Describing Motion Around Us

Chapter 4 · Science · Class 9 32 min read

Why This Matters

Imagine you are on a bus on a busy highway. The truck ahead suddenly slams its brakes. How far should your bus stay behind the truck so it does not crash? A little gap if you are crawling, a much bigger gap if you are flying along.

That single road-safety question hides all of physics’ ideas about motion. How fast are you moving? In which direction? How quickly can you slow down? How much ground will you cover before you stop?

Everything in nature moves — butterflies, trains, falling balls, planets. It looks like endless variety. But scientists tame this variety with a small set of tools: distance, displacement, speed, velocity and acceleration. Once you can describe motion with numbers, equations and graphs, you can predict it. You can work out where a moving object will be, and when.

This chapter gives you exactly those tools. By the end, you will be able to look at a moving object and describe its motion completely — not just in words, but with numbers and graphs.

The Big Idea

The Big Idea: To describe motion, we pin down two things — how far the position changes and how fast it changes. “How far” splits into distance (the whole path you walked) and displacement (the straight gap from start to end, with direction). “How fast” splits into speed (distance per second, no direction) and velocity (displacement per second, with direction). And if the velocity itself is changing, we measure that with acceleration. Graphs turn these ideas into pictures, where a slope tells you a rate of change and an area tells you how much was covered. That is the whole language of motion.

Motion is Relative — Rest and Motion Depend on the Reference Point

Before we measure anything, we must answer a sneaky question: is the object even moving?

You are sitting on a chair, reading. Are you at rest? You feel at rest. But the Earth is whizzing around the Sun, carrying you with it. So are you moving or not?

The answer is: it depends on what you compare against. This comparison object is called the reference point.

  • Compared to your chair and room, you are at rest — your position does not change.
  • Compared to the Sun, you are in motion — your position changes every second.

So motion is relative. An object is in motion if its position with respect to a chosen reference point changes with time. It is at rest if its position with respect to that reference point does not change with time. Neither answer is “more correct” — you just have to state your reference point.

To describe an object’s position, you need two things: the distance from the reference point, and the direction from it. For an object moving in a straight line, there are only two directions. We mark one side of the reference point as positive (+) and the other as negative (−). The reference point itself is called the origin, usually labelled O.

Concept check

A passenger is asleep on a moving train. Is the passenger at rest or in motion? Explain.

Distance Travelled and Displacement

Let us follow an athlete running on a straight track. We take her starting point as the origin O. She runs forward, turns back, and stops part-way. Figure 4.1 below tracks her whole trip.

She starts at O. She runs forward to point A, which is 100 m away. Then she turns and runs back to point B, which is 40 m from O. So she ran 100 m forward and then 60 m back.

Now two different questions:

  • How much path did she cover in total? That is the distance travelled = OA + AB = 100 m + 60 m = 160 m.
  • How far is her finish from her start? That is the straight gap from O to B = 40 m. This is her displacement.

Here is the key picture that makes the difference click.

An athlete runs from O to A and back to B; distance is 160 m, displacement is 40 m.
Figure 4.1 — An athlete starts at the origin O, runs forward 100 m to A (turn point, red), then runs back 60 m to B (stop, green) where she finishes. The blue arc shows the forward 100 m and the dashed red arc shows the 60 m back. Distance travelled is the whole path, 100 + 60 = 160 m, and it has no direction (blue box). Displacement is only the straight gap from start O to stop B, shown by the solid green arrow: 40 m in the positive direction (green box). Notice the two are not equal because she turned back.

So we have two new ideas:

Distance travelled is the total length of the path covered. It is just a number with a unit (here, metres). It has no direction.

Displacement is the net change in position between the start and the end. It needs both a numerical value and a direction. The numerical value of any such quantity is called its magnitude. The magnitude of displacement is simply the straight-line distance between the start position and the end position. Its direction points from the start towards the end.

Both distance and displacement are measured in the same SI unit: the metre (m).

Can distance and displacement ever be equal?

Yes — but only in a special case. If the athlete runs only forward and never turns back, then her path is the straight gap. In that case distance = magnitude of displacement.

The moment she turns back, the path becomes longer than the straight gap. So in general:

Magnitude of displacement ≤ distance travelled.

They are equal only when the object moves in one direction without turning back.

Displacement can also be zero while distance is large. Think of running one full lap of a ground and stopping exactly where you began: you covered a big distance, but your displacement is zero because your position did not change overall.

Worked example

A ball is thrown straight up from the ground O. It rises 40 cm to its highest point and falls right back to O. Find the total distance travelled and the displacement of the ball for the whole trip.

Concept check

A vehicle uses up fuel. Does the fuel used depend on the distance travelled or on the displacement?

Average Speed and Average Velocity

Knowing how far the position changed is half the story. The other half is how fast it changed. That is where speed and velocity come in.

Average speed — how fast, no direction

You already met speed in an earlier class. Average speed tells you how fast an object moves, on average, over a journey.

average speed = total distance travelled / time interval

Since distance has no direction, average speed also has no direction. It is just a number.

The SI unit of speed is metre per second (m/s or m s⁻¹). In daily life we often use kilometre per hour (km/h or km h⁻¹) — that is what a car speedometer shows.

If an object covers equal distances in equal time intervals (for every choice of interval), it is in uniform motion — constant speed. If it covers unequal distances in equal time intervals, it is in non-uniform motion — its speed is changing.

Average velocity — how fast and which way

Speed alone does not tell you where you are heading. A car at 60 km/h going north and a car at 60 km/h going south have the same speed but very different journeys. So we define a quantity that carries direction: average velocity.

average velocity = displacement / time interval

If we write average velocity as v_av, displacement as s, and time as t, then:

v_av = s / t

Velocity needs a magnitude and a direction. For straight-line motion, the direction of velocity is the same as the direction of displacement, shown by a + or − sign. The SI unit is the same as speed: m/s.

A neat way to say it: velocity is the rate of change of position. The word “rate of change” just means how much a quantity changes for each second of time.

Here is the cleanest example of why speed and velocity can be wildly different. Watch a swimmer go to the far wall and back.

A swimmer covers 25 m there and 25 m back in 50 s; average speed is 1 m/s but average velocity is 0.
Figure 4.2 — A swimmer starts at O, swims 25 m to the far wall (green arrow) and 25 m back (red arrow), taking 50 s in total and finishing where he started. The total distance is 50 m, so average speed = 50 m / 50 s = 1 m/s (green box). But his displacement is 0 m because he ends at the start, so average velocity = 0 m / 50 s = 0 m/s (red box). Same journey, two very different answers — speed cares about the whole path, velocity cares only about the net change in position.

So the swimmer’s average speed is 1 m/s, but his average velocity is 0 m/s. The numbers are different because speed uses distance (50 m) while velocity uses displacement (0 m).

Average speedAverage velocity
Uses total distance travelledUses displacement (net change in position)
Has only a magnitude (no direction)Has magnitude and a direction (+ or −)
Can never be zero if the object movedCan be zero even if the object moved a lot
= distance / time= displacement / time
Worked example

On a road trip you drive 200 km north in 3 hours, then 200 km south in 2 hours, ending back where you started. Find your average speed and average velocity for the whole trip.

For motion in one straight direction (no turning back), the average speed and the magnitude of average velocity are equal. They only differ when the object changes direction.

Average Acceleration — How Fast the Velocity Changes

Sit in a bus. When it suddenly starts, you feel a jolt backward. When it suddenly stops, you feel a jolt forward. Those jolts are your body feeling a change in velocity.

Velocity does not always stay fixed. It can grow, shrink, or flip direction. To measure how quickly velocity changes, we use average acceleration.

average acceleration = change in velocity / time interval

If the velocity changes from an initial value u to a final value v in time t, then:

a = (v − u) / t

The SI unit of acceleration is metre per second per second, written m/s² or m s⁻². (It means: how many metres-per-second the velocity gains each second.)

Like velocity, acceleration carries a direction. And here is the rule for which way it points:

  • If the magnitude of velocity is increasing (speeding up), acceleration is in the same direction as velocity.
  • If the magnitude of velocity is decreasing (slowing down), acceleration is opposite to velocity. This is often called deceleration or retardation, and shows up as a minus sign.

Figure 4.3 below shows both cases side by side.

When a car speeds up, acceleration points along velocity; when it slows down, acceleration points opposite to velocity.
Figure 4.3 — Two cars on a straight road. In panel (a) the car is speeding up: its velocity arrow (blue) and its acceleration arrow (red) point the same way, so velocity grows and acceleration is along the velocity. In panel (b) the car is slowing down: the velocity arrow (blue) still points forward but the acceleration arrow (red) points backward, opposite to velocity, which is why slowing down gives a negative acceleration. The rule: acceleration points the way velocity is *changing*, not the way the object is moving.

A very important subtlety: a fast object can have zero acceleration. Acceleration is not about how fast you move — it is about how quickly your velocity is changing. A bus cruising on a straight highway at a steady 80 km/h has a high velocity but zero acceleration, because its velocity is not changing at all.

Worked example

A bus on a straight highway has velocity 36 km/h. The driver presses the accelerator for 10 s and the velocity rises to 54 km/h. Later the driver brakes and the bus stops from 54 km/h in 5 s. Find the average acceleration (i) while accelerating and (ii) while braking.

If the velocity changes by equal amounts in equal time intervals, the acceleration is constant (also called uniform acceleration). A falling object is the classic example: near the Earth’s surface, a dropped object speeds up by about 9.8 m/s every single second. That steady 9.8 m/s² is the acceleration due to gravity, written g. In this whole chapter we will deal only with constant acceleration.

Graphical Representation of Motion

Numbers in a table are useful, but a graph lets you see motion at a glance. A graph shows how one quantity (like position or velocity) depends on another (usually time). From the shape of the line, you can instantly tell if motion is uniform or not. From its slope and the area under it, you can pull out hidden quantities.

Note: Every graph in this chapter is for motion in a straight line, in one direction only. In that special case, distance and the magnitude of displacement are equal, and speed and the magnitude of velocity are equal. So a position-time graph that starts at zero is the same as a distance-time graph.

Position-time graphs — and what their shape means

A position-time graph plots the object’s position (Y-axis) against time (X-axis). Its shape tells you the nature of the motion straight away.

A straight position-time line means constant velocity; an upward curve means increasing velocity.
Figure 4.4 — Two position-time graphs. In panel (a) the line is straight: in each equal slice of time the position jumps by the same amount (see the equal dashed steps), so the velocity is constant. In panel (b) the line curves upward: in each equal slice of time the position jumps by a bigger and bigger amount, so the velocity is increasing — the object is accelerating. The rule to remember: a straight position-time line means steady velocity, a curving one means changing velocity.

So:

  • A straight line position-time graph → constant velocity.
  • A curved position-time graph → changing velocity (accelerated motion).
  • A horizontal line (parallel to the time axis) → the object is at rest (its position is not changing).

What can you calculate from a position-time graph? The position at any instant, obviously. But also the velocity — and this is where the slope comes in.

On a position-time graph, take two points A and B on the line and make a right-angled triangle ABC. The vertical side BC is the change in position (s₂ − s₁). The horizontal side CA is the change in time (t₂ − t₁). Then:

velocity = slope = BC / CA = (s₂ − s₁) / (t₂ − t₁)

This is exactly our definition of velocity (displacement ÷ time). So the slope of a position-time graph gives the velocity. Figure 4.5 below works it out on real numbers.

The slope of a position-time graph gives velocity; here BC/CA = 40 m / 2 s = 20 m/s.
Figure 4.5 — A position-time graph (blue straight line). Two points are picked: A at (2 s, 40 m) and B at (4 s, 80 m). Dropping a horizontal and a vertical from them makes the right-angled triangle ABC. The horizontal side CA (red) is the change in time = 2 s. The vertical side BC (green) is the change in position = 40 m. The slope is BC ÷ CA = 40 m ÷ 2 s, so the velocity is 20 m/s (blue box). A steeper line would give a bigger slope, meaning a higher velocity.

This gives a quick way to compare two objects: the one whose position-time line is steeper has the higher velocity, because steeper means a bigger slope means a bigger velocity.

Concept check

On a position-time graph, what does a straight line that runs flat and parallel to the time axis tell you?

Velocity-time graphs — slope gives acceleration

A velocity-time graph plots velocity (Y-axis) against time (X-axis). Just like before, its shape reveals the motion.

A flat velocity-time line means constant velocity; sloping up means positive acceleration; sloping down means negative acceleration.
Figure 4.6 — Three velocity-time graphs. In panel (a) the line is flat: velocity stays the same, so acceleration is zero. In panel (b) the line slopes upward: velocity rises by equal amounts in equal times, so there is a steady positive acceleration. In panel (c) the line slopes downward: velocity falls by equal amounts in equal times, so there is a steady negative acceleration (the object is slowing). The steeper the slope, the bigger the acceleration.

So on a velocity-time graph:

  • A flat line → velocity constant → acceleration = 0.
  • A line sloping up → velocity increasing → positive (constant) acceleration.
  • A line sloping down → velocity decreasing → negative (constant) acceleration.

The slope of a velocity-time graph gives the acceleration. The reasoning mirrors the position-time graph. Take two points A and B; the vertical side BC is the change in velocity (v − u), the horizontal side CA is the change in time. Then:

acceleration = slope = BC / CA = (v − u) / (t₂ − t₁)

For example, a velocity-time line that rises from 5 m/s to 10 m/s between 10 s and 20 s has slope (10 − 5)/(20 − 10) = 5/10 = 0.5 m/s². A line that falls gives a negative slope, i.e. negative acceleration.

Velocity-time graphs — area gives displacement (and why)

Here is the most powerful trick of all. The area enclosed between the velocity-time line and the time axis equals the displacement. Most books just tell you this. Let us see why it is true — because once you see why, you will never forget it.

Start with the simplest case: an object moving at a constant velocity. We know displacement = velocity × time. On the graph, the velocity is the height of the line and the time is the width along the axis. So:

displacement = velocity × time = height × width = area of the rectangle.

That is the whole secret. “Velocity × time” is “height × width”, which is the area. Figure 4.7(a) shows it.

When the velocity is changing (a sloping line), the shape under the line is a rectangle plus a triangle. We just add their areas, and the total is still the displacement. Figure 4.7(b) shows this.

The area under a velocity-time graph equals displacement: a rectangle for constant velocity, a rectangle plus triangle for changing velocity.
Figure 4.7 — Why area under a velocity-time graph is displacement. In panel (a) the velocity is a constant 20 m/s for 6 s, so the area under the line is a rectangle: 20 × 6 = 120 m, which equals the displacement because velocity × time is exactly height × width. In panel (b) the velocity changes between 10 s and 20 s; the area splits into a blue rectangle (5 × 10 = 50 m) plus a yellow triangle (½ × 10 × 5 = 25 m), adding to a total displacement of 75 m. The shaded area, whatever its shape, is always the displacement.

So from a velocity-time graph you get two things:

  • slope → acceleration
  • area under the line → displacement
Worked example

From a velocity-time graph, an object's velocity rises steadily from 5 m/s at 10 s to 10 m/s at 20 s. Find its displacement between 10 s and 20 s using the area under the graph.

The Equations of Motion (Derived from a Graph)

For motion in a straight line with constant acceleration, five quantities matter:

  • u = initial velocity
  • v = final velocity
  • a = acceleration
  • t = time interval
  • s = displacement

There are three equations, the kinematic equations, that link them. The beautiful part: we can derive all three straight from a velocity-time graph, using only “slope = acceleration” and “area = displacement”. Figure 4.8 below is the graph we will read off.

A velocity-time graph with initial velocity u rising to v over time t; slope gives v = u + at and area gives s = ut + half a t squared.
Figure 4.8 — A velocity-time graph for constant acceleration, where the starting velocity u is not zero. The line goes from A (velocity u) up to B (velocity v) over time t, with the slope being the acceleration a. The right-angled triangle ABC has horizontal side AC = t and vertical side BC = (v − u). The area under the line is the blue rectangle OACD (height u, width t, area u × t) plus the yellow triangle ABC (area ½ × t × (v − u)). Reading the slope gives the first equation v = u + at, and reading the area gives the second equation s = ut + ½at², as written in the green strip at the bottom.

First equation: v = u + at (from the slope)

The line goes from velocity u (point A) up to velocity v (point B) over time t. Its slope is the acceleration:

a = (v − u) / t

Now just rearrange. Multiply both sides by t:

at = v − u

v = u + at …(first equation)

This lets you find the velocity at any later time, if you know the start velocity and the acceleration.

Second equation: s = ut + ½at² (from the area)

The displacement s is the area under the line, which is the rectangle OACD plus the triangle ABC.

  • Area of rectangle OACD = AO × DO = u × t
  • Area of triangle ABC = ½ × CA × BC = ½ × t × (v − u)

So:

s = ut + ½ × t × (v − u)

But from the first equation, (v − u) = at. Substitute that in:

s = ut + ½ × t × (at)

s = ut + ½at² …(second equation)

This gives the displacement after time t, from the start velocity and the acceleration.

Third equation: v² = u² + 2as (by combining the first two)

We can eliminate t. From the first equation, t = (v − u)/a. Put this into the second equation and simplify (the algebra is shown below), and you get a relation with no t in it:

v² = u² + 2as …(third equation)

This is the one to reach for when time is not given in a problem — it links velocity directly to displacement.

The three kinematic equations (constant acceleration only):

v = u + at

s = ut + ½at²

v² = u² + 2as

Important: these equations work only when the acceleration is constant. When the object moves in one direction, use plain magnitudes. When it moves in both directions, the + and − signs on u, v, a and s carry the directions.

Worked example

A car starts from rest and reaches 24 m/s in 6 s with constant acceleration. Find (a) the acceleration and (b) the distance travelled in these 6 s.

Worked example

A motorbike moving at 28 m/s brakes with constant acceleration and stops after travelling 98 m. Find the acceleration and the time taken to stop.

Why safe following distance matters

These equations explain the road-safety question we opened with. When brakes are applied, a vehicle does not stop instantly — it travels a “stopping distance” first. Using v² = u² + 2as with v = 0, the stopping distance is s = u² / (2 × |a|).

Notice the . If you double your speed, the stopping distance becomes four times longer (because 2² = 4). That is why a fast vehicle needs a much bigger gap from the vehicle ahead. The real stopping distance also depends on the road (wet or dry), the tyres, the brakes, and the driver’s reaction time. This is the science behind “maintain a safe distance”.

Motion in a Plane — Uniform Circular Motion

So far everything moved along a straight line (one dimension). But a kicked football, an overtaking car, or a satellite move in a plane (two dimensions). The simplest, most useful example is an object moving in a circle.

When an object moves in a circular path with constant speed, its motion is called uniform circular motion. The merry-go-round, the tip of a clock’s hand, and (roughly) a planet around the Sun are all examples.

Distance and displacement on a circle

Picture a child on a merry-go-round moving from A to B to C along the circle. The distance they travel is the curved arc ABC. The displacement is the straight chord AC. These are clearly different — the arc is longer than the chord.

Now go all the way around once (one full revolution). The distance covered equals the circumference of the circle. If the radius is R, that distance is 2πR. But the displacement after one full round is zero, because the child arrives back exactly where they started. Figure 4.9 below shows both ideas.

On a circle, distance is the arc and displacement is the chord; the velocity always points along the tangent, so its direction keeps changing.
Figure 4.9 — Uniform circular motion. In panel (a) a child moves along the circle (radius R) from A through B to C: the distance travelled is the curved red arc ABC, while the displacement is only the straight green chord AC — the two are different. Over one full revolution the distance is the circumference 2πR but the displacement is zero, since the child returns to the start. In panel (b) the blue arrows show that the velocity at every point lies along the tangent to the circle (a line touching the circle at just one point), pointing in the direction of motion. The speed stays the same all around, but the direction of velocity keeps changing — which means the motion is accelerated.

If the object takes time T to complete one revolution, its average speed is the circumference divided by the time:

v_av = 2πR / T

The average velocity over one full revolution is 0, because the displacement is zero.

Why circular motion is accelerated — even at constant speed

This is the part that surprises everyone. In uniform circular motion the speed never changes. So how can it be “accelerated”?

The answer hides in the word velocity. Velocity includes direction. At every point of the circle, the velocity points along the tangent — a straight line that just touches the circle at one point, in the direction of motion. As the object moves around, this tangent direction keeps changing, instant by instant (see the four arrows in Figure 4.9(b), all pointing different ways).

A changing direction means a changing velocity, even though the speed (the magnitude) stays fixed. And any change in velocity is an acceleration. So:

In uniform circular motion the speed is constant, but the direction of velocity is always changing. Because velocity changes, the motion is accelerated.

There is a neat way to see why the velocity is tangential. Imagine an athlete running on a square track, then a hexagon, then a track with more and more sides. With more sides, the turns get gentler and more frequent. Push the number of sides to infinity and the track becomes a perfect circle — with the direction of motion changing continuously, never stopping. And if you whirl a marble inside a ring and suddenly lift the ring away, the marble shoots off in a straight line — exactly along the tangent it had at that instant. (You will learn the deeper reason for this in a later chapter on force.)

Concept check

A girl rides a scooter at a steady speedometer reading. Can the scooter still be accelerating?

Common Mistakes

⚠️ Common mistake
What students think

Distance and displacement are just two names for the same thing.

Why it seems right

In everyday talk we use 'distance' loosely for 'how far apart' two things are, so it feels like there should be only one such quantity.

What actually happens

They are different. Distance is the total path length (no direction); displacement is the straight-line change in position (with direction). If you return to your start, distance is large but displacement is zero.

⚠️ Common mistake
What students think

Displacement can be larger than the distance travelled.

Why it seems right

Both are measured in metres and both describe 'how far', so it feels like either one could come out bigger depending on the path.

What actually happens

A straight gap can never be longer than a path that bends to connect the same two points. So the magnitude of displacement is always less than or equal to the distance, and equal only for straight one-direction motion.

⚠️ Common mistake
What students think

If an object is moving fast, it must have a large acceleration.

Why it seems right

In daily speech we say a car 'accelerates' when it goes fast, so high speed and high acceleration feel like the same thing.

What actually happens

Acceleration measures how quickly velocity is changing, not how big it is. A car at a steady 100 km/h on a straight road has high speed but zero acceleration, because its velocity is not changing.

⚠️ Common mistake
What students think

In uniform circular motion the speed is constant, so there is no acceleration.

Why it seems right

We are taught that acceleration means speeding up or slowing down, and here the speed stays fixed, so it seems there can be no acceleration.

What actually happens

Acceleration is any change in velocity, and velocity includes direction. In circular motion the direction changes continuously, so the velocity changes, so the motion is accelerated even at constant speed.

⚠️ Common mistake
What students think

The kinematic equations (v = u + at and so on) work for any motion.

Why it seems right

They are presented as 'the equations of motion', which sounds like they should cover every kind of moving object.

What actually happens

They are valid only when the acceleration is constant. For motion where the acceleration keeps changing, these particular equations do not apply.

Quick Check

An athlete runs exactly one full lap around a 400 m circular track and stops at the starting line. What is her displacement?

On a velocity-time graph for straight-line motion, what does the area between the line and the time axis represent?

A car slows down steadily from 20 m/s to a stop in 4 s. What is its acceleration?

Practice Problems

Easy

Easy

My father walks from home to a shop 250 m away on a straight road. He realises he forgot the cloth bag, walks back home to get it, then walks to the shop again and buys provisions, and finally walks back home. Find his total distance travelled and his displacement from home.

Easy

A student runs from the ground floor up to the fourth floor to collect a book, then comes down to the second floor classroom. Each floor is 3 m high. Find (i) the total vertical distance travelled and (ii) the displacement from the starting point.

Medium

Medium

A truck driver going at 54 km/h sees a speed-limit sign of 40 km/h for trucks. He slows down to 36 km/h in 36 s with constant acceleration. What distance did he travel during these 36 s?

Medium

A car starts from rest and accelerates uniformly to 20 m/s in 5 s. It then travels at 20 m/s for 10 s, and finally brakes uniformly to stop in 6 s. Find the total distance travelled. (Hint: the area under the velocity-time graph is the displacement.)

Challenge

Challenge

A bus travels at 36 km/h when the driver sees an obstacle 30 m ahead. The driver takes 0.5 s to react before pressing the brake. After braking, the bus slows with a constant acceleration of 2.5 m/s². Will the bus stop before reaching the obstacle?

Challenge

Two cars A and B start from rest with constant acceleration. Car A reaches 5 m/s in 5 s; car B reaches 3 m/s in 10 s. Find each car's acceleration, and the displacement of each in its given time interval.

Summary

You can now explain:

  • That rest and motion are relative — an object is at rest or in motion only with respect to a chosen reference point, and you must always state that reference point.
  • The difference between distance (total path, no direction) and displacement (straight-line change in position, with direction), and why displacement is always less than or equal to distance.
  • The difference between average speed (distance ÷ time, no direction) and average velocity (displacement ÷ time, with direction), and why they can differ — even giving zero velocity with non-zero speed.
  • Acceleration as the rate of change of velocity, a = (v − u)/t, which way it points when speeding up versus slowing down, and why a fast object can still have zero acceleration.
  • How to read position-time graphs (slope = velocity) and velocity-time graphs (slope = acceleration, area = displacement), and why the area equals displacement.
  • How to derive and use the three kinematic equations — v = u + at, s = ut + ½at², v² = u² + 2as — straight from a velocity-time graph.
  • Why uniform circular motion is accelerated motion even at constant speed, because the direction of velocity (along the tangent) keeps changing.

What’s Next

You have now learned to describe how objects move — with distance, velocity, acceleration, graphs and equations. But we have not yet asked what makes them move, speed up, slow down or turn. That mystery — force — comes a little later.

Next, though, we switch from physics to the world of materials. In Chapter 5 — Exploring Mixtures and their Separation, you will learn what mixtures are, how they differ from pure substances, and the clever methods scientists use to pull a mixture apart into its ingredients. Just as motion had hidden order behind its variety, so does the matter all around you.

Frequently Asked Questions

What is the difference between distance and displacement in Class 9?

Distance is the total length of the path you actually travel, and it has no direction. Displacement is the straight-line change in position from start to finish, and it has a direction. If you go and come back to the same spot, your distance is large but your displacement is zero.

Can displacement ever be greater than distance travelled?

No, never. Displacement is the shortest straight-line gap between start and end, while distance is the actual path length. The straight line can never be longer than a path that bends. So the magnitude of displacement is always less than or equal to the distance travelled, and they are equal only when the object moves in one straight direction without turning back.

Why is the area under a velocity-time graph equal to displacement?

Because displacement = velocity × time, and on the graph velocity is the height and time is the width. So velocity × time is just height × width, which is the area of the strip under the line. Adding up all such strips gives the total displacement, which is why the whole shaded area equals the displacement.

How do you derive v = u + at from a graph?

On a velocity-time graph with constant acceleration, the line is straight. Its slope is the acceleration a, and slope equals (final velocity v minus initial velocity u) divided by time t. So a = (v − u)/t. Rearranging this gives at = v − u, and therefore v = u + at.

Why is uniform circular motion called accelerated motion if the speed is constant?

Acceleration means any change in velocity, and velocity includes direction, not just speed. In uniform circular motion the speed stays the same but the direction of motion keeps changing at every point. Since the direction changes, the velocity changes, so the motion is accelerated even though the speed is constant.

Can an object be moving fast but still have zero acceleration?

Yes. Acceleration depends on how quickly the velocity is changing, not on how big the velocity is. A bus speeding along a straight highway at a steady 80 km/h has a high velocity but zero acceleration, because its velocity is not changing.