	9
\(6\tau\)	\(-36(1/2)g\tau^2\)	\(36\, y_0\)	\(11\, y_0\)	11

Table 2.2 – Positions and successive distances covered in equal time intervals under free fall.

The last column shows the astonishing result: the distances covered in successive equal intervals of time are in the ratio \(1 : 3 : 5 : 7 : 9 : 11 : \ldots\) — the odd numbers. This was one of the earliest quantitative laws of motion, first established by Galileo Galilei (1564–1642), who made the first careful measurements of free fall.

Applications to real-life motion

Solved Example 2.6

Stopping distance of vehicles. When brakes are applied to a moving vehicle, the distance it travels before coming to rest is called its stopping distance. It depends on the initial velocity \(v_0\) and the braking capacity (deceleration) \(-a\). Derive an expression for the stopping distance in terms of \(v_0\) and \(a\).

Answer

Let the stopping distance be \(d_s\). Using \(v^2 = v_0^2 + 2 a x\) and noting that the vehicle comes to rest (\(v = 0\)),

\[0 = v_0^2 + 2 a\, d_s \quad \Rightarrow \quad d_s = \frac{-v_0^2}{2 a}\]

Since the deceleration is negative, this gives a positive stopping distance. The result is striking:

\[d_s \propto v_0^2\]

The stopping distance is proportional to the square of the initial velocity. Doubling the initial speed (for the same braking) does not double the stopping distance — it makes it four times as large.

Curiosity Corner

Q. A car’s brakes bring it to a halt over 10 metres when travelling at 30 km/h. Roughly how far will the same car take to stop if it is travelling at 60 km/h — twice, or four times as far? A. Four times as far, so roughly 40 m. The stopping distance works out to \(d_s = -v_0^2/2a\), which means it is proportional to the square of the initial speed rather than to the speed itself. With the same braking, doubling the initial speed therefore multiplies the stopping distance by four.

For a car of a particular make, actual braking distances of 10 m, 20 m, 34 m and 50 m were recorded at speeds of 11, 15, 20 and 25 m/s respectively — which agree closely with the \(v_0^2\) formula.

Real-World Application

Speed limits in school zones, hospital areas, and near sharp bends are set precisely because stopping distance grows so rapidly with speed. Halving the speed limit (say, from 50 km/h to 25 km/h) shrinks the stopping distance not by half but by three-quarters. That is why speed limits are far more effective at preventing accidents than they might first appear.

Numerical 2.2

A car travelling at \(54\,\text{km h}^{-1}\) can be brought to rest in \(15\,\text{m}\) by applying brakes. What deceleration does this require, and what would the stopping distance become if the initial speed were doubled to \(108\,\text{km h}^{-1}\) (with the same braking system)?

Solved Example 2.7

Reaction time. When a situation demands our immediate action, some time passes before we actually respond. This is called the reaction time — the time a person takes to observe, think, and act. For example, if a person is driving and a boy suddenly steps into the road, the time elapsed before the driver slams the brakes is the reaction time. It depends on the complexity of the situation and on the individual.

You can measure your own reaction time by a simple experiment (Fig. 2.8). Ask a friend to hold a ruler vertically and let it drop without warning through the gap between your thumb and forefinger. Catch it as fast as you can, and note the distance \(d\) the ruler has fallen. In a particular trial, \(d\) was found to be \(21.0\,\text{cm}\). Estimate the reaction time.

[Diagram: Fig. 2.8 – A ruler held vertically by a friend at the top, being caught between the thumb and forefinger of the person below whose reaction time is being measured. Distance d is marked from the initial top of the ruler to where the person catches it.]

Answer

The ruler falls under free fall, so \(v_0 = 0\) and \(a = -g = -9.8\,\text{m s}^{-2}\). The distance fallen \(d\) and the reaction time \(t_r\) are related by the second kinematic equation (taking downward distance as positive here):

\[d = \tfrac{1}{2} g\, t_r^2 \quad \Rightarrow \quad t_r = \sqrt{\frac{2 d}{g}}\]

Substituting \(d = 21.0\,\text{cm} = 0.21\,\text{m}\) and \(g = 9.8\,\text{m s}^{-2}\),

\[t_r = \sqrt{\frac{2 \times 0.21}{9.8}} \approx 0.2\,\text{s}\]

A typical human reaction time to a visual cue is around 0.2 s. In driving, even this short delay corresponds to a considerable “blind” distance: at 60 km/h (about 17 m/s), a car covers roughly 3.4 m during the driver’s 0.2 s reaction — before the brakes have even started to act. This “reaction distance” adds to the stopping distance of Example 2.6, and together they set safe following distances between vehicles.

Numerical 2.3

A driver with a reaction time of \(0.25\,\text{s}\) is travelling at \(72\,\text{km h}^{-1}\) when a pedestrian suddenly steps onto the road \(25\,\text{m}\) ahead. If the car can decelerate at \(5\,\text{m s}^{-2}\) once the brakes are applied, will the driver be able to stop in time? (Add the reaction distance and the braking distance.)

2.5 Relative Velocity

Watch a train pull out of a station while another train alongside stands still. If you glance out of the window at just the right moment, you may not be sure for a second whether your train is moving or the other one is. Now watch two trains speed past each other on parallel tracks in opposite directions — even if each is only doing 60 km/h, they appear to whip past one another so fast that the whole other train blurs across the window in a heartbeat.

Both experiences point to the same idea: how fast something appears to move depends on who is watching. Velocity is always measured with respect to some reference — a stationary observer, the ground, another vehicle. The velocity of one object as measured from another moving object is called its relative velocity.

Defining relative velocity

Consider two objects, A and B, moving along the same straight line. Let their velocities, measured relative to the ground, be \(v_A\) and \(v_B\) respectively. The velocity of A as seen by B — that is, the rate at which A’s position appears to change from B’s frame of reference — is called the velocity of A relative to B, written \(v_{AB}\), and is given by

\[v_{AB} = v_A - v_B \tag{2.10}\]

Similarly, the velocity of B relative to A is

\[v_{BA} = v_B - v_A = -\,v_{AB}\]

So the two relative velocities are equal in magnitude but opposite in direction — which makes sense: if A moves away from B at some rate, then from A’s viewpoint B must be moving away from A at exactly the same rate, but in the opposite direction.

Definition

The relative velocity of object A with respect to object B, both moving along a straight line, is the velocity of A as measured from B’s frame of reference: \(v_{AB} = v_A - v_B\), where \(v_A\) and \(v_B\) are the velocities of A and B measured relative to the ground.

Three important cases

To see what Eq. (2.10) really tells us, consider three natural situations.

Case 1 — A and B move with the same velocity (\(v_A = v_B\)). Then

\[v_{AB} = v_A - v_B = 0\]

From B’s viewpoint, A does not move at all — it appears fixed in place, even though both are actually moving relative to the ground. This is exactly why, when your car and another car cruise at the same speed on a highway, the other car seems to hang next to your window as if the two of you were parked.

Case 2 — A and B move in the same direction with different speeds (say \(v_A > v_B > 0\)). Then

\[v_{AB} = v_A - v_B > 0\]

A slowly gains on B. From B’s viewpoint, A appears to move ahead at the small speed \(v_A - v_B\). This is why an express train overtaking a passenger train on a parallel track seems to creep past, even though both trains are moving fast over the ground.

Case 3 — A and B move in opposite directions (say \(v_A > 0\) and \(v_B < 0\)). With \(v_B\) negative, the subtraction turns into an addition of magnitudes:

\[v_{AB} = v_A - v_B = v_A + |v_B|\]

The relative speed is the sum of the two individual speeds. This is why two trains passing each other in opposite directions on parallel tracks seem to blur past — if each is doing 60 km/h, each sees the other rushing by at 120 km/h.

Curiosity Corner

Q. If two trains move in opposite directions on parallel tracks, why does each appear to rush past the other far faster than either is actually moving? A. Because what each observer sees is the relative velocity \(v_{AB} = v_A - v_B\), and for opposite directions one of the two velocities is negative, so the subtraction turns into an addition of magnitudes: \(v_{AB} = v_A + |v_B|\). The relative speed is then the sum of the two individual speeds — two trains each doing 60 km/h see one another go by at 120 km/h.

Figure to come

Fig. 2.9 – Three side-by-side position–time (x–t) graphs for two objects A and B on the same axis: (a) parallel lines of equal slope (same velocity, relative velocity zero); (b) two lines both sloping upward with A’s slope steeper than B’s (same direction, A faster than B); (c) two lines with opposite slopes (opposite directions), crossing at the point where they meet.

Reading the graphs

The three cases show up clearly on position–time graphs (Fig. 2.9):

If A and B have the same velocity, their x–t graphs are parallel lines. They never meet — their separation stays constant.
If A and B move in the same direction but A is faster, their x–t lines both slope up, but A’s slope is steeper. Sooner or later A will catch up and overtake B, and the two lines will cross at that overtaking instant.
If A and B move in opposite directions, one line slopes up while the other slopes down. They will cross when the two objects meet — after which, if the motion continues, they draw apart on either side.

Some worked applications

Solved Example 2.8

Two parallel rail tracks run east-west. Train A moves east with a speed of \(54\,\text{km h}^{-1}\), and train B moves west with a speed of \(90\,\text{km h}^{-1}\). Find (a) the velocity of B with respect to A, and (b) the velocity of a monkey running north-to-south on the roof of A with a speed of \(18\,\text{km h}^{-1}\) (with respect to A), as observed by a passenger sitting in B.

Answer

First, convert every speed to m s⁻¹ for consistency:

\[54\,\text{km h}^{-1} = 15\,\text{m s}^{-1}, \quad 90\,\text{km h}^{-1} = 25\,\text{m s}^{-1}, \quad 18\,\text{km h}^{-1} = 5\,\text{m s}^{-1}\]

Take east as the positive direction. Then \(v_A = +15\,\text{m s}^{-1}\) and \(v_B = -25\,\text{m s}^{-1}\).

Velocity of B with respect to A:

\[v_{BA} = v_B - v_A = -25 - 15 = -40\,\text{m s}^{-1}\]

The magnitude is \(40\,\text{m s}^{-1}\), and the negative sign means, as seen from A, train B moves westward at \(40\,\text{m s}^{-1}\). (Equivalently, \(144\,\text{km h}^{-1}\) — much faster than either train’s own speed over the ground.)

Along the east–west line, the monkey has no velocity (it runs north-to-south, perpendicular to the tracks). So along the east direction, the monkey’s velocity equals A’s velocity: \(+15\,\text{m s}^{-1}\). From the passenger in B’s viewpoint, along the east direction the monkey moves at

\[v_{\text{monkey},B} = v_{\text{monkey}} - v_B = 15 - (-25) = 40\,\text{m s}^{-1} \text{ (east)}\]

In the north–south direction, B is not moving, so the passenger simply sees the monkey moving south at \(5\,\text{m s}^{-1}\). (In this chapter we treat only the east–west component; the full vector picture comes in Chapter 3.)

Numerical 2.4

A car travelling north at \(20\,\text{m s}^{-1}\) is being overtaken by a truck also travelling north at \(25\,\text{m s}^{-1}\). (a) What is the velocity of the truck relative to the car? (b) At the instant they are level, the truck is 15 m behind another vehicle ahead of them. How long does the truck take to reach that vehicle, if that vehicle is stationary?

Real-World Application

Air traffic controllers, ship navigators, and even space-mission planners routinely use the idea of relative velocity. When two aircraft are on converging paths, what matters for safety is not each plane’s ground speed, but the rate at which the distance between them is shrinking — i.e., the relative velocity of one aircraft with respect to the other. The same idea determines how a spacecraft docks with a space station: the docking manoeuvre is planned so that the relative velocity between the two shrinks to nearly zero at the moment of contact.

Special case — one object at rest

If B is at rest (\(v_B = 0\)), then Eq. (2.10) gives \(v_{AB} = v_A\). In other words, when we measured \(v_A\) “relative to the ground” all along in this chapter, we were quietly using the ground as our stationary reference. Every velocity we have written so far is, secretly, a relative velocity — relative to the ground. What Section 2.5 does is generalise this idea: any moving object can serve as a reference frame, not just the ground.

Quick Question

Is there a “true” or “absolute” velocity of an object, independent of any observer?

No — this is one of the deepest lessons of classical mechanics. Every velocity is measured relative to some reference; there is no privileged frame that gives “absolute” velocities. Even the ground beneath us is spinning with the Earth, which orbits the Sun, which orbits the galactic centre. What matters in physics is usually the relative velocity between two objects, and that is well-defined regardless of the observer’s frame.

With this, we have completed the language of rectilinear kinematics: position, velocity (both instantaneous and relative), acceleration, and the equations that tie them together for constant acceleration. In the next chapter, we will extend all of these ideas from straight-line motion to motion in a plane, where velocities and accelerations must be treated as full vectors.

2.6 Summary

Motion. An object is said to be in motion if its position changes with time. The position of the object is always specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. Over any given time interval, the average speed of an object is greater than or equal to the magnitude of its average velocity.
Instantaneous velocity. Instantaneous velocity (or simply velocity) is defined as the limit of the average velocity as the time interval \(\Delta t\) becomes infinitesimally small:

\[v = \lim_{\Delta t \to 0} \bar{v} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}\]

The velocity at any particular instant equals the slope of the tangent drawn to the position–time graph at that instant.

Average acceleration. Average acceleration is the change in velocity divided by the time interval during which the change occurs:

\[\bar{a} = \frac{\Delta v}{\Delta t}\]

Instantaneous acceleration. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval \(\Delta t\) goes to zero:

\[a = \lim_{\Delta t \to 0} \bar{a} = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}\]

The acceleration at any instant equals the slope of the tangent drawn to the velocity–time graph at that instant. For uniform motion, acceleration is zero, the \(x\)–\(t\) graph is a straight line inclined to the time axis, and the \(v\)–\(t\) graph is a straight line parallel to the time axis. For motion with uniform acceleration, the \(x\)–\(t\) graph is a parabola, while the \(v\)–\(t\) graph is a straight line inclined to the time axis.

Area under the v–t curve. The area under the velocity–time curve between times \(t_1\) and \(t_2\) is equal to the displacement of the object during that interval of time.
Kinematic equations. For objects in uniformly accelerated rectilinear motion, the five quantities — displacement \(x\), time taken \(t\), initial velocity \(v_0\), final velocity \(v\), and acceleration \(a\) — are related by a set of simple equations called the kinematic equations of motion:

\[v = v_0 + a t\]

\[x = v_0 t + \tfrac{1}{2} a t^2\]

\[v^2 = v_0^2 + 2 a x\]

These forms assume that the object is at position 0 at \(t = 0\). If the object instead starts from a non-zero position \(x_0\), then \(x\) in the above equations is simply replaced by \((x - x_0)\).

Relative velocity. In a straight line, the velocity of an object A as seen from another object B is the relative velocity of A with respect to B, given by \(v_{AB} = v_A - v_B\), where both \(v_A\) and \(v_B\) are measured relative to the ground. For objects moving in opposite directions, their relative speed adds; for objects moving in the same direction with the same velocity, their relative velocity is zero.

Summary Table of Physical Quantities

Physical Quantity	Symbol	Dimensions	SI Unit	Remarks
Path length	—	\([\text{L}]\)	m	Total distance covered along the actual path.
Displacement	\(\Delta x\)	\([\text{L}]\)	m	\(= x_2 - x_1\). In one dimension, its sign indicates the direction.
Velocity — Average	\(\bar{v}\)	\([\text{L T}^{-1}]\)	m s⁻¹	\(= \dfrac{\Delta x}{\Delta t}\)
Velocity — Instantaneous	\(v\)	\([\text{L T}^{-1}]\)	m s⁻¹	\(= \lim\limits_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}\). In one dimension, its sign indicates the direction.
Speed — Average	—	\([\text{L T}^{-1}]\)	m s⁻¹	\(= \dfrac{\text{Path length}}{\text{Time interval}}\)
Speed — Instantaneous	—	\([\text{L T}^{-1}]\)	m s⁻¹	\(= \left\|\dfrac{dx}{dt}\right\|\)
Acceleration — Average	\(\bar{a}\)	\([\text{L T}^{-2}]\)	m s⁻²	\(= \dfrac{\Delta v}{\Delta t}\)
Acceleration — Instantaneous	\(a\)	\([\text{L T}^{-2}]\)	m s⁻²	\(= \lim\limits_{\Delta t \to 0} \dfrac{\Delta v}{\Delta t} = \dfrac{dv}{dt}\). In one dimension, its sign indicates the direction.

2.7 Points to Ponder

Choosing the axis comes first. The origin and the positive direction of an axis are a matter of choice — nothing in nature dictates them. You must first specify this choice clearly before you assign signs to quantities like displacement, velocity, and acceleration. A different choice of positive direction will flip the signs of these quantities, but the physics of the motion stays the same.
What “speeding up” and “slowing down” really mean. If a particle is speeding up, its acceleration is in the same direction as its velocity. If its speed is decreasing, acceleration is in the direction opposite to that of velocity. This rule is independent of the choice of origin and axis — it is a purely physical statement.
Sign of acceleration does not tell you speeding up or slowing down. By itself, the sign of \(a\) (positive or negative) only tells you the direction of acceleration relative to your chosen positive axis — nothing more. For example, if we choose upward as positive, the acceleration due to gravity is negative. A ball falling under gravity has negative acceleration but increasing speed (since velocity is also negative — both point downward). The same ball thrown upward has the same negative acceleration but decreasing speed (velocity is positive; the two point in opposite directions). Always check the directions of \(a\) and \(v\) together, not just the sign of \(a\).
Zero velocity does not mean zero acceleration. An object can be momentarily at rest and yet accelerating. The clearest example is a ball thrown vertically upward: at the topmost point of its flight its velocity is zero, but the acceleration due to gravity is still \(g\) downward, acting on it every instant. Being at rest for an instant is not the same as being at rest for a duration.
The kinematic equations are algebraic. In Eqs. (2.9), every quantity — \(x\), \(v_0\), \(v\), \(a\), and \(t\) — is an algebraic quantity that can be positive or negative. The equations are valid in all situations of one-dimensional motion with constant acceleration, provided you substitute each quantity with the correct sign as per your chosen positive direction. Most errors in kinematics problems come from careless signs, not wrong formulas.
Exact definitions vs. limited-scope equations. The definitions of instantaneous velocity and acceleration (Eqs. 2.1 and 2.3) are exact, holding for every kind of motion — uniform, non-uniform, jerky, smooth, anything. The kinematic equations (Eq. 2.9), on the other hand, are true only for motion in which the acceleration is constant in both magnitude and direction throughout the interval. If acceleration varies, these equations do not apply — you must return to calculus (as in Example 2.2) and integrate directly.

2.8 NCERT Questions

In which of the following examples of motion, can the body be considered approximately a point object:
1. a railway carriage moving without jerks between two stations.
2. a monkey sitting on top of a man cycling smoothly on a circular track.
3. a spinning cricket ball that turns sharply on hitting the ground.
4. a tumbling beaker that has slipped off the edge of a table.
The position–time (\(x\)–\(t\)) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below:
1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/different) time
5. (A/B) overtakes (B/A) on the road (once/twice).
A woman starts from her home at 9:00 am, walks with a speed of \(5\,\text{km h}^{-1}\) on a straight road up to her office \(2.5\,\text{km}\) away, stays at the office up to 5:00 pm, and returns home by an auto with a speed of \(25\,\text{km h}^{-1}\). Choose suitable scales and plot the \(x\)–\(t\) graph of her motion.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is \(1\,\text{m}\) long and requires \(1\,\text{s}\). Plot the \(x\)–\(t\) graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit \(13\,\text{m}\) away from the start.
A car moving along a straight highway with a speed of \(126\,\text{km h}^{-1}\) is brought to a stop within a distance of \(200\,\text{m}\). What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
A player throws a ball upwards with an initial speed of \(29.4\,\text{m s}^{-1}\).
1. What is the direction of acceleration during the upward motion of the ball?
2. What are the velocity and acceleration of the ball at the highest point of its motion?
3. Choose the \(x = 0\,\text{m}\) and \(t = 0\,\text{s}\) to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of the \(x\)-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
4. To what height does the ball rise, and after how long does the ball return to the player’s hands? (Take \(g = 9.8\,\text{m s}^{-2}\) and neglect air resistance.)
Read each statement below carefully and state with reasons and examples, if it is true or false. A particle in one-dimensional motion:
1. with zero speed at an instant may have non-zero acceleration at that instant.
2. with zero speed may have non-zero velocity.
3. with constant speed must have zero acceleration.
4. with positive value of acceleration must be speeding up.
A ball is dropped from a height of \(90\,\text{m}\) on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed–time graph of its motion between \(t = 0\) to \(12\,\text{s}\).
Explain clearly, with examples, the distinction between:
1. magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
2. magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval.] Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only.]
A man walks on a straight road from his home to a market \(2.5\,\text{km}\) away with a speed of \(5\,\text{km h}^{-1}\). Finding the market closed, he instantly turns and walks back home with a speed of \(7.5\,\text{km h}^{-1}\). What is the:
1. magnitude of average velocity, and
2. average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Figure 2.11 shows the \(x\)–\(t\) plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for \(t < 0\) and on a parabolic path for \(t > 0\)? If not, suggest a suitable physical context for this graph.
A police van moving on a highway with a speed of \(30\,\text{km h}^{-1}\) fires a bullet at a thief’s car speeding away in the same direction with a speed of \(192\,\text{km h}^{-1}\). If the muzzle speed of the bullet is \(150\,\text{m s}^{-1}\), with what speed does the bullet hit the thief’s car? (Note: Obtain the speed which is relevant for damaging the thief’s car.)
Suggest a suitable physical situation for each of the following graphs (Fig. 2.12).
Figure 2.13 gives the \(x\)–\(t\) plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13.) Give the signs of position, velocity, and acceleration variables of the particle at \(t = 0.3\,\text{s}, 1.2\,\text{s}, -1.2\,\text{s}\).
Figure 2.14 gives the \(x\)–\(t\) plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
Figure 2.15 gives a speed–time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of \(v\) and \(a\) in the three intervals. What are the accelerations at the points A, B, C, and D?

2.9 Check Your Concepts

An object is thrown vertically upward from the top of a tall tower and eventually falls to the ground below. Sketch the acceleration–time graph for its entire motion, from the moment it leaves the hand to the moment it hits the ground (ignore air resistance). Explain why the graph looks the way it does.
Two students argue about a car that is decelerating uniformly to a stop. Student P says: “The car’s acceleration is negative because it is slowing down.” Student Q says: “That depends on which direction we choose as positive.” Who is right, and why? Give a clear physical example to justify your reasoning.
A particle moves along a straight line and its position–time graph is a parabola opening downward. What can you say about (a) the sign of its acceleration, and (b) how its speed changes with time? Are there any instants when the particle is momentarily at rest? Explain.
Explain why the kinematic equations of motion (Eqs. 2.9) fail to describe the motion of a car whose driver keeps varying the pressure on the accelerator. What alternative method must one use in such a case?
A stone dropped from a bridge takes \(t\) seconds to hit the river below. If a second, identical stone is thrown downward from the same bridge with an initial speed \(v_0\), will it take exactly \(t/2\) seconds to reach the river if \(v_0\) equals the final velocity of the first stone divided by 2? Justify your answer without doing the full calculation, using the shape of the position–time graph.
A passenger inside a moving train drops a coin. Describe the motion of the coin as seen by (a) the passenger, and (b) a person standing on the platform. Are the two descriptions equally valid? Which frame of reference should you use to apply the kinematic equations from this chapter?
The stopping distance of a vehicle depends on the square of its initial speed. Using this fact, argue why speed limits in school zones (typically \(25\)–\(30\,\text{km h}^{-1}\)) offer far greater safety than one might naively expect compared to a \(50\,\text{km h}^{-1}\) limit. Estimate roughly how much the stopping distance shrinks in going from \(50\) to \(25\,\text{km h}^{-1}\).
A car and a truck start from rest at the same instant and move along the same straight road with constant, but different, accelerations. Their velocity–time graphs are both straight lines through the origin. Can the car overtake the truck if the truck has a greater acceleration? At what instant do they have the same velocity, and at what instant do they meet again? Answer using the geometry of the \(v\)–\(t\) graphs, not by computing.
In an \(x\)–\(t\) graph, a straight line inclined at \(45°\) to the time axis represents a certain velocity in a chosen unit system. If the units on the axes are changed (say, seconds to minutes, or metres to kilometres), does the \(45°\) inclination still represent the same velocity? Discuss.
At the top of its flight, a ball thrown vertically upward has zero velocity but non-zero acceleration. A student concludes: “So at that instant, the direction of motion is undefined.” Is this conclusion correct? Answer carefully, distinguishing between velocity, direction of motion, and acceleration.

2.10 Practice with Numericals

The position of a particle moving along the x-axis is given by \(x(t) = 2 t^3 - 15 t^2 + 36 t + 5\) (with \(x\) in metres and \(t\) in seconds). Find (a) the velocity and acceleration at \(t = 3\,\text{s}\), (b) the instants at which the particle is momentarily at rest, and (c) the total distance covered by the particle between \(t = 0\) and \(t = 5\,\text{s}\).
A stone is dropped from the top of a cliff. During the last second of its fall, it covers a distance of \(25\,\text{m}\). Taking \(g = 10\,\text{m s}^{-2}\), find the height of the cliff and the total time of fall.
A car travelling on a straight highway at \(72\,\text{km h}^{-1}\) sees a red signal \(50\,\text{m}\) ahead. The driver’s reaction time is \(0.6\,\text{s}\), after which he applies the brakes producing a uniform retardation of \(5\,\text{m s}^{-2}\). Does the car stop before reaching the signal? By what distance does it miss (or overshoot) the signal?
Two trains, each \(120\,\text{m}\) long, run on parallel tracks. Train A moves north at \(54\,\text{km h}^{-1}\), and train B moves south at \(36\,\text{km h}^{-1}\). How long does it take for the two trains to completely pass each other from the moment their fronts meet?
A ball is thrown vertically upward from ground level with an initial speed of \(25\,\text{m s}^{-1}\). Taking \(g = 10\,\text{m s}^{-2}\), find (a) the maximum height reached, (b) the total time of flight, (c) the velocity of the ball when it is at half its maximum height (both on the way up and on the way down), and (d) the average velocity of the ball for the first \(2\,\text{s}\) of its motion.
A body starts from rest and moves with uniform acceleration. It covers a distance of \(x_1\) in the first \(5\) seconds and a distance of \(x_2\) in the next \(5\) seconds. Show that \(x_2 = 3 x_1\). Then, using \(x_1 = 20\,\text{m}\), find the acceleration of the body and its velocity at \(t = 10\,\text{s}\).
A parachutist bails out of an aircraft and falls freely (no air resistance) for \(2\,\text{s}\) before opening the parachute. After the parachute opens, she decelerates uniformly at \(2\,\text{m s}^{-2}\). If she reaches the ground with a velocity of \(3\,\text{m s}^{-1}\), find (a) the height from which she bailed out, and (b) the total time of her descent. Take \(g = 10\,\text{m s}^{-2}\).
A car accelerates uniformly from rest at \(2\,\text{m s}^{-2}\) for \(10\,\text{s}\), moves at the acquired constant speed for the next \(30\,\text{s}\), and then decelerates uniformly at \(4\,\text{m s}^{-2}\) until it comes to rest. Sketch the velocity–time graph, and calculate (a) the total distance travelled, and (b) the average speed for the whole journey.
Two cars A and B are travelling on the same straight road. Car A is initially \(200\,\text{m}\) behind car B; A moves with a constant velocity of \(30\,\text{m s}^{-1}\), while B moves with a constant velocity of \(20\,\text{m s}^{-1}\) in the same direction. After what time will A catch up with B, and how far will A have travelled by then?
A ball is dropped from a height of \(45\,\text{m}\). At the same instant, another ball is thrown vertically upward from the ground with an initial speed of \(30\,\text{m s}^{-1}\). Taking \(g = 10\,\text{m s}^{-2}\), find (a) the time at which the two balls meet, and (b) the height above the ground at which they meet.
A particle’s velocity along a straight line is given by \(v(t) = 4 - 2 t\) (with \(v\) in \(\text{m s}^{-1}\) and \(t\) in \(\text{s}\)). Starting from \(x = 0\) at \(t = 0\), find (a) the time at which the particle momentarily comes to rest, (b) its position at that instant, (c) the total distance covered in the first \(4\) seconds, and (d) its displacement over the same interval.
A train of length \(150\,\text{m}\) is moving at \(54\,\text{km h}^{-1}\) when the driver applies the brakes, producing a uniform retardation of \(0.5\,\text{m s}^{-2}\). Find (a) the total distance travelled by the train before it stops, and (b) the time taken by the rear end of the train to cross a signal post that was \(200\,\text{m}\) ahead of the front of the train when the brakes were first applied.