Chapter 2: Motion In One Dimension

In the early 1600s, an Italian teacher of mathematics named Galileo Galilei wanted to settle an old argument. Everyone agreed that a falling body speeds up. Nobody could say by how much, because a falling stone passes too quickly to be timed by hand.

So Galileo slowed the fall down. He took a long wooden plank, cut a smooth straight groove along it, lined the groove, and propped one end up on a small block. Then he let a polished bronze ball roll down the groove, again and again, and measured how long each run took. For a clock he used water: a vessel of water drained through a thin pipe while the ball rolled, and he weighed the water collected. More water meant more time.

Then he began releasing the ball from different points along the groove. The result was the same every time, whatever the slope, whatever the ball. In the first stretch of time the ball covered one unit of the groove. In twice that time it covered four units. In three times that time, nine units. One, four, nine — the squares of the numbers.

That single pattern is the seed of this chapter. A ball rolling down a plank, a stone dropped from a roof and a train pulling out of a station are all doing the same thing, and it can be written in a few short relations and drawn as a few simple graphs.

Diagram

Fig. 2.0-1 — Galileo’s Inclined Plane Experiment Side view of a long wooden plank with one end raised on a small block, a straight groove running down its upper surface, and a small bronze ball part way down the groove with an arrow showing it rolling downwards; mark the release point at the top and three marks along the groove at distances of 1 unit, 4 units and 9 units from the release point, labelled ‘after 1 unit of time’, ‘after 2 units of time’, ‘after 3 units of time’; beside the plank draw a water vessel on a stand with a thin outlet pipe dripping into a small cup, labelled ‘water clock’.

Curiosity Corner

Q1. Nobody pushes the ball again once it is released, yet it covers more of the groove in each second than in the one before. What is being added to the ball’s motion, second after second? Q2. Galileo could measure only two things — how far the ball went and how long it took. From just these two, how would you work out how fast the ball was moving at one particular moment? Q3. If a heavy ball and a light ball are released together down the same groove, which reaches the bottom first — and would your answer change if they were simply dropped through empty space instead? Q4. A runner goes once round a circular track and stops exactly where she started. She is certainly tired, so something has been covered. Yet in another sense she has gone nowhere. Which two different quantities are being described here? Q5. Suppose you plot how far the ball has rolled against the time taken. What would the steepness of the line at any point be telling you about the ball?

2.1 Scalar And Vector Quantities

Physics begins with measurement. Anything about the world that can be measured — how heavy a bag is, how long a wire is, how hot a room is, how hard you pull — is called a physical quantity. A quantity that cannot be measured, such as anger or beauty, has no place in a physics equation.

Now measure two different things and compare them. If you say a bag has a mass of $5.0\ \text{kg}$, you have said everything there is to say about that mass. But if you say a body has been shifted through $5\ \text{m}$, a listener will immediately ask a further question: shifted which way? The shift is not fully described until you add a direction.

That difference splits every physical quantity into two broad families. Some are complete once you state a number and a unit. Others are complete only when you also state a direction. This is shown in Fig. 2.1-1.

Diagram

Fig. 2.1-1 — Why Some Quantities Need A Direction Two side-by-side panels. Left panel: a laboratory balance with a block on the pan and the reading ‘5.0 kg’ shown clearly; caption text inside the panel ‘a number and a unit are enough’. Right panel: a single starting point marked P with two arrows of equal length drawn from it, one pointing east labelled ‘5 m towards east’ and one pointing north labelled ‘5 m towards north’, with a compass rose showing N, E, S, W; caption text inside the panel ‘same number, same unit, different shift’.

2.1.1 Scalar Quantities

A quantity of the first family is called a scalar quantity, or simply a scalar — a quantity that is completely expressed by its magnitude alone.

Definition

Scalar Quantity A physical quantity that is expressed only by its magnitude, that is, by a numerical value together with the unit in which it is measured.

To state a scalar completely, only two things are needed: 1. the unit in which the quantity is measured, and 2. the numerical value of the measurement in that unit.

Mass, length, time, distance, density, volume, speed, temperature, work, energy, power, pressure, quantity of heat, specific heat, charge, electric power, resistance, mechanical advantage, frequency and angle are all scalars. Gravitational potential, magnetic potential and electric potential are scalars too.

There is one small exception worth noting. Some scalars are pure numbers with nothing being measured — $\pi$, or $e^2$, for instance. A pure number carries no unit at all.

How Scalars Are Handled

Scalars behave in calculations exactly as ordinary numbers do. Two scalars of the same kind can be added, subtracted, multiplied and divided by simple arithmetic. Add $3\ \text{kg}$ of rice to $2\ \text{kg}$ of rice and you have $5\ \text{kg}$ of rice, with nothing further to think about.

In symbols, a scalar is written as a plain English letter: mass as $m$, time as $t$, speed as $v$.

Side Note

The word ‘scalar’ comes from the Latin scala, a ladder or a scale. A scalar is a quantity you can read straight off a scale — a balance, a thermometer, a measuring jar — with nothing left over to describe.

2.1.2 Vector Quantities

A quantity of the second family is called a vector quantity, or simply a vector. Its meaning is incomplete until a direction is attached.

Definition

Vector Quantity A physical quantity that requires both a magnitude and a direction to be expressed completely.

To state a vector completely, three things are needed: 1. the numerical value of the quantity, 2. the unit in which it is measured, and 3. the direction in which it acts.

Take the earlier instruction again. ‘Shift the particle by $5\ \text{m}$’ cannot be carried out, because a thousand different shifts fit that description. ‘Shift the particle by $5\ \text{m}$ towards the east’ can be carried out in exactly one way.

Displacement, velocity, acceleration, momentum, force, moment of a force (also called torque), impulse, weight, temperature gradient, electric field, magnetic field and dipole moment are all vectors.

Magnitude, Sign And Direction

The numerical value of a vector together with its unit gives the magnitude of that vector — its size, leaving the direction aside. A magnitude is always positive; a force of size $-20\ \text{N}$ is not a smaller-than-nothing force.

So what does a negative sign on a vector mean? It means the reverse direction. If $\vec{F}$ is a push towards the east, then $-\vec{F}$ is a push of exactly the same size towards the west. The sign is not part of the size; it is part of the direction.

Signs and Units

Vector quantities Magnitude: always positive, and always carries a unit. Sign: tells direction only. A minus sign means ‘the same size, the opposite way’. Choose one direction as positive at the start of a problem, and keep that choice all the way through.

How Vectors Are Handled

Because a vector carries a direction, two vectors cannot simply be added like two numbers. Walk $3\ \text{m}$ east and then $4\ \text{m}$ east and you are $7\ \text{m}$ from the start; walk $3\ \text{m}$ east and then $4\ \text{m}$ north and you are only $5\ \text{m}$ from it. Vector quantities therefore follow their own rules of addition, subtraction and multiplication, different from ordinary arithmetic.

In symbols, a vector is written either as an English letter with an arrow above it, or as an English letter in bold type. Velocity is written $\vec{v}$ or $\mathbf{v}$, acceleration $\vec{a}$ or $\mathbf{a}$, and force $\vec{F}$ or $\mathbf{F}$.

The two families are set side by side in Table 2.1.

Compare Table

Table 2.1 — Scalar Quantities Compared With Vector Quantities | Basis of difference | Scalar quantity | Vector quantity | |—|—|—| | What describes it fully | Magnitude alone | Magnitude together with direction | | Parameters needed | Two: unit, and numerical value | Three: numerical value, unit, and direction | | Meaning of a negative sign | The quantity itself is less than zero | The same magnitude in the opposite direction | | Rule for combining | Ordinary arithmetic | Special rules of vector algebra | | How it is written | Plain letter, such as $m$, $t$, $v$ | Letter with an arrow, such as $\vec{F}$, or in bold type | | Examples | Mass, time, distance, speed, energy, temperature | Displacement, velocity, acceleration, force, momentum |

Example 2.1

Sorting Quantities Into Scalars And Vectors

1. Given: the quantities pressure, momentum, temperature, weight, frequency and impulse. Asked: which are scalars and which are vectors.

The test used is the definition itself. Ask of each quantity: would a direction have to be stated before the quantity is fully described? If not, it is a scalar; if yes, it is a vector.
Pressure, measured in Pa, is stated as a number alone — scalar. Momentum, in kg m s⁻¹, belongs to a body moving some particular way — vector. Temperature, in °C or K, needs no direction — scalar. Weight is the pull of gravity on a body, in N, and that pull acts vertically downwards — vector. Frequency, in Hz, is a count of events per second — scalar. Impulse, in N s, acts along the direction of the force that produced it — vector.
Scalars: pressure, temperature, frequency. Vectors: momentum, weight, impulse. The grouping is sensible, since every quantity placed in the vector list is one you could point at with your finger.

Try Yourself

Ask someone to bring you a book kept $3\ \text{m}$ away, saying only that. Then repeat, saying ‘$3\ \text{m}$ towards the window’. Notice which instruction can be carried out at once. You have just demonstrated the difference between the magnitude of a vector and the complete vector.

2.2 Rest And Motion

Look around and you will find very little standing still. Birds fly, buses move, people walk, insects crawl, animals run. The earth itself travels round the sun, carrying every one of those objects with it. The sun and the stars circle the centre of their galaxy, and the galaxies themselves are not at rest either.

So in the strictest sense, nothing in the universe is truly at rest. Yet we say a stone lying on the ground is at rest, and we are not wrong to say it — while a stone rolling down a slope is in motion. What we mean is that the stone does not change its position compared with the ground, the grass and the wall beside it.

2.2.1 Rest And Motion Are Relative

Sit on a railway platform and look at a tree nearby. The tree is at rest, you say — and it is, because it stays exactly where it is compared with the platform and with you. Now look at a train pulling out of the station. That train is in motion, because it keeps changing its position compared with you and with the platform. The situation is drawn in Fig. 2.2-1.

Diagram

Fig. 2.2-1 — Rest And Motion Judged From The Platform A railway platform scene drawn from the side. On the left, a girl standing on the platform with a tree beside her; both are labelled ‘position unchanged with respect to the observer — at rest’. On the right, a train on the track with a horizontal arrow showing it moving away, drawn in two faint positions to show it has shifted, labelled ‘position changing with respect to the observer — in motion’. Add a dashed vertical reference line from the observer to the track to show the position being compared against.

In each case the judgement is made against the things immediately around the body. That is what the definition rests on.

Definition

Rest And Motion A body is said to be at rest if it does not change its position with respect to its immediate surroundings.

A body is said to be in motion if it changes its position with respect to its immediate surroundings.

Side Note

You are never as still as you feel. The earth carries you round the sun at about $30\ \text{km s}^{-1}$, while spinning on its own axis as well. You do not notice any of it, because everything immediately around you — the room, the ground, the air — is moving along with you.

Treating A Body As A Point

Real bodies have size, and different parts of a body are at different places. That would make motion very awkward to describe. There is, however, a simple escape. If the distance a body travels in a certain time interval is much larger than the body itself, the whole body may be treated as a single point, called a point particle.

A train $200\ \text{m}$ long that runs $50\ \text{km}$ between two stations is, on that scale, a dot moving along a line. Throughout this chapter we shall describe the motion of a body by assuming it to be such a point particle.

2.2.2 One Dimensional Motion

Motion becomes simplest of all when the path is a straight line. Such motion is called one dimensional motion.

Definition

One Dimensional Motion When a body moves along a straight line path, its motion is said to be one dimensional motion. It is also called motion in a straight line, or rectilinear motion.

A train running on a straight track, a stone falling vertically downwards under gravity, and a car moving along a long straight road are all in one dimensional motion. In such a motion the body has no sideways movement at all — no movement in the lateral direction.

When a body moves on a plane along a curved path, its motion is two dimensional. When it moves through space, its motion is three dimensional. In this chapter we shall consider only one dimensional motion.

Example 2.2

Recognising One Dimensional Motion

1. Given: four motions — (a) a lift rising straight up a shaft, (b) a football curving through the air after a kick, (c) a bead sliding along a taut straight wire, (d) a boat crossing a river while drifting downstream. Asked: which of these are one dimensional.

The test used is the definition of one dimensional motion: the path traced out must be a straight line, with no sideways movement.
1. The lift shaft is straight and vertical, so the path is a straight line. (b) The football rises, curves over and falls, tracing a curved path in a vertical plane. (c) The wire is straight and the bead cannot leave it, so the path is a straight line. (d) The boat moves across the river and downstream at the same time, so its path lies over the surface of the water in two directions at once.
One dimensional: (a) and (c). Not one dimensional: (b) and (d), both of which are two dimensional. The answer is sensible, because in (a) and (c) a single measurement — how far along the shaft, or how far along the wire — fixes the position completely.

2.2.3 Representing One Dimensional Motion

Since the path of a one dimensional motion is a straight line, it can be represented by a straight line drawn parallel to the $X$-axis, with the $X$-axis taken along the direction of motion. Each point on that line represents the position of the particle at some instant.

The position of the particle at any instant $t$ is then given by stating its $x$ coordinate at that instant — that is, its distance from the chosen starting point, measured along the line. As the particle moves, its $x$ coordinate changes with time $t$. Nothing else about the particle changes, which is exactly what makes one dimensional motion easy to handle.

As an example, take a pebble falling freely and vertically downwards, and measure its position from the point where it started. The measurements are given in Table 2.2.

Table 2.2 — Position Of A Freely Falling Pebble At Different Instants

Time $t$ (in s)	0	1	2	3	4
Position $x$ (in m)	0	5	20	45	80

This motion is represented in Fig. 2.2-2. A convenient scale is chosen for $x$, and the positions are marked on a straight line drawn along the $X$-axis. Here the $X$-axis stands for the vertically downward direction, since that is the direction in which the pebble travels.

Diagram

Fig. 2.2-2 — Representation Of One Dimensional Motion A single horizontal straight line with an arrowhead at the right-hand end labelled X. Mark five points along it, spaced to a uniform scale of distance: at 0 m, 5 m, 20 m, 45 m and 80 m. Label each point above the line with its position value (x = 0, 5 m, 20 m, 45 m, x = 80 m) and below the line with the corresponding instant (at t = 0, at t = 1 s, at t = 2 s, at t = 3 s, at t = 4 s). Add a note beside the arrowhead: ‘X-axis taken along the vertically downward direction’.

Read the marks from left to right and one feature stands out at once. In the first second the pebble covers $5\ \text{m}$; in the next second it covers $15\ \text{m}$; in the third, $25\ \text{m}$; in the fourth, $35\ \text{m}$. The gaps grow steadily wider, which tells you the pebble is falling faster and faster as it goes.

Try Yourself

Sit beside a window in a moving bus or train and watch a parked vehicle outside. It seems to slide backwards. Now look at another vehicle travelling alongside at the same speed — it seems not to move at all. Note down what you are comparing each one against in the two cases.

Extra Practice 2.1

State whether each of the following is one, two or three dimensional motion, giving a reason in each case: (i) a ball rolling down a straight grooved plank, (ii) a fly buzzing about inside a room, (iii) a swimmer moving along a straight lane of a pool, (iv) a car going round a roundabout.

2.3 Distance And Displacement

Consider a body moving from a point $A$ to a point $B$ along the winding path drawn in Fig. 2.3-1. Two quite different questions can be asked about that journey. How much ground did the body cover on the way? And where did it finish, compared with where it began?

The first question is answered by the total length of the path from $A$ to $B$. That length is called the distance moved by the body. The second is answered by the straight line $AB$, taken in the direction from $A$ to $B$, shown dotted in the figure. That is called the displacement of the body.

Diagram

Fig. 2.3-1 — Distance And Displacement For A Body Moving From A To B A point A at the lower left and a point B at the upper right. Join them by a long curved, looping solid path with a small arrowhead on it partway along showing the direction of travel; label this path ‘distance — the whole length travelled’. Join A and B also by a straight dotted line with an arrowhead at B, labelled ‘displacement — the straight shift from A to B’.

2.3.1 Distance

Definition

Distance The total length of path along which a body moves is called the distance travelled by it. The distance travelled by a body depends on the path followed by the body.

It is a scalar quantity, generally represented by the letter $S$. Its SI unit is the metre (m) and its CGS unit is the centimetre (cm).

Because distance is a scalar, it carries no direction. It is simply how much ground was covered, added up over the whole journey, whichever way the body happened to turn.

2.3.2 Displacement

Definition

Displacement The shortest distance from the initial to the final position of the body is the magnitude of displacement, and its direction is from the initial position to the final position.

It is a vector quantity, represented by the symbol $\vec{S}$. Its SI unit is the metre (m) and its CGS unit is the centimetre (cm).

Notice that displacement takes no interest at all in what happened between the two ends of the journey. It looks only at where the body started and where it finished.

Signs and Units

Displacement Unit: metre (m) in SI, centimetre (cm) in CGS — the same units as distance. Sign: fix one direction as positive before you start. A displacement in that direction is then positive, and one in the opposite direction is negative. Distance, having no direction, is always positive.

2.3.3 Representing A Displacement

Being a vector, a displacement is drawn as a straight line with an arrow on it, using a convenient scale. The tip of the arrow shows the direction of the displacement, while the length of the line, read against the scale, gives its magnitude.

In Fig. 2.3-2, the arrow $\overrightarrow{PQ}$ represents a displacement of $40\ \text{m}$ in the east direction, drawn to the scale $1\ \text{cm} = 10\ \text{m}$. The point $P$, where the arrow begins, is the initial position and is called the origin; the point $Q$, where it ends, is the final position and is called the terminus.

Diagram

Fig. 2.3-2 — Representation Of A Displacement A horizontal straight line 4 cm long drawn left to right with an arrowhead at the right-hand end. Label the left-hand end P and mark ‘ORIGIN’ beneath it; label the right-hand end Q and mark ‘TERMINUS’ beneath it. Above the line write ‘40 m displacement’. Beside the figure state the scale ‘1 cm = 10 m’. To the right of the figure draw a small compass cross with N at the top, S at the bottom, E to the right and W to the left.

2.3.4 Distinction Between Distance And Displacement

Three points separate the two quantities, and each one can be seen in a simple journey.

The Magnitude Of Displacement Is Never Larger

The magnitude of displacement is either equal to or less than the distance. If the motion is along a fixed direction, the two are equal, because the path is itself the straight line joining the ends. But if the motion is along a curve or any zig-zag path, the magnitude of displacement is always less than the distance. It can never be greater.

In Fig. 2.3-1, for instance, the distance is the length of the curved path $AB$, while the displacement is along the straight dotted line — plainly the shorter of the two.

For a second example, take a boy who travels $4\ \text{km}$ towards the east and then $3\ \text{km}$ towards the north, as in Fig. 2.3-3. The total distance travelled by him is \[OA + AB = 4\ \text{km} + 3\ \text{km} = 7\ \text{km}\] but his displacement is $OB = 5\ \text{km}$ in the direction $OB$, that is, $36.9^\circ$ north of east. The figure $5\ \text{km}$ comes from the right-angled triangle $OAB$: the two shorter sides are $4\ \text{km}$ and $3\ \text{km}$, and the straight line joining the ends of such a pair measures $5\ \text{km}$.

Diagram

Fig. 2.3-3 — Displacement As A Vector A right-angled triangle OAB. From O draw a horizontal line 4 units long to the right, ending at A, labelled ‘4 km’ and marked EAST at the arrowhead. From A draw a vertical line 3 units long upwards, ending at B, labelled ‘3 km’ and marked NORTH at the arrowhead. Join O to B by a straight line with an arrowhead at B, labelled ‘5 km’. Mark the angle at O between OA and OB as 36.9°. Mark the right angle at A.

So the magnitude of a displacement is represented by the length of the straight line between the final and the initial positions.

Distance Is Always Positive

The distance is the length of path travelled by the body, so it is always positive. Displacement is the shortest length taken in the direction from the initial position to the final position, so it can be positive or negative depending on which way it points.

Displacement Can Be Zero While Distance Is Not

If a body, after travelling, comes back to its starting point, its displacement is zero, but the distance travelled by it is not zero. Two everyday cases show this clearly.

When a body is thrown vertically upwards from a point $A$ on the ground and after some time comes back to the same point $A$, the displacement of the body is zero, but the distance travelled is not zero — it is $2h$, where $h$ is the maximum height attained by the body.

When a body moving in a circular path reaches its original position after one complete round, the displacement at the end of that round is zero, but the distance travelled by it equals the circumference of the circular path, that is, $2\pi r$, where $r$ is the radius of the circular path in metre.

Curiosity Corner

Q4. A runner goes once round a circular track and stops exactly where she started. She is certainly tired, so something has been covered. Yet in another sense she has gone nowhere. Which two different quantities are being described here?

A. Distance and displacement. Her distance is the full circumference of the track, $2\pi r$ — that is the ground her legs actually covered. Her displacement is zero, because her finishing position is her starting position.

The two quantities are set side by side in Table 2.3.

Compare Table

Example 2.3

Distance And Displacement On A Straight Road

1. Given: a man walks $30\ \text{m}$ due east along a straight road, then turns round and walks $10\ \text{m}$ due west along the same road. Asked: the distance travelled and the displacement.

Distance is the whole length of path covered, so the two stretches are added. Displacement is measured from the starting point to the finishing point only, so the westward stretch is subtracted from the eastward one. Take east as the positive direction.
Distance $= 30\ \text{m} + 10\ \text{m} = 40\ \text{m}$.

Displacement $= (+30\ \text{m}) + (-10\ \text{m}) = +20\ \text{m}$.
The distance travelled is $40\ \text{m}$ and the displacement is $20\ \text{m}$ due east. The answer is sensible, since the man does finish $20\ \text{m}$ east of where he began, and the magnitude of his displacement is smaller than the distance he walked.

Try Yourself

Walk from the door of your classroom to your desk, counting your steps and taking the usual route round the furniture. Then stand at the door and estimate, in steps, the straight gap to your desk. Compare the two counts, and say which one is the distance and which the magnitude of your displacement.

Extra Practice 2.2

An athlete runs four complete rounds of a circular track of radius $35\ \text{m}$ and stops at the starting line. Find the distance travelled and the magnitude of the displacement. Take $\pi = \dfrac{22}{7}$.

2.4 Speed And Velocity

For a moving body, speed is the quantity by which we know how fast the body is moving. Velocity is the quantity by which we know how fast the body is moving in a certain direction. Speed does not tell us in which direction the body is going; velocity does.

That single difference runs through everything in this section, so it is worth holding on to from the start.

2.4.1 Speed

Cover more ground in the same time and you are moving faster. Take longer over the same ground and you are moving slower. So the quantity that measures how fast a body moves must compare the distance covered with the time taken.

Definition

Speed The speed of a body is the rate of change of distance with time. Numerically, it is the distance travelled by the body in 1 s.

It is a scalar quantity, generally represented by the letter $u$ or $v$.

If a body travels a distance $S$ in time $t$, then its speed $v$ is \[v = \frac{S}{t}\] where $S$ is the distance travelled in metre (m), $t$ is the time taken in second (s), and $v$ is the speed in metre per second (m s⁻¹).

The unit of speed follows from the formula itself: it is the unit of distance divided by the unit of time. Since the SI unit of distance is the metre and of time is the second, the SI unit of speed is metre per second, written m s⁻¹. Its CGS unit is centimetre per second, written cm s⁻¹.

2.4.1.1 Uniform Speed

A body is said to be moving with uniform speed if it covers equal distances in equal intervals of time throughout its motion. A ball rolling on a frictionless plane surface moves with uniform speed.

Once the uniform speed of a body is known, the distance it moves in any interval of time can be calculated. If a body moves with a uniform speed $v$, the distance travelled by it in time $t$ is \[S = v t\] where $v$ is the uniform speed in m s⁻¹, $t$ is the time in s, and $S$ is the distance in m. This relation holds only while the speed stays uniform; it cannot be used for a body that is speeding up or slowing down.

2.4.1.2 Non-Uniform Speed

A body is said to be moving with non-uniform speed, also called variable speed, if it covers unequal distances in equal intervals of time, or equal distances in unequal intervals of time.

The motion of a ball on a rough surface, the motion of a car in a crowded street, and the motion of a vehicle leaving or approaching a destination are all with non-uniform speed. In fact almost every real journey is of this kind.

2.4.1.3 Instantaneous Speed And Average Speed

When the speed keeps changing, one number will not describe the whole journey. Two different numbers are needed, and each answers a different question.

The first asks how fast the body is going right now. When the speed of a body keeps on changing, its speed at any instant is measured by finding the ratio of the distance travelled in a very short time interval to that time interval. This is called the instantaneous speed.

\[\text{instantaneous speed} = \frac{\text{distance travelled in a short time interval}}{\text{time interval}}\]

The interval must be short, because only then does the speed have no chance to change appreciably during it. The speedometer of a vehicle measures the instantaneous speed, as shown in Fig. 2.4-1.

Diagram

Fig. 2.4-1 — A Speedometer Reads The Instantaneous Speed A circular speedometer dial of a vehicle with graduations marked 0, 20, 40, 60, 80, 100 and 120 around the arc and the unit ‘km h⁻¹’ printed below the centre. Draw the needle pointing at 60 and label it ‘reading at this instant’. Beside the dial add a note: ‘the reading changes from moment to moment as the driver speeds up or slows down’.

Curiosity Corner

Q2. Galileo could measure only two things — how far the ball went and how long it took. From just these two, how would you work out how fast the ball was moving at one particular moment?

A. By taking a very short stretch of the groove on either side of that moment, and dividing the distance covered over it by the time taken. The shorter the stretch, the less the speed can change during it, and the closer the answer comes to the instantaneous speed at that moment.

The second question asks how the journey went as a whole. The ratio of the total distance travelled by the body to the total time of journey is called its average speed. \[\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}\] with the total distance in m and the total time in s, giving the average speed in m s⁻¹.

In case of a body moving with uniform speed, the instantaneous speed and the average speed are equal, and both are the same as that uniform speed.

Formula in Words

average speed = total distance ÷ total time Add up every metre the body covered, whatever route it took, then divide by the whole time from start to finish. It is the one steady speed at which the same journey would have taken the same time. It is not the average of the separate speeds.

Changing km h⁻¹ Into m s⁻¹

Road speeds are usually quoted in kilometre per hour, but physics calculations need metre per second, and the two must never be mixed inside one formula. The conversion is quickly worked out: \[1\ \text{km h}^{-1} = \frac{1000\ \text{m}}{60 \times 60\ \text{s}} = \frac{5}{18}\ \text{m s}^{-1}\] So to change a speed from km h⁻¹ to m s⁻¹, multiply by $\dfrac{5}{18}$; to go the other way, multiply by $\dfrac{18}{5}$.

Example 2.4

Converting A Road Speed To SI Units

1. Given: speed $= 90\ \text{km h}^{-1}$. Asked: the same speed in m s⁻¹.

A speed in km h⁻¹ is converted to m s⁻¹ by replacing the kilometre with $1000\ \text{m}$ and the hour with $60 \times 60 = 3600\ \text{s}$, since a speed is a distance divided by a time.
$90\ \text{km h}^{-1} = \dfrac{90 \times 1000\ \text{m}}{60 \times 60\ \text{s}} = \dfrac{90000\ \text{m}}{3600\ \text{s}} = 25\ \text{m s}^{-1}$
The speed is $25\ \text{m s}^{-1}$. The answer is sensible, since a speed in m s⁻¹ should come out a good deal smaller than the same speed in km h⁻¹ — here about one third of it, as the factor $\frac{5}{18}$ requires.

Example 2.5

Average Speed Over A Two-Part Journey

1. Given: a bus covers $40\ \text{km}$ at a uniform speed of $40\ \text{km h}^{-1}$, and then a further $60\ \text{km}$ at a uniform speed of $60\ \text{km h}^{-1}$, both in the same direction. Asked: the average speed for the whole journey.

Average speed is the total distance divided by the total time, so the time taken over each part must first be found separately from $t = \dfrac{S}{v}$. The two given speeds cannot simply be averaged, because the bus does not spend the same time at each of them.
$t_1 = \dfrac{40\ \text{km}}{40\ \text{km h}^{-1}} = 1\ \text{h}$ and $t_2 = \dfrac{60\ \text{km}}{60\ \text{km h}^{-1}} = 1\ \text{h}$.

Total distance $= 40\ \text{km} + 60\ \text{km} = 100\ \text{km}$; total time $= 1\ \text{h} + 1\ \text{h} = 2\ \text{h}$.

\[\text{average speed} = \frac{100\ \text{km}}{2\ \text{h}} = 50\ \text{km h}^{-1}\]
The average speed is $50\ \text{km h}^{-1}$. It lies between $40\ \text{km h}^{-1}$ and $60\ \text{km h}^{-1}$, as an average speed always must, and here it falls midway because the bus happened to spend equal times at the two speeds.

2.4.2 Velocity

Now attach a direction. If a body is shifted in a stated direction, and we ask how fast that shift is happening, we are asking for its velocity.

Definition

Velocity The rate of change of displacement of a body with time is called its velocity. Numerically, it is equal to the displacement of the body in 1 s, in a specified direction.

It is a vector quantity, represented by the symbol $\vec{u}$ or $\vec{v}$. Both its magnitude and its direction must be stated. Its SI unit is metre per second (m s⁻¹) and its CGS unit is centimetre per second (cm s⁻¹).

Two bodies are said to be moving with the same velocity only if both move with the same speed in the same direction. If two bodies move with the same speed but in different directions, or with different speeds in the same direction, they are said to be moving with different velocities.

2.4.2.1 Uniform Velocity

If a body travels equal distances in a particular direction in equal intervals of time, the body is said to be moving with a uniform velocity.

A body once started on a frictionless surface moves with uniform velocity. Raindrops also reach the earth’s surface with uniform velocity. A raindrop does speed up at first, under the force of gravity; but very soon the viscous force of the air and the upthrust together balance that gravitational pull, the net force on the drop becomes zero, and from then on the drop falls with the steady velocity called its terminal velocity.

If a body moving with a uniform velocity $\vec{v}$ has a displacement $\vec{S}$ in a time interval $t$, then by the definition of velocity $\vec{v} = \dfrac{\vec{S}}{t}$, and so \[\vec{S} = \vec{v}\, t\] where $\vec{v}$ is the uniform velocity in m s⁻¹, $t$ is the time in s, and $\vec{S}$ is the displacement in m, in the direction of $\vec{v}$. Like the corresponding relation for speed, this one holds only while the velocity stays uniform.

2.4.2.2 Variable Velocity

The velocity of a body can be variable either because its magnitude changes, or because its direction changes, or because both change.

If a body moves unequal distances in a particular direction in equal intervals of time, or if it moves equal distances in equal intervals of time but its direction of motion does not remain the same, then the velocity of the body is said to be variable, or non-uniform velocity.

The motion of a freely falling body is with variable velocity, because although the direction of motion of the body does not change, its speed continuously increases.

The second case is the more surprising one. Consider a body moving along a circular path, as in Fig. 2.4-2. At any instant its velocity is along the tangent to the circular path at that point, and as the body goes round, that tangent keeps swinging round with it. So motion along a circular path is with variable velocity even when the speed is uniform, because the direction of motion changes continuously with time. For such a body the velocity in fact changes at a uniform rate.

Diagram

Fig. 2.4-2 — Circular Path Of Constant Speed But Variable Velocity A circle with its centre marked by a dot. Mark four points on the circumference: B at the top, A at the right, D at the bottom and C at the left. At each of the four points draw an arrow of equal length along the tangent to the circle at that point, all four arrows pointing the same way round the circle, and label each arrow v. Add a note beneath: ‘every arrow is the same length — the speed is constant; every arrow points a different way — the velocity is changing’.

2.4.2.3 Instantaneous Velocity And Average Velocity

As with speed, a body moving with non-uniform velocity needs two separate quantities to describe it.

For a body moving with variable velocity, the velocity of the body at any instant is called its instantaneous velocity. It is measured by finding the ratio of the distance travelled in a sufficiently small time interval to that time interval. It is important to keep the interval small enough that the direction of motion does not change during it — otherwise the direction being reported would itself be uncertain.

If the velocity of a body moving in a particular direction changes with time, the ratio of the displacement to the time taken in the entire journey is called its average velocity. \[\text{average velocity} = \frac{\text{displacement}}{\text{total time taken}}\] with the displacement in m and the total time in s, giving the average velocity in m s⁻¹, in the direction of the displacement.

Do Not Mix Up

Average speed and average velocity. Average speed divides the total distance by the total time, so it can never be zero for a body that has moved at all. Average velocity divides the displacement by the total time, so it is zero whenever the body finishes where it started.

2.4.3 Distinction Between Speed And Velocity

Three points separate the two quantities.

First, speed is a scalar quantity, while velocity is a vector quantity. The speed of a body at a given time tells us how fast the body is moving at that time. Its velocity gives the same information, and tells us the direction of motion as well.

Second, for motion in a straight line, the magnitude of velocity is the speed. The speed is always positive, but the velocity is given a positive or a negative sign depending upon its direction of motion.

Third, the average velocity of a body can be zero even if its average speed is not zero. Two cases show this.

If a body starts its motion from a point and comes back to the same point after a certain time interval, the displacement is zero, so the average velocity is also zero; but the total distance travelled is not zero, and therefore the average speed is not zero.

If a body moves in a circular path and covers equal distances in equal intervals of time, the speed is uniform, but due to the continuous change in its direction of motion the velocity is variable. At any moment its instantaneous velocity and instantaneous speed are not zero. The displacement for one complete round is zero and therefore the average velocity for that round is also zero, but the average speed is $\dfrac{2\pi r}{T}$, where $r$ is the radius of the path in m and $T$ is the time taken in one round in s.

Compare Table

Example 2.6

Average Speed And Average Velocity Of The Same Journey

1. Given: a boy walks $600\ \text{m}$ due north along a straight road in $5\ \text{min}$, and then walks back along the same road to his starting point in $10\ \text{min}$. Asked: the average speed and the average velocity for the whole journey.

Average speed uses the total distance and average velocity uses the displacement, both divided by the same total time. The times are converted to seconds first, so that the answers come out in SI units.
Total distance $= 600\ \text{m} + 600\ \text{m} = 1200\ \text{m}$.

Total time $= 5\ \text{min} + 10\ \text{min} = 15\ \text{min} = 15 \times 60\ \text{s} = 900\ \text{s}$.

\[\text{average speed} = \frac{1200\ \text{m}}{900\ \text{s}} = 1.33\ \text{m s}^{-1}\]

The boy finishes at his starting point, so his displacement is $0\ \text{m}$, and

\[\text{average velocity} = \frac{0\ \text{m}}{900\ \text{s}} = 0\ \text{m s}^{-1}\]
The average speed is $1.33\ \text{m s}^{-1}$ and the average velocity is zero. The boy has certainly walked, so his average speed cannot be zero; but he has ended up where he began, so his displacement, and with it his average velocity, is zero.

Extra Practice 2.3

A train covers the $180\ \text{km}$ from one station to another in $3\ \text{h}$, and then returns along the same track to the first station in $2\ \text{h}$. For the whole journey out and back, find (i) the average speed in km h⁻¹ and (ii) the average velocity.

2.5 Acceleration And Retardation

Generally, bodies do not move with uniform velocities. The velocity of a body changes either in magnitude, or in direction, or in both magnitude and direction. A vehicle moves with variable velocity in a busy market, and again while leaving or approaching a destination. A planet or a satellite in an orbital path also moves with variable velocity.

Now consider motion only along a straight line path. Here there is no change in the direction of motion, so the change in velocity is only due to a change in speed. In such a case, if the velocity of the body increases with time, the motion is said to be accelerated; if the velocity of the body decreases with time, the motion is said to be decelerated, or retarded. Retardation is simply negative acceleration.

2.5.1 Acceleration

Whether a body speeds up or slows down, the physical question is the same: by how much is the velocity changing each second? The quantity that answers it is called acceleration.

Definition

Acceleration Acceleration is the rate of change of velocity with time. Numerically, it is equal to the change in velocity in 1 s.

\[\text{acceleration} = \frac{\text{change in velocity}}{\text{time interval}}\]

The unit of acceleration follows from that relation: it is the unit of velocity divided by the unit of time. The SI unit of velocity is metre per second and of time is second, so the SI unit of acceleration is \[\frac{\text{metre per second}}{\text{second}} = \text{metre per second squared, or m s}^{-2}\] Its CGS unit is centimetre per second squared, cm s⁻².

The Relation For Acceleration

Let a body be moving in a straight line in a fixed direction with an initial velocity $u$. Its velocity changes over a time interval $t$, and the final velocity becomes $v$ after that time $t$. Then the change in velocity is $(v - u)$ and the time taken is $t$, so \[a = \frac{v - u}{t} \qquad \text{or} \qquad v = u + a t\] where $u$ is the initial velocity in m s⁻¹, $v$ is the final velocity in m s⁻¹, $t$ is the time in s, and $a$ is the acceleration in m s⁻². The relation holds for motion along a straight line in which the acceleration stays uniform through the interval $t$.

The second form comes from the first in one step: multiply both sides by $t$ to get $a t = v - u$, then add $u$ to both sides.

If $v > u$, then $a$ is positive, and $a$ is the acceleration. But if $v < u$, then $a$ comes out negative, and $a$ is the retardation.

Curiosity Corner

Q1. Nobody pushes the ball again once it is released, yet it covers more of the groove in each second than in the one before. What is being added to the ball’s motion, second after second?

A. Velocity. The ball has an acceleration down the slope, and acceleration means exactly this — a fixed amount of velocity added every second. Faster ball, longer stretch of groove covered in the next second.

Two ordinary situations show the two signs.

Suppose a car initially at rest, on starting, acquires a velocity $20\ \text{m s}^{-1}$ in $10\ \text{s}$. The change in its velocity is $20\ \text{m s}^{-1} - 0\ \text{m s}^{-1} = 20\ \text{m s}^{-1}$, and this change has been brought about in $10\ \text{s}$. Therefore the acceleration of the car is \[a = \frac{20\ \text{m s}^{-1} - 0\ \text{m s}^{-1}}{10\ \text{s}} = \frac{20\ \text{m s}^{-1}}{10\ \text{s}} = 2\ \text{m s}^{-2}\]

Now suppose a car initially moving with a velocity $25\ \text{m s}^{-1}$ is brought to rest in $5\ \text{s}$ by applying the brakes. Then the acceleration of the car is \[a = \frac{0\ \text{m s}^{-1} - 25\ \text{m s}^{-1}}{5\ \text{s}} = \frac{-25\ \text{m s}^{-1}}{5\ \text{s}} = -5\ \text{m s}^{-2}\] or the retardation is $5\ \text{m s}^{-2}$, since a negative acceleration is called the retardation.

What The Sign Of Acceleration Tells You

Acceleration is a vector quantity, represented by the symbol $\vec{a}$. Its direction is the direction of the change in velocity. For motion along a straight line, the acceleration lies along that same line — pointing the same way as the motion when the body is speeding up, and the opposite way when the body is slowing down.

This makes an important point easy to miss. The acceleration of a body does not determine its direction of motion, while the velocity determines its direction of motion. The positive or negative sign of acceleration tells us whether the velocity is increasing or decreasing with time, whereas the positive or negative sign of velocity tells us the direction of motion.

A car braking while still moving forward is the everyday case: its velocity points forward, its acceleration points backward, and the car goes on moving forward all the while — just more and more slowly.

Do Not Mix Up

The sign of velocity and the sign of acceleration. The sign of velocity says which way the body is going. The sign of acceleration says only whether the velocity is growing or shrinking. Opposite signs mean a body slowing down, not a body reversing.

Example 2.7

Acceleration Of A Scooter Starting From Rest

1. Given: initial velocity $u = 0\ \text{m s}^{-1}$ (starts from rest), final velocity $v = 18\ \text{km h}^{-1}$, time $t = 2.5\ \text{s}$. Asked: the acceleration in m s⁻².

The relation $a = \dfrac{v - u}{t}$ applies, since the motion is along a straight line with a steady change of velocity. The final velocity must first be converted to m s⁻¹, or the answer will not be in SI units.
$v = 18\ \text{km h}^{-1} = 18 \times \dfrac{5}{18}\ \text{m s}^{-1} = 5\ \text{m s}^{-1}$

\[a = \frac{5\ \text{m s}^{-1} - 0\ \text{m s}^{-1}}{2.5\ \text{s}} = 2\ \text{m s}^{-2}\]
The acceleration is $2\ \text{m s}^{-2}$. The sign is positive, as it should be for a scooter picking up speed, and the size is reasonable — the machine gains $2\ \text{m s}^{-1}$ of velocity in every second.

Example 2.8

Retardation Of A Braking Train

1. Given: initial velocity $u = 20\ \text{m s}^{-1}$, final velocity $v = 0\ \text{m s}^{-1}$ (comes to rest), time $t = 8\ \text{s}$. Asked: the acceleration, and the retardation.

Again $a = \dfrac{v - u}{t}$. Here the final velocity is smaller than the initial velocity, so a negative answer is expected.
\[a = \frac{0\ \text{m s}^{-1} - 20\ \text{m s}^{-1}}{8\ \text{s}} = \frac{-20\ \text{m s}^{-1}}{8\ \text{s}} = -2.5\ \text{m s}^{-2}\]
The acceleration is $-2.5\ \text{m s}^{-2}$, and hence the retardation is $2.5\ \text{m s}^{-2}$. The minus sign is exactly what a braking train should give: the velocity is falling by $2.5\ \text{m s}^{-1}$ in every second, so after $8\ \text{s}$ the whole $20\ \text{m s}^{-1}$ has gone.

2.5.2 Uniform And Variable Acceleration

The acceleration is said to be uniform acceleration, or constant acceleration, when equal changes in velocity take place in equal intervals of time — that is, when the velocity is changing uniformly. The motion of a body under gravity, such as the free fall of a body, is an example of uniformly accelerated motion.

If the change in velocity is not the same in the same intervals of time, the acceleration is said to be variable acceleration. The motion of a vehicle on a crowded road, or on a hilly road, is with variable acceleration.

2.5.3 Acceleration Due To Gravity

Release any body near the earth’s surface and it falls, gathering speed as it goes. The earth attracts it, and that attraction produces an acceleration called the acceleration due to gravity.

Definition

Acceleration Due To Gravity When a body falls freely under gravity, the acceleration produced in the body due to the earth’s gravitational attraction is called acceleration due to gravity. It is generally denoted by the letter $g$, and is measured in m s⁻².

When a body falls down, its velocity increases with time, so the acceleration is $+g$. When the body moves vertically upwards, its velocity decreases with time, so the acceleration is $-g$, or the retardation is $g$. The two cases are drawn in Fig. 2.5-1.

Diagram

Fig. 2.5-1 — Acceleration Due To Gravity For A Falling And A Rising Body Two panels side by side, each showing the ground at the bottom. Left panel: a ball at the top with three faint images below it, spaced further and further apart as they descend, with a downward velocity arrow beside each image growing longer downwards; a separate bold downward arrow labelled ‘a = + g’; caption inside the panel ‘falling body — velocity increasing’. Right panel: a ball near the ground with three faint images above it, spaced closer and closer together as they rise, with an upward velocity arrow beside each image growing shorter; a separate bold downward arrow labelled ‘a = – g’; caption inside the panel ‘rising body — velocity decreasing’. In both panels the bold acceleration arrow must point downwards.

Signs and Units

Acceleration due to gravity, $g$ Unit: m s⁻². Its direction is always vertically downwards, whichever way the body itself is moving. Sign: take downwards as positive and $g$ is $+9.8\ \text{m s}^{-2}$ for a falling body and for a rising one alike. Fix the positive direction first, then keep it.

The Value Of g

The average value of $g$ is $9.8\ \text{m s}^{-2}$, or nearly $10\ \text{m s}^{-2}$. It actually varies from place to place: on the earth’s surface $g$ is maximum at the poles and minimum at the equator, and its value decreases with altitude and also with depth below the earth’s surface.

Thus if a body falls freely under gravity, its velocity increases at a rate of $9.8\ \text{m s}^{-2}$. Starting from rest, after $1\ \text{s}$ the velocity will be $9.8\ \text{m s}^{-1}$; after $2\ \text{s}$ it will be $2 \times 9.8 = 19.6\ \text{m s}^{-1}$; after $3\ \text{s}$ it will be $3 \times 9.8 = 29.4\ \text{m s}^{-1}$, and so on.

Similarly, if a body is projected vertically upwards, its velocity decreases at a rate of $9.8\ \text{m s}^{-2}$. If a body is projected vertically upwards with an initial velocity of $49\ \text{m s}^{-1}$, then after $1\ \text{s}$ its velocity will become $39.2\ \text{m s}^{-1}$; after $2\ \text{s}$, $29.4\ \text{m s}^{-1}$; after $3\ \text{s}$, $19.6\ \text{m s}^{-1}$, and so on. Both sets of values are collected in Table 2.5.

Table 2.5 — Velocity Of A Body Falling From Rest And Of A Body Projected Upwards At 49 m s⁻¹

Time $t$ (in s)	0	1	2	3
Velocity of the falling body (in m s⁻¹)	0	9.8	19.6	29.4
Velocity of the rising body (in m s⁻¹)	49	39.2	29.4	19.6

Read across either row and the same number appears each time as the difference between neighbours: $9.8\ \text{m s}^{-1}$, gained in the first case and lost in the second. That is what a uniform acceleration of $9.8\ \text{m s}^{-2}$ means.

Why Mass Makes No Difference

The value of $g$ does not depend on the mass of the body. Hence if two bodies of different masses are simultaneously dropped in vacuum from a height, both will reach the ground simultaneously, because in vacuum there is no effect of friction and no buoyancy due to air.

Curiosity Corner

Q3. If a heavy ball and a light ball are released together down the same groove, which reaches the bottom first — and would your answer change if they were simply dropped through empty space instead?

A. Neither reaches first, in either case. The acceleration produced by gravity does not depend on the mass of the body, so both are speeded up at exactly the same rate. In empty space they stay level all the way down, since there is no air friction and no buoyancy to hold the lighter one back.

Example 2.9

Time Taken By A Ball To Reach Its Highest Point

1. Given: a ball is thrown vertically upwards with an initial velocity $u = 19.6\ \text{m s}^{-1}$; at the highest point the final velocity $v = 0\ \text{m s}^{-1}$; the acceleration is that due to gravity, $g = 9.8\ \text{m s}^{-2}$, directed downwards. Asked: the time taken to reach the highest point.

The relation $v = u + a t$ applies, since the acceleration due to gravity is uniform. Taking the upward direction as positive, the ball’s velocity is $+19.6\ \text{m s}^{-1}$ while the acceleration opposes the motion, so $a = -9.8\ \text{m s}^{-2}$.
$0 = 19.6\ \text{m s}^{-1} + (-9.8\ \text{m s}^{-2}) \times t$

\[t = \frac{19.6\ \text{m s}^{-1}}{9.8\ \text{m s}^{-2}} = 2\ \text{s}\]
The ball takes $2\ \text{s}$ to reach its highest point. The answer is sensible, since the ball loses $9.8\ \text{m s}^{-1}$ of velocity every second, and it began with almost twice that much.

Extra Practice 2.4

A stone is dropped from rest at the top of a tower and falls freely. Taking $g = 9.8\ \text{m s}^{-2}$, find its velocity after $4\ \text{s}$. If a second stone of twice the mass is dropped from the same point at the same instant, what would its velocity be after $4\ \text{s}$? Ignore air resistance.

2.6 Displacement–Time Graph

If a body moves in a straight line path, its motion is in one dimension and is called linear or rectilinear motion. Such a motion can be studied and analysed very well with the help of three kinds of graph: 1. the distance–time and displacement–time graph, 2. the speed–time and velocity–time graph, and 3. the acceleration–time graph.

For motion in one direction along a straight line, if the body does not return — that is, it never moves in the opposite direction — then the distance–time and displacement–time graphs are the same, and the velocity–time and speed–time graphs are the same too. But when the body returns, the distance keeps on increasing while the displacement does not, and different graphs are then needed to represent them. The graphs also differ for two and three dimensional motions.

We begin with the first of the three.

In the displacement–time graph, the time is taken on the $X$-axis and the displacement of the body is taken on the $Y$-axis. From this graph we can determine the velocity of the body.

Why The Slope Gives The Velocity

Velocity is the ratio of displacement to time. On the graph, moving from one point of the line to another, the rise measured up the $Y$-axis is a displacement and the run measured along the $X$-axis is the time taken over it. Their ratio — the rise divided by the run — is exactly what we call the slope, or gradient, of the line. So

the slope of a displacement–time graph gives the velocity.

If the slope is positive, it implies that the body is moving away from the starting or reference point. If the slope is negative, the body is returning towards the starting or reference point.

Formula in Words

slope of a displacement–time graph = velocity Pick two points on the line. Find how much the displacement rose between them, in metre, and how much time passed, in second. Divide the first by the second. A steeper line means more metres in the same second, so a greater velocity.

Let us now consider the situations that can arise.

2.6.1 Graph For A Stationary Body

If the position of a body does not change with time, the body is said to be stationary. Its displacement measured from the origin is the same at all instants as it was at $t = 0$, so the displacement–time graph is a straight line parallel to the time axis, as shown in Fig. 2.6-1. Here $OP$ is the distance of the body from the origin $O$, and it remains the same at each instant.

Diagram

Fig. 2.6-1 — Displacement–Time Graph For A Stationary Object Graph axes with the horizontal axis labelled TIME (s), marked 0 to 9 in steps of 1, and the vertical axis labelled DISPLACEMENT (m), marked 0 to 40 in steps of 10. Draw a horizontal straight line at a displacement of 25 m running the full width of the graph, parallel to the time axis. Mark the point where this line meets the vertical axis as P, and the origin as O. Label the vertical gap OP.

The slope of this line is zero, and so is the velocity — which is exactly right for a body that is not moving.

2.6.2 Graph For A Body Moving With Uniform Velocity

If a body is moving with uniform velocity, its displacement increases by the same amount in each second, and so the displacement–time graph, also called the distance–time graph in this case, is a straight line inclined to the time axis. The velocity of the body can be obtained by finding the slope of the straight line.

As an example, take a car moving on a straight path in a given direction with a uniform speed. Table 2.6 gives its displacement — that is, its distance from the starting point — at different instants.

Table 2.6 — Displacement Of A Car Moving With Uniform Velocity

Time (in s)	0	1	2	3	4	5
Displacement (in m)	0	10	20	30	40	50

The displacement–time graph for this data is a straight line $OP$ inclined to the time axis, as drawn in Fig. 2.6-2. It shows the linear relationship between the displacement and time — that is, the displacement of the car is directly proportional to the time of travel, or the car travels equal distances in equal intervals of time in a certain direction. Thus, for a body moving with uniform velocity, the displacement is directly proportional to the time, $S \propto t$.

Diagram

Fig. 2.6-2 — Displacement–Time Graph With A Uniform Velocity Graph axes with the horizontal axis labelled TIME (s), marked 0 to 5 in steps of 1, and the vertical axis labelled DISPLACEMENT (m), marked 0 to 50 in steps of 10. Plot the points (0, 0), (1, 10), (2, 20), (3, 30), (4, 40) and (5, 50) as small circles and join them by a single straight line rising from the origin. Label the origin O and the topmost plotted point P. Draw light dashed guide lines from each plotted point across to the vertical axis and down to the horizontal axis.

The velocity of the car can be obtained by finding the slope of the straight line $OP$. For this, take any two points $A$ and $B$ on the line $OP$ as shown in Fig. 2.6-3, and draw the perpendiculars $AD$ and $BE$ on the $Y$-axis and $AF$ and $BG$ on the $X$-axis from these points.

Diagram

Fig. 2.6-3 — Finding The Velocity From A Displacement–Time Graph The same straight line OP as in Fig. 2.6-2, on axes labelled TIME (s) 0 to 5 and DISPLACEMENT (m) 0 to 50. Mark a point A on the line at (2, 20) and a point B on the line at (4, 40). From A draw a horizontal dashed line left to the vertical axis meeting it at D (20 m), and from B a horizontal dashed line left to the vertical axis meeting it at E (40 m). From A draw a vertical dashed line down to the time axis meeting it at F (2 s), and from B a vertical dashed line down to the time axis meeting it at G (4 s). Extend the horizontal line from A rightwards to meet the vertical line through B at the point C, so that the right-angled triangle ABC is formed with CA horizontal and BC vertical. Label BC and CA clearly.

Interpret Data

Fig. 2.6-3 — Finding The Velocity From A Displacement–Time Graph 1. Read the axes first. The horizontal axis carries time in second, from 0 s to 5 s. The vertical axis carries displacement in metre, from 0 m to 50 m. Every point on the line therefore states one thing: where the car was at one instant. 2. Pick two points on the line — here $A$ at $t = 2\ \text{s}$ and $B$ at $t = 4\ \text{s}$. 3. Read off the rise. The car’s displacement went from $20\ \text{m}$ at $A$ to $40\ \text{m}$ at $B$, so $BC = ED = (40 - 20)\ \text{m} = 20\ \text{m}$. 4. Read off the run. The time went from $2\ \text{s}$ to $4\ \text{s}$, so $CA = GF = (4 - 2)\ \text{s} = 2\ \text{s}$. 5. Divide the rise by the run. This gives the slope, which for a displacement–time graph is the velocity: \[\text{velocity of the car} = \frac{BC}{CA} = \frac{ED}{GF} = \frac{(40 - 20)\ \text{m}}{(4 - 2)\ \text{s}} = \frac{20\ \text{m}}{2\ \text{s}} = 10\ \text{m s}^{-1}\] 6. Check that the answer means something. A velocity of $10\ \text{m s}^{-1}$ says the car covers $10\ \text{m}$ every second, and Table 2.6 shows exactly that. Any other pair of points on this line would have given the same answer, because the line is straight.

Obviously, the greater the slope — that is, the more inclined the straight line — the higher is the velocity.

Curiosity Corner

Q5. Suppose you plot how far the ball has rolled against the time taken. What would the steepness of the line at any point be telling you about the ball?

A. How fast it was moving at that point. The steepness, or slope, of a displacement–time graph is the displacement gained divided by the time taken over it, and that ratio is the velocity. A steeper line means a faster ball.

2.6.3 Graph For A Body That Returns

Now take a ball which, while moving on a perfectly smooth floor, hits a wall at $t = 6\ \text{s}$ and then comes back along the same line. Its displacement — the distance of the ball from the starting point — and the distance covered by it at different instants are given in Table 2.7.

Table 2.7 — Displacement And Distance For A Ball That Rebounds From A Wall

Time (in s)	0	2	4	6	8	10	12	14
Displacement (in m)	0	20	40	60	45	30	15	0
Distance (in m)	0	20	40	60	75	90	105	120

Read the two rows across and the difference between the quantities stands out. The displacement climbs to $60\ \text{m}$ and then falls back to zero as the ball returns to where it began, while the distance simply keeps on growing to $120\ \text{m}$.

The displacement–time graph for this motion is drawn in Fig. 2.6-4.

Diagram

Fig. 2.6-4 — Displacement–Time Graph With Positive And Negative Slopes Graph axes with the horizontal axis labelled TIME (s), marked 0 to 14 in steps of 2, and the vertical axis labelled DISPLACEMENT (m), marked 0 to 60 in steps of 10. Plot the points (0, 0), (2, 20), (4, 40) and (6, 60) as small circles and join them by a straight line rising from the origin; label the peak point at (6, 60) as A. Then plot (8, 45), (10, 30), (12, 15) and (14, 0) as small circles and join them by a straight line falling from A; label the final point at (14, 0) as B. Add a light dashed horizontal line from A across to the vertical axis at 60 m.

\[\text{slope of the line } OA = \frac{60\ \text{m} - 0}{6\ \text{s} - 0} = 10\ \text{m s}^{-1}\] \[\text{slope of the line } AB = \frac{0 - 60\ \text{m}}{14\ \text{s} - 6\ \text{s}} = -\frac{60\ \text{m}}{8\ \text{s}} = -7.5\ \text{m s}^{-1}\]

Thus the slope of the line $OA$ is positive, $10\ \text{m s}^{-1}$, which represents the uniform motion of the ball with velocity $10\ \text{m s}^{-1}$ towards the wall, that is, away from the starting point. The slope of the line $AB$ is negative, $-7.5\ \text{m s}^{-1}$, which represents the uniform motion of the ball with velocity $7.5\ \text{m s}^{-1}$ in the opposite direction, that is, towards the starting point after hitting the wall.

Fig. 2.6-5 shows the distance–time graph for the same motion. Here the slope of the line gives the speed of the body.

Diagram

Fig. 2.6-5 — Distance–Time Graph For The Same Motion Graph axes with the horizontal axis labelled TIME (s), marked 0 to 14 in steps of 2, and the vertical axis labelled DISTANCE (m), marked 0 to 120 in steps of 20. Plot the points (0, 0), (2, 20), (4, 40), (6, 60), (8, 75), (10, 90), (12, 105) and (14, 120) as small circles. Join the first four by a straight line from the origin and label the point at (6, 60) as A; join the remaining points by a second straight line of gentler slope, rising to the point at (14, 120), which is labelled B. The whole graph rises throughout — it never comes down.

\[\text{speed during the first six seconds} = \text{slope of } OA = \frac{60\ \text{m} - 0}{6\ \text{s} - 0} = 10\ \text{m s}^{-1}\] \[\text{speed during the last eight seconds} = \text{slope of } AB = \frac{120\ \text{m} - 60\ \text{m}}{14\ \text{s} - 6\ \text{s}} = 7.5\ \text{m s}^{-1}\]

The average speed can also be calculated for the whole journey, which is $\dfrac{120\ \text{m}}{14\ \text{s}} = 8.6\ \text{m s}^{-1}$.

Do Not Mix Up

The displacement–time graph and the distance–time graph. They are the same line only while the body keeps going one way. The moment it turns back, the displacement graph comes down towards zero while the distance graph goes on climbing, because distance never decreases.

One shape is impossible. The displacement–time graph can never be a straight line parallel to the displacement axis, because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time — that is, the velocity of the body is infinite, which is impossible.

2.6.4 Graph For A Body Moving With Variable Velocity

If a body moves with varying speed in a fixed direction, that is, with variable velocity, the displacement–time graph is not a straight line but a curve. The velocity at any instant can then be obtained by finding the slope, or the gradient, of the tangent drawn on the curve at that instant of time.

A tangent is simply the straight line that just touches the curve at that one point and matches its steepness there. Once it is drawn, it is treated exactly like the straight-line graphs above.

As an example, Fig. 2.6-6 represents the displacement–time graph of a body for which the displacement — the distance from the starting point — at different instants is given in Table 2.8.

Table 2.8 — Displacement Of A Body Moving With Variable Velocity

Time (in second)	0	2	5	6
Displacement (in metre)	0	10	30	47

Diagram

Fig. 2.6-6 — Displacement–Time Graph With A Variable Velocity Graph axes with the horizontal axis labelled TIME (s), marked 0 to 7 in steps of 1, and the vertical axis labelled DISPLACEMENT (m), marked 0 to 60 in steps of 10. Plot the points (0, 0), (2, 10), (5, 30) and (6, 47) as small circles and join them by a smooth curve that becomes steeper as it rises; label the lower end of the curve P and the upper end Q. Mark the point on the curve at (5, 30) as A. Through A draw a straight tangent line touching the curve only at A, extending from a point D at (3, 10) to a point B at (7, 50); mark D and B on this tangent. From B drop a vertical dashed line and from D draw a horizontal dashed line so that they meet at C, forming the right-angled triangle BCD with BC vertical and CD horizontal. Label BC and CD.

The displacement–time graph is a curve $PQ$. The velocity of the body at time $t = 5\ \text{s}$, that is, when the displacement $S = 30\ \text{m}$, is obtained by finding the slope of the tangent $BD$ drawn to the curve at the point $A$ corresponding to $t = 5\ \text{s}$ or $S = 30\ \text{m}$. The slope of the tangent $BD$ is \[\text{slope} = \frac{BC}{CD} = \frac{(50 - 10)\ \text{m}}{(7 - 3)\ \text{s}} = \frac{40\ \text{m}}{4\ \text{s}} = 10\ \text{m s}^{-1}\] that is, the velocity of the body at $t = 5\ \text{s}$, or at $S = 30\ \text{m}$, is $10\ \text{m s}^{-1}$.

Note that in this case the slope of the curve is different at different points, so the velocity is also different at different instants; that is, the body is moving with a variable speed or velocity.

2.6.5 What The Displacement–Time Graph Tells Us

The whole of this section can be gathered into three conclusions.

If the displacement–time graph of an object is a straight line parallel to the time axis, the object is stationary. If the graph is a straight line inclined to the time axis, the motion is with uniform velocity. If the graph is a curve, the motion is with non-uniform velocity.
In the displacement–time graph, the slope of the straight line, or of the tangent to the curve at an instant, gives the velocity of the object at that instant. If the slope is positive, it represents motion away from the origin or reference point; if the slope is negative, it represents motion towards the origin.
Knowing the velocity of the object at different instants from the displacement–time graph, the velocity–time graph can be drawn.

Example 2.10

Reading Velocities From A Displacement–Time Graph

1. Given: the displacement–time graph of a cyclist on a straight road consists of two straight parts — from $t = 0$ to $t = 4\ \text{s}$ the displacement rises steadily from $0\ \text{m}$ to $24\ \text{m}$, and from $t = 4\ \text{s}$ to $t = 10\ \text{s}$ the graph is a horizontal line at $24\ \text{m}$. Asked: the velocity in each part, and the nature of the motion in each.

The slope of a displacement–time graph gives the velocity, so each part is handled by dividing the rise in displacement by the run in time. Both parts are straight lines, so within each part the velocity is uniform.
First part: rise $= (24 - 0)\ \text{m} = 24\ \text{m}$, run $= (4 - 0)\ \text{s} = 4\ \text{s}$, so

\[v_1 = \frac{24\ \text{m}}{4\ \text{s}} = 6\ \text{m s}^{-1}\]

Second part: rise $= (24 - 24)\ \text{m} = 0\ \text{m}$, run $= (10 - 4)\ \text{s} = 6\ \text{s}$, so

\[v_2 = \frac{0\ \text{m}}{6\ \text{s}} = 0\ \text{m s}^{-1}\]
The cyclist moves with a uniform velocity of $6\ \text{m s}^{-1}$ away from the starting point for the first $4\ \text{s}$, and is then at rest for the next $6\ \text{s}$. The second answer is sensible, because a horizontal line means the displacement is not changing at all — the cyclist has stopped, but has not come back.

Extra Practice 2.5

A body moving in a straight line has the displacements given below.

| Time (in s) | 0 | 1 | 2 | 3 | 4 | 5 |

|—|—|—|—|—|—|—|

| Displacement (in m) | 0 | 8 | 16 | 16 | 16 | 8 |

Draw the displacement–time graph and use it to find the velocity of the body during each of the three parts of its journey, stating clearly what the sign of each answer means.

2.7 Velocity–Time Graph

In the velocity–time graph, time is taken on the $X$-axis and the velocity is taken on the $Y$-axis.

Since velocity is a vector quantity, the positive velocity means that the body is moving in a certain direction away from its initial position, and the negative velocity means that the body is moving in the opposite direction, that is, towards the initial position.

This graph is worth more than the last one, because two things can be read from it instead of one. From the velocity–time graph we can determine the displacement of the body in a certain time interval, and the acceleration of the body at any instant.

2.7.1 Displacement From The Velocity–Time Graph

Since velocity $\times$ time $=$ displacement, the area enclosed between the velocity–time sketch and the $X$-axis, that is, the time axis, gives the displacement of the body.

To see why, look at what an area on this graph is made of. A height measured up the graph is a velocity in m s⁻¹, and a width measured across it is a time in s. Multiply the two, as finding an area requires, and the result is a velocity multiplied by a time — which is a displacement in metre.

The area enclosed above the time axis represents the positive displacement, that is, the distance travelled away from the starting point, while the area enclosed below the time axis represents the negative displacement, that is, the distance travelled towards the starting point. The total displacement is obtained by adding them numerically with proper sign. But the total distance travelled by the body is their arithmetic sum, taken without sign.

Consider the velocity–time graph of a body in motion shown in Fig. 2.7-1.

Diagram

Fig. 2.7-1 — Velocity–Time Graph Of A Body In Motion Graph axes with the horizontal axis labelled Time (s), marked 1 to 14 in steps of 1, and the vertical axis labelled Velocity (m s⁻¹), marked +5 above the origin and –5 below it, with the time axis drawn through zero so that the graph can go below it. Draw the line as a single unbroken path through these points, labelling each named point: a at (0, 0); rising straight to b at (3, 5); falling straight through the time axis to c at (4, 0) and continuing down to d at (5, –5); running horizontally to e at (8, –5); rising straight through the time axis at f (9, 0) and continuing up to g at (10, 5); running horizontally to h at (13, 5); then falling to i at (14, 0). Shade lightly the three enclosed regions: triangle abc above the axis, trapezium cdef below the axis, and trapezium fghi above the axis.

In Fig. 2.7-1, the area of the triangle $abc$ is \[\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 4\ \text{s} \times 5\ \text{m s}^{-1} = 10\ \text{m}\] the area of the trapezium $cdef$ is \[\tfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \tfrac{1}{2} \times (5 + 3)\ \text{s} \times 5\ \text{m s}^{-1} = 20\ \text{m}\] and the area of the trapezium $fghi$ is \[\tfrac{1}{2} \times (4 + 2)\ \text{s} \times 5\ \text{m s}^{-1} = 15\ \text{m}\]

In each trapezium the two parallel sides are the two horizontal time-stretches, measured in second, and the height is the vertical stretch of velocity, measured in m s⁻¹.

Then the displacement of the body is \[\text{area of } \triangle abc - \text{area of trapezium } cdef + \text{area of trapezium } fghi = 10\ \text{m} - 20\ \text{m} + 15\ \text{m} = 5\ \text{m}\] but the total distance travelled by the body is \[10\ \text{m} + 20\ \text{m} + 15\ \text{m} = 45\ \text{m}\]

Formula in Words

area under a velocity–time graph = displacement Every strip of area is a velocity multiplied by a time, and that is a displacement. Areas above the time axis count as positive, areas below as negative. Add them with their signs for displacement; add them ignoring signs for distance.

2.7.2 Acceleration From The Velocity–Time Graph

Since acceleration is equal to the ratio of change in velocity to time taken, the slope, or gradient, of the velocity–time sketch gives the acceleration.

Take the same graph, Fig. 2.7-1. For the part $ab$ of the motion, \[\text{slope} = \frac{\text{change in velocity}}{\text{change in time}} = \frac{(5 - 0)\ \text{m s}^{-1}}{(3 - 0)\ \text{s}}\] \[\therefore\ \text{acceleration} = \frac{5}{3} = 1.67\ \text{m s}^{-2}\]

In the part $bd$, \[\text{slope} = \frac{[(-5) - 5]\ \text{m s}^{-1}}{(5 - 3)\ \text{s}} = -5\ \text{m s}^{-2}\] \[\therefore\ \text{acceleration} = -5\ \text{m s}^{-2}\]

In the part $de$, the slope is zero, and therefore the acceleration is zero.

Since the slope is positive in the part $ab$, it is an accelerated motion. The slope is negative in the part $bd$, so it represents decelerated or retarded motion. In the part $de$ the slope is zero, so the motion is with constant velocity. The parts $eg$, $gh$ and $hi$ are with positive, zero and negative acceleration respectively.

Interpret Data

Fig. 2.7-1 — Velocity–Time Graph Of A Body In Motion 1. Read the axes. Time runs along the horizontal axis in second; velocity runs up the vertical axis in m s⁻¹, and the axis continues below zero because velocity can be negative. 2. Read the sign of the line’s position. Between $a$ and $c$ the line is above the time axis, so the body is moving in the chosen positive direction. Between $c$ and $f$ it is below, so the body is moving the opposite way. 3. Read the steepness for acceleration. Between $a$ and $b$ the line climbs $5\ \text{m s}^{-1}$ in $3\ \text{s}$, giving $1.67\ \text{m s}^{-2}$. Between $b$ and $d$ it drops $10\ \text{m s}^{-1}$ in $2\ \text{s}$, giving $-5\ \text{m s}^{-2}$. Along $de$ it is flat, so the acceleration is zero there — but note that the velocity along $de$ is not zero, it is a steady $-5\ \text{m s}^{-1}$. 4. Read the enclosed areas for displacement. Above the axis: $10\ \text{m}$ and $15\ \text{m}$. Below the axis: $20\ \text{m}$. 5. Combine them the right way. Displacement $= 10\ \text{m} - 20\ \text{m} + 15\ \text{m} = 5\ \text{m}$; distance $= 10\ \text{m} + 20\ \text{m} + 15\ \text{m} = 45\ \text{m}$. 6. Check the sense of it. The body went forward, came back further than it had gone, then went forward again, finishing just $5\ \text{m}$ ahead of its starting point after covering $45\ \text{m}$ of ground.

Do Not Mix Up

The slope and the area of a velocity–time graph. The slope tells you how quickly the velocity is changing — it gives the acceleration, in m s⁻². The area tells you how far the body has got — it gives the displacement, in m. A flat line has zero slope but can still enclose a large area.

We can now consider the cases that arise one by one.

2.7.3 Graph For Uniform Velocity

If a body is in motion with uniform velocity, that is, the velocity remains constant with time, the velocity–time graph is a straight line parallel to the time axis.

In Fig. 2.7-2, a straight line $AB$ represents the velocity–time graph of a body moving with a uniform velocity $4\ \text{m s}^{-1}$ for $5\ \text{s}$. The slope of the straight line $AB$ is zero, therefore its acceleration is zero.

Diagram

Fig. 2.7-2 — Velocity–Time Graph For Uniform Velocity Graph axes with the horizontal axis labelled TIME (s), marked 0 to 5 in steps of 1, and the vertical axis labelled VELOCITY (m s⁻¹), marked 0 to 5 in steps of 1. Draw a horizontal straight line at a velocity of 4 m s⁻¹ from time 0 to time 5 s. Label its left-hand end A, on the velocity axis, and its right-hand end B. Drop a vertical line from B to the time axis, meeting it at C at 5 s. Label the origin O and shade the whole rectangle OABC with light hatching.

\[\text{displacement in 5 second} = \text{area of the rectangle } OABC = OC \times OA = 5\ \text{s} \times 4\ \text{m s}^{-1} = 20\ \text{m}\]

2.7.4 Graph For Uniform Acceleration

If the body is in motion with uniform acceleration, that is, equal changes in velocity take place in equal intervals of time, the velocity–time graph is a straight line inclined to the time axis. The slope of the line gives the acceleration.

Table 2.9 represents the velocity of a body at different instants, starting from rest.

Table 2.9 — Velocity Of A Body Moving With Uniform Acceleration From Rest

Time (in s)	0	1	2	3	4	5	6	7	8
Velocity (in m s⁻¹)	0	10	20	30	40	50	60	70	80

Obviously, the velocity is increasing by an equal amount in each second, that is, the body is moving with uniform acceleration. The velocity–time graph is a straight line $OP$ inclined to the time axis, as shown in Fig. 2.7-3.

Diagram

Fig. 2.7-3 — Velocity–Time Graph With Velocity Increasing By An Equal Amount Each Second Graph axes with the horizontal axis labelled TIME (s), marked 0 to 8 in steps of 1, and the vertical axis labelled VELOCITY (m s⁻¹), marked 0 to 80 in steps of 10. Plot the points (0, 0), (1, 10), (2, 20), (3, 30), (4, 40), (5, 50), (6, 60), (7, 70) and (8, 80) as small circles and join them by a single straight line from the origin. Label the origin O and the top point P. Drop a vertical line from P to the time axis meeting it at Q at 8 s. Shade the triangle OPQ with light hatching, and add faint dashed horizontal guide lines from each plotted point to the velocity axis.

\[\text{distance travelled by the body in 8 second} = \text{area of the triangle } OPQ = \tfrac{1}{2} \times OQ \times QP\] \[= \tfrac{1}{2} \times 8\ \text{s} \times 80\ \text{m s}^{-1} = 320\ \text{m}\]

\[\text{acceleration of the body} = \text{slope of the line } OP = \frac{PQ}{QO} = \frac{(80 - 0)\ \text{m s}^{-1}}{(8 - 0)\ \text{s}} = \frac{80\ \text{m s}^{-1}}{8\ \text{s}} = 10\ \text{m s}^{-2}\]

2.7.5 Graph For Uniform Retardation

If the motion is with uniform retardation, that is, the velocity decreases by an equal amount in each second, the velocity–time graph will be a straight line inclined to the time axis with a negative slope.

In Fig. 2.7-4, the straight line $AB$ represents the velocity–time graph for a body initially moving with a velocity $40\ \text{m s}^{-1}$ which comes to a stop in $4\ \text{s}$ with uniform retardation.

Diagram

Fig. 2.7-4 — Velocity–Time Graph With Uniform Retardation Graph axes with the horizontal axis labelled TIME (s), marked 0 to 4 in steps of 1, and the vertical axis labelled VELOCITY (m s⁻¹), marked 0 to 40 in steps of 10. Draw a straight line falling from the point A at (0, 40) on the velocity axis down to the point B at (4, 0) on the time axis. Label A, B and the origin O, and shade the triangle AOB lightly.

\[\text{distance travelled by the body in 4 s} = \text{area of triangle } AOB = \tfrac{1}{2} \times OB \times OA\] \[= \tfrac{1}{2} \times 4\ \text{s} \times 40\ \text{m s}^{-1} = 80\ \text{m}\]

\[\text{retardation of the body} = -\,\text{slope of the line } AB = -\frac{OA}{BO} = -\frac{(0 - 40)\ \text{m s}^{-1}}{(4 - 0)\ \text{s}} = \frac{40}{4}\ \text{m s}^{-2} = 10\ \text{m s}^{-2}\]

Obviously, on a velocity–time graph, the larger the slope — that is, the more inclined the straight line — the higher is the acceleration or retardation.

One shape is again impossible. The velocity–time graph can never be a straight line parallel to the velocity axis, because such a line would mean that the velocity increases without any increase in time, that is, the acceleration is infinite, which is impossible.

2.7.6 Graph For A Body Not Starting From Rest

If the body is initially moving with some constant velocity and then it accelerates, the velocity–time sketch for the accelerated motion will start from the point on the velocity axis corresponding to the initial velocity of the body — not from the origin.

In Fig. 2.7-5, the straight line $AB$ represents the velocity–time graph of the accelerated motion of a car initially moving with velocity $10\ \text{m s}^{-1}$. Its velocity from the instant of acceleration at different times is given in Table 2.10.

Table 2.10 — Velocity Of A Car That Was Already Moving When It Began To Accelerate

Time (in second)	0	1	2	3	4	5
Velocity (in m s⁻¹)	10	15	20	25	30	35

Diagram

Fig. 2.7-5 — Velocity–Time Graph When The Body Is Initially Not At Rest Graph axes with the horizontal axis labelled TIME (s), marked 0 to 5 in steps of 1, and the vertical axis labelled VELOCITY (m s⁻¹), marked 0 to 40 in steps of 10. Plot the points (0, 10), (1, 15), (2, 20), (3, 25), (4, 30) and (5, 35) as small circles and join them by a straight line. Label the starting point on the velocity axis A and the end point B. Drop a vertical line from B to the time axis, meeting it at D at 5 s, and mark the point C on this vertical line level with A. Label the origin O and shade the trapezium ABDO with light hatching.

The displacement of the car in 5 second is the area of the trapezium $ABDO$: \[\tfrac{1}{2}(OA + DB) \times OD = \tfrac{1}{2} \times (10 + 35)\ \text{m s}^{-1} \times 5\ \text{s} = \tfrac{1}{2} \times 45 \times 5 = 112.5\ \text{m}\]

The acceleration of the car is the slope of the line $AB$: \[\frac{BC}{CA} = \frac{(35 - 10)\ \text{m s}^{-1}}{(5 - 0)\ \text{s}} = \frac{25\ \text{m s}^{-1}}{5\ \text{s}} = 5\ \text{m s}^{-2}\]

2.7.7 Graph For A Body Falling And Rising Again

Consider the motion of a body released from a height. It is initially at rest and then falls down vertically with velocity increasing uniformly for $5\ \text{s}$, acquiring a final velocity of $50\ \text{m s}^{-1}$. Assume that after hitting the ground it rises vertically upwards to the same height, with velocity decreasing uniformly.

Here a choice of sign has to be made before the graph can be drawn, and the choice is ours. Taking the downward direction as positive, the velocity–time graph is as shown in Fig. 2.7-6, in which the part $AB$ shows the downward journey of the body with increasing velocity, while the part $CD$ shows the upward journey with decreasing velocity. Conventionally the direction away from the earth is taken as positive instead; you may like to draw the graph for the same motion following that convention yourself.

Diagram

Fig. 2.7-6 — Velocity–Time Graph For The Free Fall And Rise Of A Body Graph axes with the horizontal axis labelled Time (in s), marked 1 to 10 in steps of 1, and the vertical axis labelled Velocity (in m s⁻¹), marked from +50 down to –50 in steps of 10, with the time axis drawn through zero so the graph extends below it. Mark A at the origin (0, 0). Draw a straight line rising from A to B at (5, 50) and label this part ‘Downward motion’. Drop a dashed vertical line from B down to the point C at (5, –50). From C draw a straight line rising to D at (10, 0), and label this part ‘Upward motion’. Mark the point E on the time axis at 5 s, so that triangle ABE lies above the axis and triangle CDE lies below it.

In the part $AB$, \[\text{acceleration} = \text{slope of line } AB = \frac{50\ \text{m s}^{-1} - 0\ \text{m s}^{-1}}{5\ \text{s}} = 10\ \text{m s}^{-2}\]

In the part $CD$, \[\text{retardation} = -\,\text{slope of line } CD = -\frac{0\ \text{m s}^{-1} - 50\ \text{m s}^{-1}}{(10 - 5)\ \text{s}} = 10\ \text{m s}^{-2}\]

Wait a moment over the second of these. The line $CD$ rises on the graph, yet the body is slowing down. That is because the velocity along $CD$ is negative throughout — the body is moving upwards while downwards was chosen as positive — and a negative velocity shrinking towards zero is a body coming to a halt.

\[\text{total distance travelled in 10 s} = \text{area of } \triangle ABE + \text{area of } \triangle CDE\] \[= \tfrac{1}{2} \times 5\ \text{s} \times 50\ \text{m s}^{-1} + \tfrac{1}{2} \times 5\ \text{s} \times 50\ \text{m s}^{-1} = 250\ \text{m}\]

The displacement in $10\ \text{s}$ is $0$, since the initial and final positions are the same.

2.7.8 What The Velocity–Time Graph Tells Us

The section gathers into five conclusions.

For motion with a uniform velocity, the velocity–time graph is a straight line parallel to the time axis. If the velocity–time graph is a straight line inclined to the time axis, the motion is with uniform acceleration. If the velocity–time graph is a curve, the motion is with non-uniform acceleration.
The slope of the straight line, or of the tangent to the curve at an instant, gives the acceleration at that instant. A positive slope means velocity increasing with time, that is, accelerated motion. A negative slope means velocity decreasing with time, that is, retarded motion. A zero slope implies motion with constant velocity.
Knowing the acceleration, that is, the slope, at different instants from the velocity–time graph, we can draw the acceleration–time graph.
The area enclosed between the velocity–time sketch and the time axis for a certain time interval gives the displacement in that interval of time. The area above the time axis gives the positive displacement, while the area below the time axis gives the negative displacement.
Knowing the distance, or displacement, in different time intervals from the velocity–time graph, we can draw the distance–time, or displacement–time, graph.

Example 2.11

Reading A Three-Part Velocity–Time Graph

1. Given: the velocity–time graph of a bus on a straight road is made of three straight parts — the velocity rises uniformly from $0\ \text{m s}^{-1}$ to $15\ \text{m s}^{-1}$ between $t = 0$ and $t = 5\ \text{s}$; stays at $15\ \text{m s}^{-1}$ from $t = 5\ \text{s}$ to $t = 15\ \text{s}$; then falls uniformly to $0\ \text{m s}^{-1}$ at $t = 20\ \text{s}$. Asked: the acceleration in the first part, the retardation in the last part, and the total distance travelled.

The slope of each part gives its acceleration, and the total area under the whole graph gives the distance. The graph encloses a triangle, then a rectangle, then a second triangle, all above the time axis.
Acceleration in the first part $= \dfrac{(15 - 0)\ \text{m s}^{-1}}{(5 - 0)\ \text{s}} = 3\ \text{m s}^{-2}$.

Slope of the last part $= \dfrac{(0 - 15)\ \text{m s}^{-1}}{(20 - 15)\ \text{s}} = -3\ \text{m s}^{-2}$, so the retardation is $3\ \text{m s}^{-2}$.

Area of the first triangle $= \tfrac{1}{2} \times 5\ \text{s} \times 15\ \text{m s}^{-1} = 37.5\ \text{m}$.

Area of the rectangle $= 10\ \text{s} \times 15\ \text{m s}^{-1} = 150\ \text{m}$.

Area of the last triangle $= \tfrac{1}{2} \times 5\ \text{s} \times 15\ \text{m s}^{-1} = 37.5\ \text{m}$.

Total distance $= 37.5\ \text{m} + 150\ \text{m} + 37.5\ \text{m} = 225\ \text{m}$.
The bus accelerates at $3\ \text{m s}^{-2}$, retards at $3\ \text{m s}^{-2}$, and covers $225\ \text{m}$ in all. The figure is sensible: had the bus held $15\ \text{m s}^{-1}$ for the whole $20\ \text{s}$ it would have covered $300\ \text{m}$, and it must cover less than that because it spent time speeding up and slowing down.

Extra Practice 2.6

A train moving along a straight track has the velocities given below.

| Time (in s) | 0 | 10 | 20 | 30 | 40 |

|—|—|—|—|—|—|

| Velocity (in m s⁻¹) | 5 | 15 | 25 | 25 | 25 |

Draw the velocity–time graph and use it to find (i) the acceleration during the first $20\ \text{s}$, (ii) the acceleration during the last $20\ \text{s}$, and (iii) the total distance travelled in $40\ \text{s}$.

2.8 Acceleration–Time Graph

In the acceleration–time graph, time is taken on the $X$-axis and acceleration is taken on the $Y$-axis. From this graph we can find the change in speed in a certain interval of time.

For linear motion, acceleration $\times$ time $=$ change in speed. So, just as with the velocity–time graph, an area on this graph carries a meaning: a height on it is an acceleration in m s⁻² and a width is a time in s, and their product is a change of speed in m s⁻¹. Therefore, from the area enclosed between the acceleration–time sketch and the time axis, we get the change in speed of the body for the given time interval.

Formula in Words

area under an acceleration–time graph = change in speed Multiply how much velocity is being added each second by the number of seconds, and you have the total velocity added. It gives the change in speed over that interval, not the speed itself — the starting speed must still be added on.

Let us consider the following cases.

When The Acceleration Is Zero

If the body is stationary, or if it is moving with a uniform velocity, the acceleration is zero. The acceleration–time graph in such a case is a straight line coinciding with the time axis, as shown in Fig. 2.8-1.

Diagram

Fig. 2.8-1 — Acceleration–Time Graph For Motion With Uniform Velocity Graph axes with the horizontal axis labelled TIME and an arrowhead at the right marked X, and the vertical axis labelled ACCELERATION with an arrowhead at the top marked Y. Mark the origin O. Draw a bold straight line lying exactly along the horizontal time axis, running from the origin to the right-hand end, to show that the acceleration is zero at every instant.

Do Not Mix Up

Zero acceleration and zero velocity. Zero acceleration means the velocity is not changing — the body may still be racing along at a steady speed. Zero velocity means the body is not moving at that instant. A car cruising at $60\ \text{km h}^{-1}$ has the first and not the second.

When The Acceleration Is Constant

If the velocity of a body in motion increases uniformly with time, the acceleration is constant, that is, the motion is uniformly accelerated. In such a case, the acceleration–time graph is a straight line parallel to the time axis on the positive side of the acceleration axis. In Fig. 2.8-2, the straight line $PQ$ represents the acceleration–time graph of a body moving with a constant acceleration equal to $OP$.

Diagram

Fig. 2.8-2 — Acceleration–Time Graph For Uniform Acceleration Graph axes with the horizontal axis labelled TIME with an arrowhead at the right marked X, and the vertical axis labelled ACCELERATION with an arrowhead at the top marked Y. Mark the origin O. Draw a bold horizontal straight line well above the time axis, running the full width of the graph; label its left-hand end P, where it meets the acceleration axis, and its right-hand end Q. Mark the vertical gap OP on the acceleration axis.

When The Retardation Is Constant

If the velocity of a body decreases at a constant rate, the retardation is constant, that is, the motion is uniformly retarded. The acceleration–time graph is then a straight line parallel to the time axis on the negative side of the acceleration axis. In Fig. 2.8-3, the straight line $PQ$ represents the acceleration–time graph for a body moving with a constant retardation equal to $OP$.

Diagram

Fig. 2.8-3 — Acceleration–Time Graph For Uniform Retardation Graph axes with the horizontal time axis drawn across the middle of the figure, labelled TIME with an arrowhead at the right marked X, and a vertical axis through the origin O, labelled ACCELERATION above the origin and RETARDATION below it, with –Y marked at the bottom. Draw a bold horizontal straight line well below the time axis, running the full width of the graph; label its left-hand end P, where it meets the vertical axis, and its right-hand end Q.

When The Acceleration Is Variable

If the velocity of a body changes in an irregular manner, the acceleration is variable. The acceleration–time graph will then be a curve of any shape.

Example 2.12

Change In Speed From An Acceleration–Time Graph

1. Given: the acceleration–time graph of a body moving in a straight line is a horizontal line at $4\ \text{m s}^{-2}$, drawn from $t = 0$ to $t = 6\ \text{s}$; the body’s speed at $t = 0$ is $2\ \text{m s}^{-1}$. Asked: the change in speed during these $6\ \text{s}$, and the speed at the end of them.

The area enclosed between the acceleration–time sketch and the time axis gives the change in speed. Here that area is a rectangle, since the acceleration stays constant throughout.
Change in speed $=$ area $= 4\ \text{m s}^{-2} \times 6\ \text{s} = 24\ \text{m s}^{-1}$.

Speed at the end $= 2\ \text{m s}^{-1} + 24\ \text{m s}^{-1} = 26\ \text{m s}^{-1}$.
The speed rises by $24\ \text{m s}^{-1}$, ending at $26\ \text{m s}^{-1}$. The answer is sensible, since the body gains $4\ \text{m s}^{-1}$ in every one of the six seconds. Note that the area gave only the gain — the starting speed had to be added to it.

Extra Practice 2.7

A body starts from rest and its acceleration–time graph is a horizontal line at $2\ \text{m s}^{-2}$ for the first $5\ \text{s}$, then a horizontal line along the time axis for the next $5\ \text{s}$. Find the speed of the body at $t = 5\ \text{s}$ and at $t = 10\ \text{s}$, and state what kind of motion the body has during the second interval.

2.9 Motion Under Gravity

A body falling freely under gravity moves with a uniform acceleration of $9.8\ \text{m s}^{-2}$, or nearly $10\ \text{m s}^{-2}$. For a body moving vertically upwards, there is a uniform retardation of $9.8\ \text{m s}^{-2}$. Thus motion under gravity is an example of uniformly accelerated or uniformly retarded motion.

Here we shall consider the motion of a freely falling body under gravity, and we shall use the acceleration–time graph to obtain the velocity–time graph, and then the velocity–time graph to obtain the displacement–time graph. The route is set out in FC 2.1. In the graphs that follow, the value of the acceleration due to gravity is taken as $10\ \text{m s}^{-2}$ throughout.

Flowchart

FC 2.1 — Building The Three Graphs Of A Freely Falling Body Acceleration–time graph (a straight line parallel to the time axis) → area under it for each 1 s interval → velocity at the end of each second → velocity–time graph (a straight line through the origin) → area under it up to each instant → displacement at each instant → displacement–time graph (a curve)

2.9.1 From The Acceleration–Time Graph To The Velocity–Time Graph

The acceleration–time graph for such a motion is a straight line parallel to the time axis, because the acceleration keeps the same value $10\ \text{m s}^{-2}$ at every instant. In Fig. 2.9-1, the straight line $AF$ represents the acceleration–time graph for a body falling freely with this uniform acceleration.

Diagram

Fig. 2.9-1 — Acceleration–Time Graph For A Freely Falling Body Graph axes with the horizontal axis labelled TIME (s), marked 0 to 5 in steps of 1, and the vertical axis labelled ACCELERATION (m s⁻²), marked 0 to 20 in steps of 5. Draw a bold horizontal straight line at 10 m s⁻² running from the acceleration axis across to 5 s. Label the point where it meets the acceleration axis A, and mark points on the line vertically above each second: B at 1 s, C at 2 s, D at 3 s, E at 4 s and F at 5 s. From each of these points drop a dashed vertical line to the time axis, meeting it at P (1 s), Q (2 s), R (3 s), S (4 s) and T (5 s). Mark the origin O.

This graph can be used to obtain the velocity–time graph, by finding the area enclosed between the straight line and the time axis for each interval of time of $1\ \text{s}$. Let the initial velocity at $t = 0$ be zero. Then

velocity after $1 = $ area $OABP = 10 \times 1 = 10\ \text{m s}^{-1}$
velocity after $2 = $ area $OACQ = 10 \times 2 = 20\ \text{m s}^{-1}$
velocity after $3 = $ area $OADR = 10 \times 3 = 30\ \text{m s}^{-1}$
velocity after $4 = $ area $OAES = 10 \times 4 = 40\ \text{m s}^{-1}$
velocity after $5 = $ area $OAFT = 10 \times 5 = 50\ \text{m s}^{-1}$

The velocity–time graph from the above data is shown in Fig. 2.9-2. It is a straight line $OA$ inclined to the time axis and having a slope of $10\ \text{m s}^{-2}$, which is equal to the acceleration of the body.

Diagram

Fig. 2.9-2 — Velocity–Time Graph For A Freely Falling Body Graph axes with the horizontal axis labelled TIME (s), marked 0 to 5 in steps of 1, and the vertical axis labelled VELOCITY (m s⁻¹), marked 0 to 50 in steps of 10. Plot the points (0, 0), (1, 10), (2, 20), (3, 30), (4, 40) and (5, 50) as small circles and join them by a straight line rising from the origin. Label the origin O and the top point A. Label the plotted points along the line a, b, c, d and e in order, and mark the corresponding points on the time axis as p (1 s), q (2 s), r (3 s), s (4 s) and t (5 s), joining each to its plotted point by a dashed vertical line.

2.9.2 From The Velocity–Time Graph To The Displacement–Time Graph

The velocity–time graph of Fig. 2.9-2 can in turn be used to obtain the displacement–time graph, by finding the area enclosed by the straight line $OA$ with the time axis at each interval of time of $1\ \text{s}$. Each of these areas is a triangle, whose base is the time elapsed and whose height is the velocity reached by then.

displacement in $1 = $ area of $\triangle\ oap = \tfrac{1}{2} \times 1 \times 10 = 5\ \text{m}$
displacement in $2 = $ area of $\triangle\ obq = \tfrac{1}{2} \times 2 \times 20 = 20\ \text{m}$
displacement in $3 = $ area of $\triangle\ ocr = \tfrac{1}{2} \times 3 \times 30 = 45\ \text{m}$
displacement in $4 = $ area of $\triangle\ ods = \tfrac{1}{2} \times 4 \times 40 = 80\ \text{m}$
displacement in $5 = $ area of $\triangle\ oet = \tfrac{1}{2} \times 5 \times 50 = 125\ \text{m}$

The displacement–time graph from the above data is shown in Fig. 2.9-3, which is a curve $OA$, in fact a parabola.

Diagram

Fig. 2.9-3 — Displacement–Time Graph For A Freely Falling Body Graph axes with the horizontal axis labelled TIME (s), marked 0 to 5 in steps of 1, and the vertical axis labelled DISPLACEMENT (m), marked 0 to 140 in steps of 20. Plot the points (0, 0), (1, 5), (2, 20), (3, 45), (4, 80) and (5, 125) as small circles and join them by a smooth upward-curving line that becomes steeper and steeper. Label the origin O and the top of the curve A. Draw faint dashed guide lines from each plotted point across to the displacement axis and down to the time axis.

Notice how the curve steepens as it climbs. The slope of a displacement–time graph is the velocity, so a steepening curve is precisely a body whose velocity is growing — which is what a falling body does.

2.9.3 The Displacement–Square Of Time Graph

It may be noted from the numbers above that for a freely falling body, the displacement is directly proportional to the square of time, $S \propto t^2$. Read them again: at $1\ \text{s}$, $2\ \text{s}$, $3\ \text{s}$ the displacements are $5\ \text{m}$, $20\ \text{m}$, $45\ \text{m}$ — that is, $5\ \text{m}$ multiplied by $1$, $4$ and $9$.

Formula in Words

$S \propto t^2$ for a freely falling body Double the time of fall and the body does not fall twice as far — it falls four times as far. Treble the time and it falls nine times as far. This holds only for a body starting from rest and falling freely, with air resistance ignored.

Table 2.11 represents the square of time, $t^2$, and the displacement, $S$, from the data above.

Table 2.11 — Displacement Of A Freely Falling Body Against The Square Of Time

Square of time $t^2$ (in s²)	1	4	9	16	25
Displacement $S$ (in m)	5	20	45	80	125

Now a graph plotted by taking the displacement $S$ on the $Y$-axis and the square of time $t^2$ on the $X$-axis is a straight line $OA$, as shown in Fig. 2.9-4.

Diagram

Fig. 2.9-4 — The S–t² Graph For A Freely Falling Body Graph axes with the horizontal axis labelled SQUARE OF TIME (s²), marked 0 to 25 in steps of 5, and the vertical axis labelled DISPLACEMENT (m), marked 0 to 140 in steps of 20. Plot the points (1, 5), (4, 20), (9, 45), (16, 80) and (25, 125) as small circles and join them by a single straight line passing through the origin. Label the origin O and the top point A. Draw faint dashed guide lines from the points at (4, 20) and (16, 80) across to the displacement axis and down to the horizontal axis.

Interpret Data

Fig. 2.9-4 — The S–t² Graph For A Freely Falling Body 1. Read the axes. The horizontal axis is not time but the square of time, measured in s². The vertical axis is displacement in metre. Plotting $t^2$ rather than $t$ is what has turned the curve of Fig. 2.9-3 into a straight line. 2. Note what a straight line through the origin means here: $S$ is directly proportional to $t^2$. 3. Take two points on the line and find the slope, dividing the rise in displacement by the run in $t^2$: \[\text{slope} = \frac{(80 - 20)\ \text{m}}{(16 - 4)\ \text{s}^2} = \frac{60}{12}\ \text{m s}^{-2} = 5\ \text{m s}^{-2}\] 4. Interpret the slope. The slope is half the acceleration due to gravity. Thus the value of the acceleration due to gravity, $g$, can be obtained by doubling the slope of the $S$–$t^2$ graph for a freely falling body. 5. Check it. Doubling $5\ \text{m s}^{-2}$ gives $10\ \text{m s}^{-2}$, which is exactly the value of $g$ used to produce the data in the first place.

This is worth pausing on, because it is a genuine measurement. A body falling in air is far too quick to time by hand, but the displacements and times of a slow, gentle version of the same motion can be measured comfortably — and the $S$–$t^2$ graph turns those measurements straight into a value of $g$.

Example 2.13

Finding g From The Slope Of An S–t² Graph

1. Given: for a body falling freely from rest, the graph of displacement $S$ in metre against the square of time $t^2$ in s² is a straight line through the origin, and two points on it are $(2\ \text{s}^2,\ 9.8\ \text{m})$ and $(6\ \text{s}^2,\ 29.4\ \text{m})$. Asked: the value of the acceleration due to gravity at that place.

For a freely falling body the slope of the $S$–$t^2$ graph is half the acceleration due to gravity, so $g$ is found by measuring the slope and doubling it.
\[\text{slope} = \frac{(29.4 - 9.8)\ \text{m}}{(6 - 2)\ \text{s}^2} = \frac{19.6\ \text{m}}{4\ \text{s}^2} = 4.9\ \text{m s}^{-2}\]

\[g = 2 \times 4.9\ \text{m s}^{-2} = 9.8\ \text{m s}^{-2}\]
The acceleration due to gravity at that place is $9.8\ \text{m s}^{-2}$. The answer is exactly the average value of $g$ on the earth’s surface, so the measurement is a sensible one.

Try Yourself

Hold a small stone in one hand and a crumpled ball of paper in the other, at the same height, and release both together. Watch which lands first. Now flatten the paper into an open sheet and repeat. Note what changed between the two trials — and remember that it is not the mass.

Extra Practice 2.8

A body falls freely from rest and covers $45\ \text{m}$ in the first $3\ \text{s}$. Taking the displacement to be directly proportional to the square of the time, find the distance it will have fallen in the first $6\ \text{s}$, and state the value of $g$ that the data implies.

2.10 Equations Of Uniformly Accelerated Motion

Five quantities describe a straight-line journey with uniform acceleration: the initial velocity $u$, the final velocity $v$, the acceleration $a$, the time of journey $t$, and the distance travelled $S$. Given any three of them, the other two can be found — and the tools for doing so are three short relations.

For motion of a body moving with a uniform acceleration, the following three equations give the relationship between initial velocity $u$, final velocity $v$, acceleration $a$, time of journey $t$ and distance travelled $S$: \[v = u + at\] \[S = \tfrac{1}{2}(u + v)\,t = ut + \tfrac{1}{2}at^2\] \[v^2 = u^2 + 2aS\] where $u$ and $v$ are in m s⁻¹, $a$ is in m s⁻², $t$ is in s and $S$ is in m. All three hold only for motion along a straight line in which the acceleration remains uniform throughout the interval considered. They must never be applied to a journey in which the acceleration changes partway; such a journey is split into stages, and the equations are used on each stage separately.

2.10.1 Derivation From The Velocity–Time Graph

Consider the linear motion of a body with an initial velocity $u$. The body accelerates uniformly, and in time $t$ it acquires the final velocity $v$. The velocity–time graph is then a straight line $AB$, as shown in Fig. 2.10-1.

Diagram

Fig. 2.10-1 — Graph Showing The Linear Motion With Uniform Acceleration Graph axes with the horizontal axis labelled TIME (s) with an arrowhead at the right marked X, and the vertical axis labelled VELOCITY (m s⁻¹) with an arrowhead at the top marked Y. Mark the origin O. On the velocity axis mark A at a height u, low down, and C higher up at a height v. Draw a straight line rising from A to the point B, which lies vertically above the point D on the time axis at time t, and level with C. Draw the dashed horizontal line from A across to the point E on the line BD, so that AE is horizontal and EB is the vertical part of BD above E. Draw the dashed horizontal line from C across to B. Label the points A, B, C, D, E and O, and mark u against OA, v against OC and t against OD.

It is evident from the graph that \[\text{initial velocity (at } t = 0) = OA = u, \qquad \text{final velocity (at time } t) = OC = v\]

The First Equation

Acceleration is the slope of the line $AB$: \[a = \frac{EB}{AE} = \frac{AC}{OD} = \frac{OC - OA}{OD} = \frac{v - u}{t}\] \[\text{or}\qquad at = v - u\] \[\therefore\qquad v = u + at\]

The Second Equation

The distance $S$ travelled in time $t$ is the area of the trapezium $OABD$. That trapezium can be cut into two simpler pieces — the rectangle $OAED$ beneath the dashed line, and the triangle $ABE$ above it. So \[S = \text{area of rectangle } OAED + \text{area of triangle } ABE\] \[S = OA \times OD + \tfrac{1}{2} \times BE \times AE = u \times t + \tfrac{1}{2} \times (v - u) \times t\] But from the first equation, $v - u = at$. Substituting this for $(v - u)$, \[S = ut + \tfrac{1}{2} \times at \times t\] \[\therefore\qquad S = ut + \tfrac{1}{2}at^2\]

The Third Equation

The distance $S$ travelled in time $t$ is the area of the same trapezium $OABD$, and the area of a trapezium can also be written directly as half the sum of the two parallel sides multiplied by the distance between them: \[S = \tfrac{1}{2}(OA + DB) \times OD\] \[S = \tfrac{1}{2}(u + v) \times t\] From the first equation, $t = \dfrac{v - u}{a}$. Substituting this for $t$, \[S = \tfrac{1}{2}(u + v) \times \left(\frac{v - u}{a}\right) = \frac{1}{2}\left(\frac{v^2 - u^2}{a}\right)\] since $(u + v)(v - u) = v^2 - u^2$. Multiplying both sides by $2a$, \[2aS = v^2 - u^2\] \[\therefore\qquad v^2 = u^2 + 2aS\]

Formula in Words

$S = ut + \frac{1}{2}at^2$ The distance covered is made of two parts. The first, $ut$, is how far the body would have gone had it simply kept its starting velocity. The second, $\frac{1}{2}at^2$, is the extra distance the acceleration adds. It holds only while the acceleration stays uniform.

2.10.2 Derivation By The Algebraic Method

The same three equations can be reached without drawing a graph at all. This route lies outside the syllabus, but it is short and it shows that the equations do not depend on the picture.

By definition, \[\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}} = \frac{\text{final velocity} - \text{initial velocity}}{\text{time taken}}\] \[a = \frac{v - u}{t} \qquad \text{or} \qquad at = v - u \qquad \text{or} \qquad v = u + at\]

Next, since the velocity changes uniformly, the average velocity over the interval is simply the mean of the velocities at its two ends: \[\text{distance travelled} = \text{average velocity} \times \text{time} = \left(\frac{\text{initial velocity} + \text{final velocity}}{2}\right) \times \text{time}\] \[S = \frac{u + v}{2} \times t\] Putting $v = u + at$ into this, \[S = \frac{u + (u + at)}{2} \times t = \left(\frac{2u + at}{2}\right) \times t \qquad \text{or} \qquad S = ut + \tfrac{1}{2}at^2\]

Finally, starting again from $S = \dfrac{u + v}{2} \times t$ and putting $t = \dfrac{v - u}{a}$ into it, \[S = \frac{u + v}{2} \times \frac{v - u}{a} = \frac{v^2 - u^2}{2a} \qquad \text{or} \qquad v^2 - u^2 = 2aS \qquad \text{or} \qquad v^2 = u^2 + 2aS\]

These are the same three equations as before.

2.10.3 Special Cases Of The Equations

Two situations come up so often that the equations are worth writing out in their shortened forms.

When The Body Starts From Rest

When a body starts from rest, the initial velocity is zero, $u = 0$, and the equations become \[v = at, \qquad S = \tfrac{1}{2}at^2, \qquad v^2 = 2aS\]

When The Body Is Uniformly Retarded

If a body is moving with a uniform retardation, $a$ will be negative. The equations then take the form \[v = u - at, \qquad S = ut - \tfrac{1}{2}at^2, \qquad v^2 = u^2 - 2aS\] where $a$ now stands for the magnitude of the retardation in m s⁻².

Signs and Units

Using the equations of motion Units: convert every quantity to m, s, m s⁻¹ and m s⁻² before substituting. Never mix km h⁻¹ with metre. Signs: fix one direction as positive first. A body slowing down then has $a$ negative; a body falling with downwards positive has $a = +g$.

Choosing The Right Equation

Each of the three equations leaves one quantity out. That is the quickest way to pick one: look for the quantity the question neither gives you nor asks for, and use the equation in which it does not appear. FC 2.2 sets out the route, and Table 2.12 lists the three side by side.

Flowchart

FC 2.2 — Choosing Which Equation Of Motion To Use List the quantities given and the quantity asked for → find which of $u$, $v$, $a$, $t$, $S$ is neither given nor asked → if the missing one is $S$, use $v = u + at$ → if the missing one is $v$, use $S = ut + \frac{1}{2}at^2$ → if the missing one is $t$, use $v^2 = u^2 + 2aS$ → if the missing one is $a$, use $S = \frac{1}{2}(u+v)t$ → substitute in SI units and solve

Compare Table

Table 2.12 — The Equations Of Uniformly Accelerated Motion Compared | Basis of difference | $v = u + at$ | $S = ut + \frac{1}{2}at^2$ | $v^2 = u^2 + 2aS$ | |—|—|—|—| | Quantities linked | $u$, $v$, $a$, $t$ | $S$, $u$, $a$, $t$ | $v$, $u$, $a$, $S$ | | Quantity left out | The distance $S$ | The final velocity $v$ | The time $t$ | | Use it when | The time is known and the distance is not needed | The distance in a stated time is wanted | The distance is known but the time is not | | Form when the body starts from rest | $v = at$ | $S = \frac{1}{2}at^2$ | $v^2 = 2aS$ |

Example 2.14

A Body Starting From Rest

1. Given: initial velocity $u = 0\ \text{m s}^{-1}$, uniform acceleration $a = 2\ \text{m s}^{-2}$, time $t = 6\ \text{s}$. Asked: the distance travelled and the velocity acquired.

For the distance, the quantity not involved is the final velocity, so $S = ut + \frac{1}{2}at^2$ is used. For the velocity, $v = u + at$ is used. Both apply because the acceleration is uniform and the motion is along a straight line.
\[S = 0 \times 6\ \text{s} + \tfrac{1}{2} \times 2\ \text{m s}^{-2} \times (6\ \text{s})^2 = \tfrac{1}{2} \times 2 \times 36 = 36\ \text{m}\]

\[v = 0 + 2\ \text{m s}^{-2} \times 6\ \text{s} = 12\ \text{m s}^{-1}\]
The body travels $36\ \text{m}$ and reaches $12\ \text{m s}^{-1}$. The two answers agree with each other: the average velocity over the interval is $\frac{0 + 12}{2} = 6\ \text{m s}^{-1}$, and $6\ \text{m s}^{-1} \times 6\ \text{s}$ is indeed $36\ \text{m}$.

Example 2.15

Retardation Over A Known Distance

1. Given: a car moving with $u = 20\ \text{m s}^{-1}$ is braked uniformly and comes to rest, $v = 0\ \text{m s}^{-1}$, after travelling $S = 50\ \text{m}$. Asked: the retardation produced by the brakes.

The time is neither given nor asked for, so the equation without $t$ in it is the one to use: $v^2 = u^2 + 2aS$. Take the direction of motion as positive.
\[(0)^2 = (20\ \text{m s}^{-1})^2 + 2 \times a \times 50\ \text{m}\]

\[0 = 400\ \text{m}^2\text{s}^{-2} + 100\ \text{m} \times a\]

\[a = -\frac{400\ \text{m}^2\text{s}^{-2}}{100\ \text{m}} = -4\ \text{m s}^{-2}\]
The acceleration is $-4\ \text{m s}^{-2}$, so the retardation is $4\ \text{m s}^{-2}$. The minus sign is expected for a braking car, and the size is reasonable — the car loses $4\ \text{m s}^{-1}$ of velocity each second, and so stops in $5\ \text{s}$.

Example 2.16

Final Velocity After A Given Distance

1. Given: a train moving with $u = 10\ \text{m s}^{-1}$ accelerates uniformly at $a = 2\ \text{m s}^{-2}$ over a distance $S = 75\ \text{m}$. Asked: its final velocity.

Again the time is not involved, so $v^2 = u^2 + 2aS$ is chosen.
\[v^2 = (10\ \text{m s}^{-1})^2 + 2 \times 2\ \text{m s}^{-2} \times 75\ \text{m} = 100 + 300 = 400\ \text{m}^2\text{s}^{-2}\]

\[v = \sqrt{400\ \text{m}^2\text{s}^{-2}} = 20\ \text{m s}^{-1}\]
The final velocity is $20\ \text{m s}^{-1}$. It is larger than the initial velocity, as it must be for an accelerating train, and the units work out correctly: the square root of m² s⁻² is m s⁻¹.

Example 2.17

Depth Of A Well From The Time Of Fall

1. Given: a stone is dropped from rest at the mouth of a well, so $u = 0\ \text{m s}^{-1}$, and it reaches the water surface after $t = 2\ \text{s}$. The acceleration is that due to gravity, $g = 9.8\ \text{m s}^{-2}$. Asked: the depth of the water surface below the mouth.

Free fall is uniformly accelerated motion, so the equations apply. The distance in a stated time is wanted, so $S = ut + \frac{1}{2}at^2$ is used, with the downward direction taken as positive so that $a = +9.8\ \text{m s}^{-2}$. Air resistance is ignored.
\[S = 0 \times 2\ \text{s} + \tfrac{1}{2} \times 9.8\ \text{m s}^{-2} \times (2\ \text{s})^2 = \tfrac{1}{2} \times 9.8 \times 4 = 19.6\ \text{m}\]
The water surface is $19.6\ \text{m}$ below the mouth of the well. The figure is a believable depth for a well, and it agrees with the $S \propto t^2$ pattern of Section 2.9 — in half the time, one quarter of the distance.

Extra Practice 2.9

A motorcycle moving at $8\ \text{m s}^{-1}$ accelerates uniformly at $1.5\ \text{m s}^{-2}$ for $6\ \text{s}$. Find (i) its velocity at the end of the $6\ \text{s}$, (ii) the distance it covers in that time, and (iii) the further distance it would take to stop if the brakes then produced a uniform retardation of $5\ \text{m s}^{-2}$.

2.11 Summary

A physical quantity is one that can be measured. Physical quantities are of two kinds: scalars, which are expressed by magnitude alone, and vectors, which need both a magnitude and a direction.
A scalar needs two things to state it fully — the unit and the numerical value. A vector needs three — the numerical value, the unit and the direction.
The magnitude of a vector is always positive. A negative sign on a vector means the same magnitude in the opposite direction.
A body is at rest if it does not change its position with respect to its immediate surroundings, and in motion if it does change it. Rest and motion are therefore relative.
Motion along a straight line path is one dimensional, or rectilinear, motion. Motion on a plane along a curved path is two dimensional, and motion in space is three dimensional.
The total length of path along which a body moves is the distance travelled by it. It is a scalar quantity, it depends on the path followed, and its SI unit is the metre (m).
The shortest distance from the initial to the final position is the magnitude of the displacement, its direction being from the initial position to the final position. It is a vector quantity and its SI unit is the metre (m).
The magnitude of displacement is equal to or less than the distance, and can never be greater. The displacement can be zero even when the distance is not zero.
Speed is the rate of change of distance with time. It is a scalar quantity and its SI unit is m s⁻¹. \[v = \frac{S}{t}\] where $S$ is the distance in m, $t$ is the time in s and $v$ is the speed in m s⁻¹.
A body has uniform speed if it covers equal distances in equal intervals of time, and non-uniform speed otherwise. For uniform speed, $S = vt$, with $S$ in m, $v$ in m s⁻¹ and $t$ in s.
The speed at any instant is the instantaneous speed; a speedometer reads it. Over a whole journey, \[\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}\] both distances in m and times in s, giving m s⁻¹.
Velocity is the rate of change of displacement with time. It is a vector quantity and its SI unit is m s⁻¹.
A body has uniform velocity if it travels equal distances in a particular direction in equal intervals of time. Velocity is variable if its magnitude changes, or its direction changes, or both — so motion in a circular path with constant speed is still motion with variable velocity.
\[\text{average velocity} = \frac{\text{displacement}}{\text{total time taken}}\] with displacement in m and time in s. The average velocity can be zero while the average speed is not.
Acceleration is the rate of change of velocity with time. It is a vector quantity and its SI unit is m s⁻². \[a = \frac{v - u}{t} \qquad \text{or} \qquad v = u + at\] where $u$ is the initial velocity in m s⁻¹, $v$ the final velocity in m s⁻¹, $t$ the time in s and $a$ the acceleration in m s⁻².
A negative acceleration is called retardation. The sign of the acceleration says whether the velocity is growing or shrinking; the sign of the velocity says which way the body is going.
Acceleration is uniform when equal changes in velocity take place in equal intervals of time, and variable otherwise.
The acceleration produced in a freely falling body by the earth’s attraction is the acceleration due to gravity, $g$, whose average value is $9.8\ \text{m s}^{-2}$, or nearly $10\ \text{m s}^{-2}$. It is $+g$ for a falling body and $-g$ for a rising one. Its value does not depend on the mass of the body, so in vacuum all bodies fall together.
The value of $g$ is maximum at the poles and minimum at the equator, and decreases with altitude and with depth.
In a displacement–time graph, the slope gives the velocity. A line parallel to the time axis means a stationary body; a straight line inclined to the time axis means uniform velocity; a curve means variable velocity, whose value at an instant is the slope of the tangent there. A positive slope means motion away from the starting point, a negative slope motion back towards it.
In a velocity–time graph, the slope gives the acceleration and the area enclosed with the time axis gives the displacement. Area above the time axis is positive displacement and area below it is negative; the total distance is their sum taken without sign.
A line parallel to the time axis on a velocity–time graph means uniform velocity and zero acceleration; a straight line inclined to it means uniform acceleration or uniform retardation; a curve means non-uniform acceleration.
In an acceleration–time graph, the area enclosed with the time axis gives the change in speed in that interval. The graph lies along the time axis for zero acceleration, and is a straight line parallel to the time axis for uniform acceleration or uniform retardation.
For a freely falling body the velocity–time graph is a straight line through the origin, the displacement–time graph is a curve, and the displacement is directly proportional to the square of time, $S \propto t^2$. The slope of the $S$–$t^2$ graph is half of $g$, so doubling that slope gives $g$ in m s⁻².
For motion in a straight line with uniform acceleration, the three equations of motion are \[v = u + at, \qquad S = ut + \tfrac{1}{2}at^2, \qquad v^2 = u^2 + 2aS\] and also $S = \tfrac{1}{2}(u + v)t$, where $u$ and $v$ are in m s⁻¹, $a$ in m s⁻², $t$ in s and $S$ in m. They hold only while the acceleration remains uniform.
For a body starting from rest, $u = 0$, and the equations become $v = at$, $S = \frac{1}{2}at^2$ and $v^2 = 2aS$. For a body under uniform retardation, $a$ is negative, giving $v = u - at$, $S = ut - \frac{1}{2}at^2$ and $v^2 = u^2 - 2aS$.

Flowchart

FC 2.3 — The Spine Of The Chapter Position of a body → change of position gives distance and displacement → rate of change of distance gives speed, rate of change of displacement gives velocity → rate of change of velocity gives acceleration → displacement–time graph (slope = velocity) → velocity–time graph (slope = acceleration, area = displacement) → acceleration–time graph (area = change in speed) → for uniform acceleration these relations condense into the three equations of motion

2.12 Key Terms

Acceleration — the rate of change of velocity with time; SI unit m s⁻².
Acceleration due to gravity — the acceleration produced in a body by the earth’s gravitational attraction, denoted $g$, of average value $9.8\ \text{m s}^{-2}$; SI unit m s⁻².
Average speed — the total distance travelled divided by the total time taken; SI unit m s⁻¹.
Average velocity — the displacement divided by the total time taken; SI unit m s⁻¹.
Displacement — the shortest distance from the initial to the final position of a body, taken in that direction; a vector; SI unit m.
Distance — the total length of path along which a body moves; a scalar; SI unit m.
Instantaneous speed — the speed of a body at a particular instant, found from the distance covered in a very short interval divided by that interval; SI unit m s⁻¹.
Instantaneous velocity — the velocity of a body at a particular instant; SI unit m s⁻¹.
Magnitude — the size of a vector quantity, given by its numerical value with its unit, always positive.
Motion — the state of a body that changes its position with respect to its immediate surroundings.
Non-uniform speed — speed of a body that covers unequal distances in equal intervals of time; SI unit m s⁻¹.
Non-uniform velocity — velocity whose magnitude, or direction, or both, change with time; SI unit m s⁻¹.
One dimensional motion — motion of a body along a straight line path, also called rectilinear motion.
Physical quantity — any quantity about the world that can be measured.
Point particle — a body treated as a single point, allowed when the distance it travels is much larger than its own size.
Rest — the state of a body that does not change its position with respect to its immediate surroundings.
Retardation — a negative acceleration, that is, the decrease in velocity per second; SI unit m s⁻².
Scalar quantity — a physical quantity expressed by its magnitude alone.
Slope — the rise of a graph divided by the corresponding run; it gives velocity on a displacement–time graph and acceleration on a velocity–time graph.
Speed — the rate of change of distance with time; a scalar; SI unit m s⁻¹.
Terminal velocity — the steady velocity a falling body reaches when the viscous force and upthrust of the air together balance its weight; SI unit m s⁻¹.
Uniform acceleration — acceleration in which equal changes of velocity take place in equal intervals of time; SI unit m s⁻².
Uniform speed — speed of a body that covers equal distances in equal intervals of time throughout its motion; SI unit m s⁻¹.
Uniform velocity — velocity of a body that travels equal distances in a particular direction in equal intervals of time; SI unit m s⁻¹.
Variable acceleration — acceleration whose value is not the same in equal intervals of time; SI unit m s⁻².
Vector quantity — a physical quantity that needs both a magnitude and a direction to be expressed completely.
Velocity — the rate of change of displacement with time; a vector; SI unit m s⁻¹.

2.13 Points To Ponder

A displacement–time graph can never be a straight line parallel to the displacement axis, and a velocity–time graph can never be a straight line parallel to the velocity axis. Both shapes are ruled out for the same kind of reason. What is that reason, and can you describe any other graph shape that physics forbids for a similar reason?
The average speed of a journey is not the average of the separate speeds along it. Under what condition would the two happen to give the same answer — and how often is that condition met in a real journey?
At the highest point of its flight, a ball thrown straight up has zero velocity but its acceleration is still $9.8\ \text{m s}^{-2}$. What does this tell you about the relationship between the two quantities, and how would you argue against someone who says a body with zero velocity must have zero acceleration?
The acceleration due to gravity does not depend on the mass of a body, so a heavy stone and a light one fall together. Yet the earth surely pulls the heavier stone harder. How can both of these statements be true at the same time?
Which direction we call positive is entirely our own choice. If the choice were reversed for the body that falls and then rises in Section 2.7, which of the quantities read from the graph — velocity, acceleration, displacement, distance — would change sign, and which would not? What does that tell you about which quantities are physically real and which are bookkeeping?
We can find the velocity of a body at an instant from the slope of a tangent, and the distance covered from an area under a graph. Both are geometrical ideas, yet they gave physical answers with proper units. Why should shapes drawn on paper be able to tell us anything true about a moving train?

2.14 Objective Questions

Which one of the following is a vector quantity? (1 mark)
1. Work
2. Momentum
3. Frequency
4. Specific heat
A body covers equal distances in equal intervals of time along a curved path. Its motion is with: (1 mark)
1. uniform velocity
2. uniform speed and uniform velocity
3. uniform speed but variable velocity
4. variable speed and uniform velocity
The SI unit of retardation is: (1 mark)
1. m s⁻¹
2. m s⁻²
3. s m⁻¹
4. m² s⁻¹
The area enclosed between a velocity–time graph and the time axis gives the: (1 mark)
1. acceleration
2. speed
3. displacement
4. retardation
$72\ \text{km h}^{-1}$ is equal to: (1 mark)
1. $7.2\ \text{m s}^{-1}$
2. $20\ \text{m s}^{-1}$
3. $25\ \text{m s}^{-1}$
4. $36\ \text{m s}^{-1}$
For a freely falling body starting from rest, the displacement is directly proportional to: (1 mark)
1. $t$
2. $t^2$
3. $\sqrt{t}$
4. $\dfrac{1}{t}$
A body starts from rest and moves with uniform acceleration $a$. The equation that gives its distance in time $t$ is: (1 mark)
1. $S = at$
2. $S = \dfrac{1}{2}at^2$
3. $S = 2at^2$
4. $S = at^2$
The acceleration–time graph of a body moving with uniform velocity is: (1 mark)
1. a straight line coinciding with the time axis
2. a straight line parallel to the acceleration axis
3. a straight line inclined to the time axis
4. a curve of any shape

For each of the following, choose from the codes given below. (1 mark each) a. Both A and R are true, and R is the correct explanation of A b. Both A and R are true, but R is not the correct explanation of A c. A is false but R is true d. A is true but R is false

Assertion (A): A body moving round a circular track at constant speed has a variable velocity. Reason (R): The direction of motion of the body changes continuously with time. (1 mark)
Assertion (A): The displacement–time graph of a body at rest is a straight line parallel to the time axis. Reason (R): The slope of a displacement–time graph gives the acceleration of the body. (1 mark)
Assertion (A): Two bodies of different masses dropped together in vacuum reach the ground at the same instant. Reason (R): The value of the acceleration due to gravity does not depend on the mass of the falling body. (1 mark)
Match each quantity in Column A with the correct entry in Column B. (4 marks)

Column A	Column B
a. Slope of a displacement–time graph	i. Change in speed
b. Slope of a velocity–time graph	ii. Displacement
c. Area under a velocity–time graph	iii. Velocity
d. Area under an acceleration–time graph	iv. Acceleration

Fill in the blanks.
1. The magnitude of displacement can never be __________ than the distance travelled. (1 mark)
2. A negative sign attached to a vector quantity indicates its __________. (1 mark)
3. The value of $g$ is maximum at the __________ and minimum at the __________ of the earth. (1 mark)
State whether the following statements are true or false.
1. The average velocity of a body can be zero while its average speed is not zero. (1 mark)
2. A body with a negative acceleration must be moving backwards. (1 mark)
3. The velocity–time graph of a body can be a straight line parallel to the velocity axis. (1 mark)
Name the following.
1. The instrument fitted in a vehicle that reads its instantaneous speed. (1 mark)
2. The quantity obtained by dividing the displacement of a body by the time taken. (1 mark)
3. The steady velocity with which a raindrop finally reaches the ground. (1 mark)
Give one word or phrase for each of the following.
1. Motion of a body along a straight line path. (1 mark)
2. The decrease in velocity per second of a moving body. (1 mark)
State the SI unit of the following. (2 marks)
1. displacement
2. speed
3. acceleration
4. acceleration due to gravity.
Pick the odd one out and give a reason for your choice.
1. Momentum, weight, distance, displacement — pick the odd one out and give a reason. (1 mark)
2. Metre, second, metre per second squared, kilogram — pick the odd one out and give a reason. (1 mark)

2.15 Check Your Concepts

Define speed and state its SI unit. (2 marks)
When is a body said to be at rest? Explain why rest and motion are said to be relative. (2 marks)
State the two parameters needed to express a scalar quantity completely, and the three needed for a vector quantity. (2 marks)
Give a reason for the following: the magnitude of the displacement of a body can never be greater than the distance travelled by it. (2 marks)
What does the slope of a velocity–time graph represent? What does the area enclosed between the graph and the time axis represent? (2 marks)
Differentiate between distance and displacement, stating the basis of difference in each case. (3 marks)
Differentiate between uniform acceleration and variable acceleration, giving one example of each. (3 marks)
Define acceleration due to gravity. State its average value, and state two factors on which its value at a place depends. (3 marks)
Why does a body moving in a circular path with a constant speed have a variable velocity? Explain with the help of the direction of its velocity at two different points. (3 marks)
‘A body may have zero velocity and yet a non-zero acceleration at the same instant.’ Explain this statement with the example of a ball thrown vertically upwards. (3 marks)
State the three equations of uniformly accelerated motion, naming each symbol with its SI unit, and state the condition under which they may be used. (4 marks)
What would happen to the shape of a displacement–time graph if a body, after moving away from its starting point for a while, turned round and came back to it? Describe the graph and explain the sign of the slope in each part. (4 marks)
Explain how the acceleration–time graph of a freely falling body can be used to obtain first its velocity–time graph and then its displacement–time graph, stating what is measured on the graph at each stage. (5 marks)

2.16 Diagrams, Graphs & Constructions

Draw the displacement–time graph for (i) a stationary body and (ii) a body moving with a uniform velocity, and state how the velocity is obtained from each. (3 marks)
In Fig. 2.6-4, the graph rises from $O$ to $A$ and then falls from $A$ to $B$. State what the body is doing in each part, and explain what the sign of the slope tells you in each. (3 marks)
Study Fig. 2.7-1. State the intervals in which the body has (i) positive acceleration, (ii) zero acceleration, and (iii) negative acceleration, and state the interval in which it is moving in the negative direction. (4 marks)
Sketch the velocity–time graph you would expect for a bus that starts from rest, speeds up uniformly, runs at a steady velocity, and then is braked uniformly to rest. Mark on your sketch the region whose area gives the total distance travelled. (3 marks)
Draw the acceleration–time graph for a body moving with (i) uniform velocity and (ii) uniform retardation, labelling both axes with their units. (3 marks)
Fig. 2.9-4 is the $S$–$t^2$ graph for a freely falling body. Describe how you would use such a graph to determine the acceleration due to gravity at a place, naming the quantity you would measure from the graph and the arithmetic you would do to it. (4 marks)
Fig. 2.10-1 is the velocity–time graph used to derive the equations of motion. Redraw it, label the initial velocity, the final velocity and the time of journey on it, and show which area on the graph represents the distance travelled. (4 marks)

2.17 Practice With Numericals

Express a speed of $54\ \text{km h}^{-1}$ in m s⁻¹. (2 marks)
A body moving with a uniform speed covers $120\ \text{m}$ in $8\ \text{s}$. Find its speed. (2 marks)
The velocity of a car increases uniformly from $5\ \text{m s}^{-1}$ to $25\ \text{m s}^{-1}$ in $8\ \text{s}$. Find its acceleration. (2 marks)
A body starts from rest and moves with a uniform acceleration of $3\ \text{m s}^{-2}$. Find (i) its velocity and (ii) the distance travelled by it, at the end of $4\ \text{s}$. (3 marks)
A stone is dropped from rest from the top of a tower. Taking $g = 9.8\ \text{m s}^{-2}$ and ignoring air resistance, find (i) its velocity and (ii) the distance fallen, after $3\ \text{s}$. (3 marks)
A train moving at $30\ \text{m s}^{-1}$ is retarded uniformly at $1.5\ \text{m s}^{-2}$ until it comes to rest. Find (i) the time it takes to stop and (ii) the distance it covers in stopping. (3 marks)
A ball is thrown vertically upwards with a velocity of $19.6\ \text{m s}^{-1}$. Taking $g = 9.8\ \text{m s}^{-2}$, find (i) the time it takes to reach the highest point and (ii) the maximum height reached. (4 marks)
A cyclist covers $4\ \text{km}$ at a uniform speed of $12\ \text{km h}^{-1}$ and the next $6\ \text{km}$ at a uniform speed of $18\ \text{km h}^{-1}$, both in the same direction. Find (i) the total time of the journey and (ii) the average speed in km h⁻¹. (4 marks)
A bus moving at $15\ \text{m s}^{-1}$ accelerates uniformly at $0.5\ \text{m s}^{-2}$ for $20\ \text{s}$. Find (i) its final velocity and (ii) the distance travelled in that time. (4 marks)
A body moving in a straight line starts from rest, accelerates uniformly to $20\ \text{m s}^{-1}$ in $5\ \text{s}$, travels at that velocity for the next $10\ \text{s}$, and is then brought uniformly to rest in a further $5\ \text{s}$. Draw the velocity–time graph and use it to find (i) the total distance travelled and (ii) the average speed for the whole journey. (5 marks)
A body moving with uniform acceleration covers $32\ \text{m}$ in the first $4\ \text{s}$ and $80\ \text{m}$ in the next $4\ \text{s}$. Find (i) its initial velocity and (ii) its acceleration. (5 marks)
A car travelling at $72\ \text{km h}^{-1}$ is braked and comes to rest after covering $100\ \text{m}$. Find (i) the retardation produced by the brakes and (ii) the time taken by the car to stop. (5 marks)