Describing Motion Around Us
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.
A passenger is asleep on a moving train. Is the passenger at rest or in motion? Explain.
Both answers are correct, depending on the reference point. Compared to the train seat and the other passengers, the sleeping passenger is at rest — their position relative to the seat does not change. But compared to a tree or a station outside, the passenger is in motion, because their position relative to the ground changes with time. You must always state the reference point before saying “at rest” or “in motion”.
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.
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.
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.
- List the journey. The ball goes up 40 cm, then comes down 40 cm back to where it started.
- Distance is the total path length. Up 40 cm + down 40 cm = 80 cm. Distance is just a number, so distance travelled = 80 cm.
- Displacement is the straight gap between start and end. The ball ends exactly at O, the same place it started. So the start and end positions are the same point.
- That means the change in position is zero. So the total distance travelled is 80 cm, but the displacement is 0 cm. This shows displacement can be zero even when a real journey happened.
A vehicle uses up fuel. Does the fuel used depend on the distance travelled or on the displacement?
Fuel used depends on the distance travelled, not the displacement. The engine works for every metre of the actual path, whether the path bends, loops or doubles back. Even if you return to your exact starting spot (displacement = 0), you still burn fuel for the whole path you covered. So distance is what matters for fuel.
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.
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 speed | Average velocity |
|---|---|
| Uses total distance travelled | Uses displacement (net change in position) |
| Has only a magnitude (no direction) | Has magnitude and a direction (+ or −) |
| Can never be zero if the object moved | Can be zero even if the object moved a lot |
| = distance / time | = displacement / time |
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.
- Find the total distance. North 200 km + south 200 km = 400 km. Find the total time: 3 h + 2 h = 5 h.
- Average speed = total distance / total time = 400 km / 5 h = 80 km/h. Speed has no direction.
- Find the displacement. You drove 200 km north then 200 km back south, so you end exactly where you started. The net change in position is zero, so displacement = 0 km.
- Average velocity = displacement / time = 0 km / 5 h = 0 km/h. So the average speed is 80 km/h but the average velocity is 0 km/h, because you returned to your start. Final answer: average speed = 80 km/h, average velocity = 0 km/h.
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.
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.
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.
- Convert to m/s first (divide km/h by 3.6). 36 km/h = 10 m/s, 54 km/h = 15 m/s. A stop means final velocity 0 m/s.
- Part (i), pressing accelerator: u = 10 m/s, v = 15 m/s, t = 10 s. Use a = (v − u)/t = (15 − 10)/10 = 5/10 = 0.5 m/s². It is positive, so acceleration is in the direction of velocity (speeding up).
- Part (ii), braking: u = 15 m/s, v = 0 m/s, t = 5 s. Use a = (v − u)/t = (0 − 15)/5 = −15/5 = −3 m/s².
- The minus sign means the acceleration is opposite to the velocity, because the bus is slowing down. Final answers: (i) acceleration = 0.5 m/s² along the motion; (ii) acceleration = −3 m/s², opposite to the motion.
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.
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.
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.
On a position-time graph, what does a straight line that runs flat and parallel to the time axis tell you?
A flat, horizontal position-time line means the position is not changing as time passes. The object is staying in the same place, so it is at rest. The slope of a flat line is zero, which matches a velocity of zero.
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.
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.
So from a velocity-time graph you get two things:
- slope → acceleration
- area under the line → displacement
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 region under the sloping line between 10 s and 20 s is a rectangle sitting under a triangle. Split it into these two simple shapes.
- The rectangle has the lower velocity 5 m/s as its height and the time gap (20 − 10 = 10 s) as its width. Area of rectangle = 5 × 10 = 50 m.
- The triangle sits on top. Its base is the time gap = 10 s, and its height is the rise in velocity (10 − 5 = 5 m/s). Area of triangle = ½ × base × height = ½ × 10 × 5 = 25 m.
- Add the two areas: 50 m + 25 m = 75 m. So the displacement between 10 s and 20 s is 75 m.
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.
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.
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.
- Write down what is given. Starts from rest means u = 0 m/s. Final velocity v = 24 m/s. Time t = 6 s.
- For acceleration use v = u + at. So 24 = 0 + a × 6, giving a = 24 / 6 = 4 m/s².
- For distance use s = ut + ½at². Here u = 0, so s = 0 + ½ × 4 × 6². And 6² = 36.
- So s = ½ × 4 × 36 = 2 × 36 = 72 m. Final answers: acceleration = 4 m/s² and distance travelled = 72 m.
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.
- Given: u = 28 m/s, v = 0 m/s (it stops), s = 98 m. Time is not given, so use the third equation v² = u² + 2as.
- Substitute: 0² = 28² + 2 × a × 98. So 0 = 784 + 196a.
- Solve for a: 196a = −784, so a = −784 / 196 = −4 m/s². The minus sign shows it is slowing down.
- Now find time with v = u + at: 0 = 28 + (−4) × t, so 4t = 28, giving t = 7 s. Final answers: acceleration = −4 m/s² and time = 7 s.
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 u². 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.
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.)
A girl rides a scooter at a steady speedometer reading. Can the scooter still be accelerating?
Yes. A steady speedometer reading means the speed is constant, but speed is not the whole story. If the scooter is going around a bend, the direction of its velocity is changing even though the speed is the same. A changing direction is a changing velocity, and any change in velocity is acceleration. So a scooter turning at constant speed is indeed accelerating.
Common Mistakes
Distance and displacement are just two names for the same thing.
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.
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.
Displacement can be larger than the distance travelled.
Both are measured in metres and both describe 'how far', so it feels like either one could come out bigger depending on the path.
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.
If an object is moving fast, it must have a large acceleration.
In daily speech we say a car 'accelerates' when it goes fast, so high speed and high acceleration feel like the same thing.
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.
In uniform circular motion the speed is constant, so there is no acceleration.
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.
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.
The kinematic equations (v = u + at and so on) work for any motion.
They are presented as 'the equations of motion', which sounds like they should cover every kind of moving object.
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
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.
Trace the trips, each 250 m long: home → shop (250 m), shop → home (250 m), home → shop (250 m), shop → home (250 m). That is four trips of 250 m each.
Total distance = 4 × 250 = 1000 m (or 1 km).
Displacement: he ends at home, exactly where he started, so the net change in position is zero. Displacement = 0 m.
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.
Going up from the ground floor (floor 0) to the fourth floor is 4 floors = 4 × 3 = 12 m up.
Coming down from the fourth floor to the second floor is 2 floors = 2 × 3 = 6 m down.
(i) Total vertical distance = 12 m + 6 m = 18 m.
(ii) Displacement: the student ends on the second floor, which is 2 floors = 2 × 3 = 6 m above the starting ground floor. So displacement = 6 m upward.
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?
First convert to m/s (divide by 3.6): u = 54/3.6 = 15 m/s, v = 36/3.6 = 10 m/s, t = 36 s.
Find the acceleration: a = (v − u)/t = (10 − 15)/36 = −5/36 m/s² (negative, since slowing).
Use s = ut + ½at²: s = 15 × 36 + ½ × (−5/36) × 36².
Now 36² = 1296, and ½ × (−5/36) × 1296 = (−5/36) × 648 = −90.
So s = 540 + (−90) = 450 m.
(You could also use the average velocity: average of 15 and 10 is 12.5 m/s, and 12.5 × 36 = 450 m. Same answer.)
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.)
Sketch the velocity-time graph in your head: it rises from 0 to 20 m/s, stays flat at 20 m/s, then falls back to 0. Find the area of each part.
Phase 1 (speeding up, a triangle): base = 5 s, height = 20 m/s. Area = ½ × 5 × 20 = 50 m.
Phase 2 (steady, a rectangle): width = 10 s, height = 20 m/s. Area = 20 × 10 = 200 m.
Phase 3 (braking, a triangle): base = 6 s, height = 20 m/s. Area = ½ × 6 × 20 = 60 m.
Total distance = 50 + 200 + 60 = 310 m.
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?
Convert speed: 36 km/h = 36/3.6 = 10 m/s.
Reaction phase (0.5 s before braking): the bus moves at a steady 10 m/s. Distance = speed × time = 10 × 0.5 = 5 m.
Braking phase: u = 10 m/s, v = 0, a = −2.5 m/s². Use v² = u² + 2as (time not needed): 0² = 10² + 2 × (−2.5) × s 0 = 100 − 5s 5s = 100, so s = 20 m.
Total stopping distance = reaction distance + braking distance = 5 + 20 = 25 m.
Since 25 m is less than the 30 m gap, the bus stops with 5 m to spare. Yes, it stops in time. This shows why reaction time matters: those extra 5 m came from the half-second of reacting before the brake even engaged.
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.
First the accelerations, using a = (v − u)/t with u = 0:
Car A: a = 5/5 = 1 m/s². Car B: a = 3/10 = 0.3 m/s².
Now the displacements, using s = ut + ½at² with u = 0 (so s = ½at²):
Car A: s = ½ × 1 × 5² = ½ × 1 × 25 = 12.5 m in 5 s. Car B: s = ½ × 0.3 × 10² = ½ × 0.3 × 100 = 15 m in 10 s.
So even though car A accelerates harder, car B covers more ground because it travels for a longer time. (On a velocity-time graph, each displacement is just the area of the triangle under that car’s line — A’s triangle is ½ × 5 × 5 = 12.5 m, B’s is ½ × 10 × 3 = 15 m.)
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.