Class 9 Science Chapter 4: Describing Motion Around Us

A fixed point chosen so that positions can be described relative to it. Marked as the origin O on a line, it makes the question where is the object answerable at all, since position means distance and direction measured from somewhere.

The distance and direction of an object from the reference point at a particular instant of time. Both parts are needed; a distance alone leaves the object anywhere on a circle around the reference point.

An instant of time is one reading of a clock. A time interval is the duration between two readings. Questions about position refer to instants; questions about how long something took refer to intervals.

The net change in an object's position between two instants. Its magnitude is the distance between those two positions and its direction runs from the first toward the second, so it ignores whatever path was taken in between.

The numerical value of a physical quantity together with its units. Displacement needs a magnitude and a direction to be fully described, whereas total distance travelled needs only the magnitude.

Position means the distance and the direction of the object measured from some fixed point, so without a reference point neither quantity has any meaning. The reference point is marked as the origin, and positions are then given relative to it.

An object is in motion if its position with respect to the reference point changes with time. It is at rest if that position does not change with time. Both are defined relative to the reference point rather than being properties of the object alone.

An instant of time is a single reading of a clock at a given point of time. A time interval is the duration between two instants, that is, between two readings of a clock. A position belongs to an instant, while a journey occupies an interval.

Take where a body is at one instant and where it is at another: the difference between those positions is its displacement. Describing it fully requires a magnitude, meaning a numerical value with units equal to the distance between the two positions, and a direction running from the earlier position toward the later one.

The total distance travelled is the whole path, 100 m plus 60 m, which is 160 m. Her displacement is the net change in position, and since she finishes 40 m from the origin it is 40 m in the positive direction.

For motion in a straight line they are equal when the object moves without turning back, that is, when it keeps travelling in one direction throughout. As soon as it reverses, part of the path cancels out of the displacement while still counting toward the distance, and the two figures separate.

Fuel is burnt over every metre actually covered, whatever route is taken, so it follows the total distance travelled. Displacement measures only the net change in position, and a vehicle returning to where it started has a displacement of zero while having used a great deal of fuel.

It is correct that the magnitude of displacement can never exceed the total distance travelled, but the word always is too strong. For an object moving one way along a straight line without turning back the two are exactly equal, so the correct statement is that the magnitude of displacement is less than or equal to the distance travelled.

Describing position begins with a choice. A fixed point must be specified as the reference point, because position means the distance and direction of the object measured from somewhere.

Both parts of that are necessary. A distance alone would leave the object anywhere on a circle around the reference point; it is the direction that narrows it to a single place.

For an object confined to a straight line the problem simplifies, because only two directions are available, forward and backward. These are written using plus and minus signs, with positions to the right of the origin generally taken as positive and those to the left as negative, so a single signed number carries both how far and which way.

Once position is available, rest and motion follow immediately. An object is in motion if its position with respect to the reference point changes with time, and at rest if that position does not change with time.

It is worth noticing that both definitions are stated relative to the reference point. Motion is not simply a property an object has on its own, but a statement about how its position relates to the point chosen to measure from.

Total distance travelled measures the whole path an object covers, and needs only a numerical value with units. Displacement is the net change in position between two given instants, and needs both a magnitude and a direction.

Take an athlete on a straight track. She starts at O, runs 100 m to A, then turns and runs 60 m back to B.

Her total distance travelled adds up the legs of the journey: 100 m plus 60 m, giving 160 m. Her displacement asks a narrower question, namely how far she is from where she began and in which direction, and since she finishes 40 m from the origin it is 40 m in the positive direction.

The SI unit of both quantities is the metre. They differ not in what they are measured in but in whether a direction must accompany the number.

As for when they agree: for motion in a straight line the magnitude of displacement equals the total distance travelled only when the object moves without turning back. The moment it reverses, part of the path begins cancelling out of the displacement while still counting toward the distance, and the two separate.

The ball rises from its starting point, reaches a highest point, and falls back to where it began. Although a diagram may show the upward and downward paths side by side for clarity, in reality both occur along the same straight line, so this is still motion in a straight line.

The total distance travelled is the whole path covered, which is twice the height reached, once going up and once coming down.

The displacement over the whole flight is zero, because displacement is the net change in position and the ball finishes exactly where it started.

This single case settles several claims. It shows that displacement is not never zero, since here it is zero. It shows that displacement need not always be strictly less than the distance travelled in every case, since one-way motion gives equality elsewhere.

Of the four possible statements, only one survives: the magnitude of displacement is less than or equal to the total distance travelled. Its magnitude can never exceed the distance covered, because the shortest route between two positions cannot be longer than the path actually taken.

The question is ambiguous because two distinct quantities answer it. Total distance travelled reports the whole path covered, while displacement reports only the net change in position between the starting and finishing instants.

For the same journey these can differ enormously. An athlete running out and back along a track may cover 160 m while ending only 40 m from her starting point, and a taxi returning to its depot covers a great distance while its displacement is exactly zero.

Which answer is wanted depends entirely on the purpose. Fuel consumption follows the total distance travelled, because fuel is burnt over every metre actually covered whatever route is taken. That is why a taxi with zero displacement still empties its tank.

Displacement is the useful figure when the question concerns where something has ended up. To know how far a delivery has got from the warehouse, the winding route taken is beside the point and only the net change in position matters.

So neither quantity is more correct than the other. Each answers a different question, and stating which one you mean is part of giving a proper answer at all.

The total distance travelled divided by the time interval taken to cover it. Since distance carries no direction, neither does average speed: a numerical value with units describes it completely.

The displacement divided by the time interval, written $v_{av} = \frac{s}{t}$. It describes how fast the position changes and in which direction, so unlike speed it needs a direction stated, shown by a plus or minus sign on a straight line.

Motion in which equal distances are covered in equal intervals of time, for every possible choice of interval. The object moves at constant speed. If unequal distances are covered in equal intervals, the motion is non-uniform.

How much one quantity shifts, set against how much time passed while it shifted. Average velocity is the average rate of change of position with respect to time, which is why it is found by dividing displacement by a time interval.

The change in velocity divided by the time interval over which that change happens, given by $a = \frac{v - u}{t_2 - t_1}$. Its SI unit is $\text{m s}^{-2}$, and it may arise from a change in the magnitude of velocity, in its direction, or in both.

Average speed is all the ground covered, divided by however long the covering took. Since distance travelled has only a numerical value and no direction, the average speed calculated from it also has no direction and is described by a numerical value with units alone.

In uniform motion the object covers equal distances in equal intervals of time, for every possible choice of interval, and so moves at constant speed. In non-uniform motion it covers unequal distances in equal intervals, so its speed is increasing, decreasing, or a combination of the two.

Average velocity is the displacement divided by the time interval in which that displacement occurs, written $v_{av} = \frac{s}{t}$. Its direction is the same as the direction of the displacement, and for motion in a straight line this is indicated by a plus or a minus sign.

His total distance travelled is 50 m, so his average speed is $\frac{50\ \text{m}}{50\ \text{s}} = 1\ \text{m s}^{-1}$. His displacement is zero because he finishes at the end he started from, so his average velocity is $\frac{0\ \text{m}}{50\ \text{s}} = 0\ \text{m s}^{-1}$.

A rate of change is the ratio of a change in one quantity to the corresponding change in time. Average velocity is calculated as the ratio of the change in position to the time taken, so average velocity is the average rate of change of position with respect to time.

Average acceleration is the change in velocity divided by the time interval over which the change occurs. If the velocity changes from an initial value $u$ at time $t_1$ to a final value $v$ at time $t_2$, then $a = \frac{v - u}{t_2 - t_1}$. Its SI unit is $\text{m s}^{-2}$.

If the magnitude of the velocity is increasing over the interval, the average acceleration is in the same direction as the velocity. If the magnitude of the velocity is decreasing, the average acceleration is in the direction opposite to the velocity.

Average acceleration can result from a change in the magnitude of velocity, a change in its direction, or both. If only the direction changes while the magnitude stays the same, the velocity has still changed, so there is an acceleration. An object moving round a circle at a steady rate is exactly this case.

Average speed is the total distance travelled divided by the time interval over which that distance was covered. Average velocity is the displacement divided by the same kind of interval, written $v_{av} = \frac{s}{t}$.

The difference in what each is built from settles everything else. Distance travelled has only a numerical value and no direction, so average speed inherits none. Displacement has both a magnitude and a direction, so average velocity inherits a direction too, shown on a straight line by a plus or minus sign.

The two share the same SI unit, the metre per second, so units are no guide. Only the numerator distinguishes them.

A swimmer makes the point. Sarang covers 50 m of water in 50 s, swimming one length and back. His average speed is $\frac{50\ \text{m}}{50\ \text{s}} = 1\ \text{m s}^{-1}$.

His displacement, however, is zero, because he finishes at the end he started from. So his average velocity is $\frac{0\ \text{m}}{50\ \text{s}} = 0\ \text{m s}^{-1}$. He was moving throughout, yet his average velocity is exactly zero, which is not a contradiction but a sign that the two quantities answer different questions.

Acceleration concerns how quickly velocity itself changes. You feel it directly as the jolt when a vehicle pulls away from rest, and again when it stops suddenly; both sensations are a change in velocity making itself felt.

Average acceleration over a time interval is the change in velocity divided by that interval, which can also be written as the final velocity minus the initial velocity, all divided by the time taken.

In symbols, if velocity runs from an initial value $u$ at time $t_1$ to a final value $v$ at time $t_2$, then $a = \frac{v - u}{t_2 - t_1}$. The SI unit is $\text{m s}^{-2}$, the second appearing squared because acceleration is a change of velocity per unit time and velocity was already a distance per unit time.

Acceleration needs a direction as well as a magnitude. Along a straight line, a body gathering speed has its acceleration pointing the same way it travels; a body losing speed has it pointing back against the travel. A braking car therefore has an acceleration directed backward while the car still travels forward.

Finally, acceleration may arise from a change in the magnitude of velocity, from a change in its direction, or from both. An object moving round a circle at steady speed has a changing direction and so is accelerating, despite its speed never altering.

An object moving in a straight line is in uniform motion if it travels equal distances in equal intervals of time. In that case it moves at constant speed.

The definition carries an important qualification: this must hold for every possible choice of time interval, not merely for some. Without that condition, motion which happened to average out over certain chosen intervals could be mistaken for uniform motion even though the object was speeding up and slowing down within them.

Non-uniform motion is the opposite case. The object covers unequal distances in equal intervals of time, so its speed is increasing, decreasing, or a combination of both at different stages.

The pattern of successive distances tells you which. If the distances travelled in successive equal intervals are increasing, the speed is increasing; if they are falling, the speed is decreasing.

The distinction matters because a single average speed describes uniform motion completely, whereas for non-uniform motion an average conceals what happened within the interval, which is why velocity-time graphs and the idea of instantaneous velocity become necessary later.

While the car accelerates from rest its velocity is increasing in magnitude. The average acceleration over this stage is therefore directed along the velocity, in the direction the car is travelling.

During the steady cruise the car covers equal distances in equal intervals of time, which is uniform motion in a straight line at constant speed. Since the velocity is not changing, the average acceleration over this stage is zero.

While braking, the magnitude of the velocity is decreasing. There is still an acceleration, because acceleration measures any change in velocity, but it now points opposite to the direction of travel, which is why passengers are thrown forward against their seatbelts.

Over the whole journey a single average acceleration could be computed from the initial and final velocities, and since the car begins and ends at rest that average would be zero. That figure would be perfectly correct and would conceal the entire story.

The same is true of average velocity, which uses only the displacement between the first and last instants. This is exactly why averages over long intervals are not enough, and why the chapter moves on to graphs, which show what happened within an interval rather than only across it.

A graph showing how the position of an object changes with time. Reading a value off it gives the position at any instant, and reading its shape or steepness gives information about the velocity.

Indicates constant velocity. Equal intervals of time carry equal displacements, so the object covers the same ground in each equal stretch of time.

Indicates that velocity is not constant, so the object is in accelerated motion. The displacements covered in equal intervals of time are changing rather than staying the same.

Represents a stationary object. Its position keeps the same value however much time passes, so the object is at rest at that position. In this case the position-time graph is not the same as the distance-time graph.

The steeper the line, the greater the displacement covered in a given time, and so the greater the velocity. Comparing two lines on the same axes therefore compares the two velocities directly.

It gives a visual picture of how position, velocity and acceleration change with time. In particular it makes it easy to compare the motion of two objects, to calculate physical quantities, and to identify whether motion is uniform or non-uniform.

All the graphs describe motion in a straight line in one direction only, and only cases of constant acceleration are considered. A consequence is that distance travelled equals the magnitude of displacement, and speed equals the magnitude of velocity, so that if the position is zero at time zero the position-time graph is also the distance-time graph.

A straight line indicates that the object is moving with constant velocity, because equal intervals of time carry equal displacements. A curve indicates that the velocity is not constant, so the object is in accelerated motion.

It represents a stationary object, whose position keeps the same value however much time passes. It is worth mentioning separately because it is easily misread as slow steady motion, and because in this case the position-time graph is not the same as the distance-time graph.

Take two points A and B on the graph and draw lines parallel to the axes from each, extending the horizontal line from A to form a right-angled triangle ABC. The vertical side BC gives the change in position and the horizontal side AC the change in time. Dividing the first by the second gives $v = \frac{s_2 - s_1}{t_2 - t_1}$.

The change in position is $80 - 40 = 40$ m and the change in time is $4 - 2 = 2$ s. The velocity is $\frac{40\ \text{m}}{2\ \text{s}} = 20\ \text{m s}^{-1}$.

By comparing the steepness of their lines. Drawing lines parallel to the axes shows which object covers the greater displacement in the same interval of time, and the steeper line belongs to that object, which therefore has the greater velocity.

Reading a value directly off the graph gives the position of the object at any instant of time. Reading the steepness of the line, by means of the triangle construction, gives the magnitude of the velocity. So the one drawing supplies both position and velocity.

The shape of a position-time graph can be read at a glance, and each shape follows from a single test: whether equal intervals of time carry equal displacements.

A sloping straight line indicates constant velocity. If an object covers 20 m between 2 and 3 s, and 20 m again between 5 and 6 s, then equal intervals of time are carrying equal displacements and the magnitude of the velocity is constant. A straight line is what that draws.

A curve indicates that the velocity is not constant, so the object is in accelerated motion. Here the displacements covered in equal intervals are changing, as when more ground is covered between 10 and 12 s than between 4 and 6 s, so the magnitude of velocity is increasing.

A straight line drawn parallel to the time axis represents a stationary object. Its position keeps the same value however much time passes, so its velocity is zero. A vehicle whose position reads 40 m and does not change is simply at rest 40 m from the origin.

That third case deserves care for two reasons. It is easily misread as slow steady motion, when in fact nothing is moving at all. And in that case the position-time graph is no longer the distance-time graph, since the vehicle sits at 40 m while having travelled nothing.

A position-time graph gives positions directly, but it also contains the velocity, and a simple construction extracts it.

Take any part of the graph, say the stretch between two points A and B. From A, draw a line parallel to the X-axis and another parallel to the Y-axis, and do the same from B. Extending the horizontal line from A forms a right-angled triangle ABC.

The two sides of that triangle carry the information needed. The side BC represents the change in position, $s_2 - s_1$, and the side AC represents the change in time, $t_2 - t_1$.

Since average velocity is displacement divided by the time interval, dividing one side by the other performs that calculation graphically, giving $v = \frac{BC}{CA} = \frac{s_2 - s_1}{t_2 - t_1}$.

As an example, suppose the graph shows the object at 40 m when the time is 2 s and at 80 m when the time is 4 s. Then the velocity is $\frac{80 - 40}{4 - 2} = \frac{40}{2} = 20\ \text{m s}^{-1}$. So by finding the positions of an object at two instants of time, the average velocity can always be calculated from the graph.

Two assumptions govern every graph in the chapter. All of them describe motion in a straight line in one direction only, and only cases of constant acceleration are considered.

The first assumption has a direct consequence. When an object moves one way without turning back, the distance travelled equals the magnitude of the displacement, and the speed equals the magnitude of the velocity. The two pairs of quantities collapse into one another.

From this it follows that if the position is zero at time zero, the position-time graph is the same drawing as the distance-time graph, while the velocity-time graph is the same as the speed-time graph. That is why the names are sometimes used interchangeably here.

Without the assumption the equivalence breaks. An object that reverses would travel more distance than its displacement, so the two graphs would diverge, and a graph labelled one way could not be read as the other.

The stationary vehicle shows even the stated conditions have limits. Sitting at 40 m from the origin, its position-time graph is a flat line at 40 m, while its distance travelled is nil, so that particular graph is not a distance-time graph at all. The second assumption, constant acceleration, plays a different role: it keeps the curves simple enough to be read by eye rather than requiring more advanced treatment.

The comparison can be made without calculating either velocity, provided the right feature is compared.

Draw lines parallel to the axes from points on each graph, covering the same interval of time. This shows how much displacement each object covered during that interval.

If object B covers more displacement than object A over the same time interval, then the line for B is steeper than the line for A. Since a greater displacement in the same time is precisely what a greater velocity means, B has the higher velocity.

The feature to compare is therefore steepness, not height. Height on a position-time graph gives the position of the object, which is a different quantity altogether, so a line that begins higher up the axes tells you only where that object started.

This matters because a line can start high and rise gently while another starts low and rises steeply. The second object is the faster one, even though its line spends the whole graph below the first. Reading height as speed would give exactly the wrong answer.

A graph representing the change in an object's velocity with time. It is plotted by the same procedure as a position-time graph, with velocity on the vertical axis.

A line running level with the X-axis shows the velocity is not changing at all. The velocity is constant, and the acceleration is zero.

Shows velocity changing by equal amounts in equal intervals of time, so the acceleration is constant. Rising means the acceleration runs with the velocity; falling means it runs against it.

How fast the velocity itself is changing with time is exactly what acceleration means, so dividing the change in velocity by the change in time reads the acceleration off the graph.

The area enclosed between the line and the time axis over a chosen interval equals the displacement during that interval. A level line gives a rectangle; a sloping line gives a rectangle plus a triangle.

It represents the change in an object's velocity with time. It is plotted in a manner similar to a position-time graph, following the same steps, with velocity taking the place of position on the vertical axis.

A line that never rises or falls tells you the velocity is holding steady, so there is no acceleration at all. A line climbing evenly tells you the velocity is growing at a steady rate, with the acceleration pointing the way the object moves. A line dropping evenly tells you the velocity is shrinking at a steady rate, with the acceleration pointing against the motion.

It gives the acceleration. The steepness measures how fast the velocity is changing with time, and that is precisely what acceleration means, so dividing the change in velocity by the change in time reads the acceleration directly off the graph.

The change in velocity is $10 - 5 = 5\ \text{m s}^{-1}$. The change in time is $20 - 10 = 10$ s, so the acceleration is $\frac{5}{10} = 0.5\ \text{m s}^{-2}$.

It indicates that the direction of the acceleration is opposite to the direction of the velocity. That is what is happening whenever the velocity is decreasing, so the sign carries physical information rather than signalling an error.

The height of the region is the magnitude of the velocity and its width is the time interval, so their product is velocity multiplied by time. Since average velocity multiplied by a time interval is a displacement, the area equals the displacement over that interval.

The region is a rectangle whose height is the velocity and whose width is the time interval, so the displacement is $20\ \text{m s}^{-1} \times 6\ \text{s} = 120\ \text{m}$. That is how far the car has moved from the origin in six seconds.

The region is no longer a plain rectangle, so it is split into a rectangle sitting on the time axis plus a triangle capping it, and the two areas are added. Between 10 s and 20 s this gives $(5 \times 10) + \frac{1}{2}(10 \times 5) = 50 + 25 = 75$ m. Leaving out the triangle would lose a third of the answer.

Three cars make the point between them, and each gives a different line.

The first moves in the same direction along a straight stretch of highway at a steady 72 kilometres per hour, which is 20 metres per second. Since its velocity is not changing with time, the line runs level with the X-axis, and the acceleration works out at zero.

The second starts from rest and its velocity climbs: 2.5 metres per second at 5 seconds, 5.0 at 10 seconds, on up to 15.0 at 30 seconds. In equal intervals of time the velocity increases by equal amounts, so the acceleration is constant and points in the direction of the velocity. The graph is a straight line climbing steadily.

The third begins at 15.0 metres per second and falls away by 2.5 every 5 seconds, reaching zero at 30 seconds. Here the velocity decreases by equal amounts in equal intervals, so the acceleration is again constant, but this time it points opposite to the direction of the velocity. The graph is a straight line falling steadily.

What matters is that both sloping cases are constant acceleration. Only the direction differs, and it is the sign of the answer that records which way it runs.

On a position-time graph the steepness of the line gave the velocity. On a velocity-time graph the same feature gives something else: how fast the velocity itself is changing with time, which is the acceleration.

For a level line the case is settled at once. There is no change in the height of the line above the X-axis, so its steepness is zero, the velocity is constant and the acceleration is zero.

For a sloping line, take a part of it from A to B. At the earlier instant $t_1$, at point A, call the velocity $u$. At the later instant $t_2$, at point B, call it $v$. Dividing the change in velocity by the change in time gives $a = \frac{v - u}{t_2 - t_1}$.

Substituting from the rising graph, between 10 seconds and 20 seconds the velocity goes from 5 to 10 metres per second, so $a = \frac{10 - 5}{20 - 10} = \frac{5}{10} = 0.5\ \text{m s}^{-2}$.

Following the same steps on the falling graph gives an average acceleration of $-0.5\ \text{m s}^{-2}$. The minus sign indicates the direction of the acceleration is opposite to the direction of the velocity, which is what a decreasing velocity means.

Shade the region between the line and the time axis, and that area turns out to represent a physical quantity.

Take the level line first. Its height is the magnitude of the velocity and its width is the time interval, so the area of the rectangle is velocity multiplied by time interval. Since the velocity is constant it equals the average velocity, and average velocity multiplied by a time interval is a displacement. So the area of the rectangle is the displacement.

Substituting the values, the displacement between 0 seconds and 6 seconds is $20\ \text{m s}^{-1} \times 6\ \text{s} = 120$ metres, which is how far the car has moved from the origin in those six seconds.

The general rule follows: whatever area is shut in between the velocity curve and the time axis across any stretch you choose is the displacement over that stretch.

Where the line slopes the region is no longer a plain rectangle, so it is split in two: a rectangle sitting on the time axis plus a triangle capping it. Adding them, displacement $= (CD \times DE) + \frac{1}{2}(CA \times BC)$. Between 10 and 20 seconds this gives $(5 \times 10) + \frac{1}{2}(10 \times 5) = 50 + 25 = 75$ metres. Leaving the triangle out would give 50 metres, losing a third of the answer.

The two graphs look alike but answer different questions, and the same visual feature means different things on each.

On a position-time graph, reading a value gives the position at an instant, and the steepness of the line gives the velocity.

On a velocity-time graph, reading a value gives the velocity at an instant. The steepness now gives the acceleration, because it measures how fast the velocity is changing. And the area enclosed with the time axis gives the displacement, a reading the position-time graph does not offer.

Care is needed because a shape can mean opposite things on the two graphs. A line level with the time axis on a position-time graph means the object is at rest, since its position never changes. The same shape on a velocity-time graph means the object is moving steadily, since it is the velocity that is not changing and the acceleration that is zero.

The habit that prevents the error is simple: check which quantity sits on the vertical axis before reading any shape or steepness. For motion with constant acceleration, the velocity-time graph then yields acceleration from its steepness and displacement from its area.

The set $v = u + at$, $s = ut + \frac{1}{2}at^2$ and $v^2 = u^2 + 2as$, relating displacement, time, the two velocities and the acceleration for straight-line motion at constant acceleration.

How far a vehicle runs on after the brakes are applied. It grows with the square of the speed, and also depends on the road surface, the braking power and the driver's reaction time.

Motion in two dimensions, such as one vehicle overtaking another, the flight of a kicked ball, or a satellite on a circular path.

Motion along a circular path at constant speed. The speed has the same value everywhere on the circle, but the direction of the velocity changes continuously.

A straight line meeting a circle at one point only. The velocity of an object on a circular path lies along the tangent at its position, pointing the way it is travelling.

The acceleration must be constant. Because it never changes, its value at each instant is equal to the average acceleration over any time interval chosen, and that is what allows a single general formula to be written for the whole motion.

Average acceleration is the change in velocity divided by the time interval, so $a = \frac{v - u}{t}$, where $u$ is the initial velocity at $t = 0$ and $v$ the final velocity at time $t$. Multiplying across gives $at = v - u$, and rearranging gives $v = u + at$.

The displacement is the area enclosed under the velocity-time line, which here splits into a rectangle of area $ut$ and a triangle of area $\frac{1}{2} \times t \times (v - u)$. Replacing $(v - u)$ with $at$ from the first equation gives $s = ut + \frac{1}{2}at^2$.

They are $v = u + at$, $s = ut + \frac{1}{2}at^2$ and $v^2 = u^2 + 2as$. Five quantities are involved: how far the body goes $s$, how long it takes $t$, how fast it began $u$, how fast it ended $v$, and the acceleration $a$.

They are valid only when the acceleration is constant. For straight-line motion in one direction the distance travelled equals the magnitude of displacement and the speed equals the magnitude of velocity, while for motion in both directions the sign of each of $u$, $v$, $a$ and $s$ tells you its direction.

Using $v^2 = u^2 + 2as$ with $v = 0$ gives $0 = u^2 - 8s$, so $s = \frac{u^2}{8}$. Substituting $u = 15\ \text{m s}^{-1}$ gives $s = \frac{225}{8} = 28.1$ m.

The distance travelled in one revolution is the circumference, $2\pi R$. The displacement is zero, since the child returns to the original position. If the revolution takes time $T$, the average speed is $v_{av} = \frac{2\pi R}{T}$, while the average velocity over that interval is zero.

Acceleration is non-zero whenever the velocity of an object changes, and the velocity changes if either its magnitude or its direction, or both, changes. In uniform circular motion the magnitude stays constant at every point, but the direction of the velocity changes continuously, so the motion is accelerated.

The derivation rests on one condition. Since the acceleration is constant, its value at each instant equals the average acceleration over any interval, so a single formula can describe the whole motion.

Begin with the definition of average acceleration, $a = \frac{v - u}{t}$, where $u$ is the initial velocity at $t = 0$, $v$ the final velocity at time $t$, and the interval is $t$. Multiplying across gives $at = v - u$, and therefore $v = u + at$.

For the second, draw the velocity-time graph with a non-zero starting value. The displacement is the area under the line, which is a rectangle of area $ut$ together with a triangle of area $\frac{1}{2} \times t \times (v - u)$. Replacing $(v - u)$ with $at$ from the first equation gives $s = ut + \frac{1}{2}at^2$.

Those two are the primary equations, and combining them yields others. Eliminating the time by writing $t = \frac{v - u}{a}$ and substituting into the second, then simplifying, gives $v^2 = u^2 + 2as$.

Together these connect five quantities: displacement, time interval, initial velocity, final velocity and acceleration. They describe mathematically how motion changes with time, which is what makes a position or velocity at some later moment predictable.

Take the data first. The acceleration is $-4\ \text{m s}^{-2}$ and the final velocity is zero, since the car stops. Converting the two starting velocities, 54 km per hour is 15 metres per second and 108 km per hour is 30 metres per second.

The time is neither given nor wanted, so use $v^2 = u^2 + 2as$. Substituting gives $0 = u^2 + 2 \times (-4) \times s$, that is $0 = u^2 - 8s$, so $s = \frac{u^2}{8}$.

Putting the numbers in, the first case gives 28.1 metres and the second gives 112.5 metres.

The velocity doubled, but the stopping distance went up four times over, because the distance depends on the square of the velocity rather than on the velocity itself.

That result is the case for keeping a safe distance from the vehicle ahead, and for adjusting that distance to the speed being carried. How far a vehicle runs on also depends on the road surface, whether wet or dry, on the braking capacity of the vehicle, and on the driver's reaction time. Vehicle-to-vehicle communication technology, now being developed in many countries including India, lets vehicles exchange signals and warn drivers of a possible collision.

Start from what acceleration means. It is non-zero whenever the velocity of an object changes, and the velocity changes if either its magnitude or its direction, or both, changes. That second possibility is the one everyday speech overlooks.

In uniform circular motion the magnitude never changes, since the speed is constant at every point on the circle. But the direction of the velocity changes continuously, and that alone makes the motion accelerated.

The many-sided track shows why the direction cannot settle. An athlete on a rectangular track changes direction four times a lap, and on a hexagonal track six times. As the number of sides increases the turns come more and more frequently, and in the limit each side shrinks to a point, the track becomes a circle, and the direction of the velocity changes without pause.

The marble shows which way the velocity points. Set spinning inside a ring and then released when the ring is lifted, it moves off in a straight line, carrying on in the direction it had at that instant. That direction is the tangent to the circle at that point, a straight line meeting the circle at one point only.

So the velocity is always tangential, always turning, and the motion is therefore accelerated throughout.

Take a child on a moving merry-go-round travelling from one point round to another. The distance travelled is the curved path the child actually follows, while the displacement is the straight line joining the starting and finishing positions. The two are not equal.

Now go all the way round once. The distance travelled is the circumference of the path, which for radius $R$ is $2\pi R$. The displacement, however, is zero, because the child comes back to the original position.

From this the two averages part company. If one revolution takes time $T$, the average speed is $v_{av} = \frac{2\pi R}{T}$, since average speed uses the distance. The average velocity over that same interval is zero, since average velocity uses the displacement.

When the speed along such a path is constant, the motion is called uniform circular motion, and the speed then has the same value at every point on the circle.

It is called an idealised model because the two conditions it requires, a constant speed and a circular path, are often not met in the real world. It earns its place all the same, being what the harder cases are built on, from a planet going round the Sun to a car taking a bend.

Position means the distance and the direction of the object measured from some fixed point, so without a reference point neither quantity has any meaning. The reference point is marked as the origin, and positions are then given relative to it.

An object is in motion if its position with respect to the reference point changes with time. It is at rest if that position does not change with time. Both are defined relative to the reference point rather than being properties of the object alone.

An instant of time is a single reading of a clock at a given point of time. A time interval is the duration between two instants, that is, between two readings of a clock. A position belongs to an instant, while a journey occupies an interval.

Average speed is all the ground covered, divided by however long the covering took. Since distance travelled has only a numerical value and no direction, the average speed calculated from it also has no direction and is described by a numerical value with units alone.

In uniform motion the object covers equal distances in equal intervals of time, for every possible choice of interval, and so moves at constant speed. In non-uniform motion it covers unequal distances in equal intervals, so its speed is increasing, decreasing, or a combination of the two.

Average velocity is the displacement divided by the time interval in which that displacement occurs, written $v_{av} = \frac{s}{t}$. Its direction is the same as the direction of the displacement, and for motion in a straight line this is indicated by a plus or a minus sign.

It gives a visual picture of how position, velocity and acceleration change with time. In particular it makes it easy to compare the motion of two objects, to calculate physical quantities, and to identify whether motion is uniform or non-uniform.

All the graphs describe motion in a straight line in one direction only, and only cases of constant acceleration are considered. A consequence is that distance travelled equals the magnitude of displacement, and speed equals the magnitude of velocity, so that if the position is zero at time zero the position-time graph is also the distance-time graph.

A straight line indicates that the object is moving with constant velocity, because equal intervals of time carry equal displacements. A curve indicates that the velocity is not constant, so the object is in accelerated motion.

It represents the change in an object's velocity with time. It is plotted in a manner similar to a position-time graph, following the same steps, with velocity taking the place of position on the vertical axis.