Think about the last time you ran for a bus. You started slowly, sped up, maybe slowed down as you got close, then stopped. Every moment of that run was a story of motion — changing speed, changing direction, changing position. Now think of a car on a highway, a rocket leaving the Earth, a ball thrown into the air. All of these are examples of linear motion — motion in a straight line.
Linear motion is the foundation of mechanics. It is where we learn the language of motion: distance, displacement, speed, velocity, and acceleration. Once you understand these ideas, you can describe and predict how anything moves — from a falling leaf to a planet orbiting the Sun. In this chapter, we are going to build that language together, step by step. Let us begin.
I once asked a class, "What is the difference between speed and velocity?" Most students said they were the same thing. Then I asked, "If you run around a circular track and end up where you started, what is your average velocity?" They were shocked to learn it was zero. That moment changed how they saw motion. By the end of this chapter, you will understand why.By the end of this chapter, you should be able to:
Distance and displacement sound similar, but they are very different.
Think of walking around a block. If you walk 100 m east, 100 m north, 100 m west, and 100 m south, you end up where you started. Your distance is 400 m. Your displacement is zero — because you are back at the starting point.
| Quantity | Type | Definition | Example |
|---|---|---|---|
| Distance | Scalar | Total path length | 400 m |
| Displacement | Vector | Straight-line distance + direction | 0 m (if you return to start) |
Speed and velocity are also related but different.
Average speed: speed = total distance / total time
Average velocity: velocity = total displacement / total time
Think of a car driving around a circular track at a constant speed of 20 m/s. Its speed is constant. But its velocity is constantly changing because its direction is constantly changing. This is why velocity is a vector and speed is not.
Instantaneous speed vs average speed:
Acceleration is the rate of change of velocity. It tells you how quickly velocity is changing.
Acceleration: a = (v − u) / t
a = acceleration (m/s²), v = final velocity (m/s), u = initial velocity (m/s), t = time (s)
The SI unit of acceleration is metres per second squared (m/s²).
Positive acceleration: velocity is increasing (speeding up).
Negative acceleration (deceleration): velocity is decreasing (slowing down).
Think of a car. When you press the accelerator, the car accelerates. When you press the brake, the car decelerates. When you hold the speed steady, the car has zero acceleration.
| Situation | Velocity Change | Acceleration |
|---|---|---|
| Speeding up | Increasing | Positive |
| Slowing down | Decreasing | Negative (deceleration) |
| Constant speed in a straight line | Constant | Zero |
When an object moves with constant acceleration in a straight line, we can use a set of equations to describe its motion. These are called the equations of uniformly accelerated motion.
First equation: v = u + at
Second equation: s = ut + ½at²
Third equation: v² = u² + 2as
Fourth equation: s = ((u + v) / 2) × t
Where: u = initial velocity (m/s), v = final velocity (m/s), a = acceleration (m/s²), t = time (s), s = displacement (m)
These four equations are the workhorses of linear motion. Learn them well. With them, you can solve almost any problem involving constant acceleration.
How to choose the right equation:
| Known Quantities | Unknown | Equation to Use |
|---|---|---|
| u, v, t | a | v = u + at |
| u, a, t | s | s = ut + ½at² |
| u, v, a | s | v² = u² + 2as |
| u, v, t | s | s = ((u + v) / 2) × t |
When an object falls freely under gravity, it accelerates downward at a constant rate. This acceleration is called the acceleration due to gravity, denoted by g.
On Earth, g ≈ 10 m/s² (more precisely, 9.8 m/s², but we will often use 10 for simplicity).
Key points about free fall:
Using the equations for free fall:
A displacement-time graph shows how displacement changes with time.
| Shape of Graph | Meaning |
|---|---|
| Horizontal line | Object is at rest (zero velocity) |
| Straight line with constant gradient | Constant velocity (uniform motion) |
| Curve with increasing gradient | Accelerating |
| Curve with decreasing gradient | Decelerating |
Think of a displacement-time graph as a story of position. The steeper the line, the faster the object is moving. A flat line means the object is standing still.
The gradient of a displacement-time graph gives velocity. If the graph is a straight line, the velocity is constant. If it curves, the velocity is changing — the object is accelerating or decelerating. This is one of the most important graph skills in physics.A velocity-time graph shows how velocity changes with time.
| Shape of Graph | Meaning |
|---|---|
| Horizontal line | Constant velocity (zero acceleration) |
| Straight line with positive gradient | Uniform acceleration |
| Straight line with negative gradient | Uniform deceleration |
| Curve | Non-uniform acceleration |
The area under a velocity-time graph is a powerful tool. It gives you the displacement directly. For a straight-line graph, the area is a triangle or a trapezium. For a curve, you can estimate the area by counting squares.
Velocity-time graphs are the most information-rich graphs in kinematics. From one graph, you can read velocity, acceleration, and displacement. Master them, and you master motion.Relative motion is the motion of one object as observed from another moving object. Your perception of motion depends on your frame of reference.
Think of two cars travelling side by side at the same speed. To each driver, the other car appears stationary. But to a person on the roadside, both cars are moving at high speed. Both views are correct — they just use different frames of reference.
Examples of relative motion:
Average speed: speed = total distance / total time
Average velocity: velocity = total displacement / total time
Acceleration: a = (v − u) / t
Equations of motion:
1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as
4. s = ((u + v) / 2) × t
Acceleration due to gravity: g = 10 m/s² (approximately)
Graphs: gradient of displacement-time graph = velocity; gradient of velocity-time graph = acceleration; area under velocity-time graph = displacement
Question: A runner completes one lap of a 400 m circular track in 50 seconds. Calculate her average speed and average velocity.
Answer:
Average speed = total distance / total time = 400 m / 50 s = 8 m/s
Average velocity = total displacement / total time = 0 m / 50 s = 0 m/s (because she returns to her starting point)
Units check: the answer is in m/s. Notice that the average velocity is zero even though the runner is moving at 8 m/s. This is the difference between speed and velocity — one is scalar, the other is vector.Question: A car accelerates uniformly from rest to 30 m/s in 10 seconds. Calculate its acceleration and the distance travelled.
Answer:
Given: u = 0 m/s, v = 30 m/s, t = 10 s
Find a: v = u + at → 30 = 0 + a × 10 → a = 3 m/s²
Find s: s = ut + ½at² → s = 0 × 10 + ½ × 3 × (10)² = 150 m
Units check: acceleration in m/s², distance in m. Notice that we used two equations. Always write down what you know and what you need before choosing an equation.Question: A car travelling at 25 m/s brakes and comes to rest in 5 seconds. Calculate its deceleration and the distance travelled during braking.
Answer:
Given: u = 25 m/s, v = 0 m/s, t = 5 s
Find a: a = (v − u) / t = (0 − 25) / 5 = −5 m/s² (deceleration of 5 m/s²)
Find s: s = ((u + v) / 2) × t = ((25 + 0) / 2) × 5 = 62.5 m
Units check: acceleration is negative because the car is slowing down. The distance is in metres. Notice that deceleration is just negative acceleration.Question: A stone is dropped from a height of 45 m. Calculate the time it takes to reach the ground and its velocity just before impact. Use g = 10 m/s².
Answer:
Given: u = 0 m/s, s = 45 m, a = g = 10 m/s²
Find t: s = ut + ½at² → 45 = 0 + ½ × 10 × t² → 45 = 5t² → t² = 9 → t = 3 s
Find v: v = u + at = 0 + 10 × 3 = 30 m/s
Units check: time in seconds, velocity in m/s. Notice that the stone reaches 30 m/s — about 108 km/h. That is why falling objects can be dangerous.Question: A ball is thrown vertically upward with an initial velocity of 20 m/s. Calculate the maximum height reached and the total time of flight. Use g = 10 m/s².
Answer:
Given: u = 20 m/s, v = 0 m/s (at highest point), a = −g = −10 m/s²
Find height: v² = u² + 2as → 0 = (20)² + 2 × (−10) × s → 0 = 400 − 20s → 20s = 400 → s = 20 m
Find time to highest point: v = u + at → 0 = 20 − 10t → t = 2 s
Total time of flight = 2 × 2 = 4 s
Units check: height in metres, time in seconds. Notice that the time up equals the time down. This is a beautiful symmetry of free fall motion.Graphs are essential for understanding linear motion.
Displacement-time graphs:
Velocity-time graphs:
Think of a graph as a map of motion. The shape tells you what is happening, and the gradient and area give you the numbers.
Apparatus: A ticker-tape timer, a tape, a trolley, a runway, a pulley, and weights.
Method:
Calculation: If the distance between dots increases by 2 cm every 0.1 s, the acceleration is 20 cm/s² = 0.2 m/s².
Precautions:
Likely source of error: Friction in the pulley and timer can affect the results. Keep the equipment clean and well-maintained.
The ticker-tape timer is a classic tool in physics. It makes the invisible visible — you can actually see the acceleration as the dots get farther apart. It is a beautiful way to connect theory with reality.Distance is the total path length (scalar). Displacement is the straight-line distance from start to finish, with direction (vector).
Speed is the rate of change of distance (scalar). Velocity is the rate of change of displacement (vector).
Acceleration is the rate of change of velocity. Its SI unit is metres per second squared (m/s²).
1. v = u + at; 2. s = ut + ½at²; 3. v² = u² + 2as; 4. s = ((u + v)/2) × t.
a = (v − u)/t = (30 − 10)/5 = 4 m/s².
s = ut + ½at² → 20 = 0 + ½ × 10 × t² → 20 = 5t² → t² = 4 → t = 2 s.
Velocity.
Displacement.
(a) Distinguish between distance and displacement. [4]
(b) A runner completes a 400 m lap of a track in 50 s. Calculate her average speed and average velocity. [4]
(c) Explain why the answers are different. [2]
(a) Distance is the total path length (scalar). Displacement is the straight-line distance from start to finish, with direction (vector). (4 marks.)
(b) Average speed = 400/50 = 8 m/s. Average velocity = 0/50 = 0 m/s. (2 marks each.)
(c) The runner ends where she started, so her displacement is zero. Speed depends on distance; velocity depends on displacement. (2 marks.)
(a) Define acceleration and state its SI unit. [2]
(b) A car accelerates uniformly from rest to 25 m/s in 8 s. Calculate its acceleration and the distance travelled. [5]
(c) The car then brakes and stops in 4 s. Calculate its deceleration. [3]
(a) Acceleration is the rate of change of velocity. SI unit: m/s². (2 marks.)
(b) a = (25 − 0)/8 = 3.125 m/s². s = ut + ½at² = 0 + ½ × 3.125 × 64 = 100 m. (5 marks.)
(c) a = (0 − 25)/4 = −6.25 m/s² (deceleration of 6.25 m/s²). (3 marks.)
(a) Describe how to determine the acceleration of a trolley using a ticker-tape timer. [5]
(b) A ball is thrown vertically upward with a velocity of 15 m/s. Calculate the maximum height reached and the total time of flight. Use g = 10 m/s². [5]
(a) Set up the trolley with tape through the timer. Release and let it accelerate. Measure the distance between dots. Calculate velocities and plot a velocity-time graph. The gradient is the acceleration. (5 marks.)
(b) At highest point, v = 0. v² = u² + 2as → 0 = 225 − 20s → s = 11.25 m. Time up: v = u + at → 0 = 15 − 10t → t = 1.5 s. Total time = 3 s. (5 marks.)
A car travelling at 20 m/s passes a stationary police car. The police car immediately accelerates at 4 m/s². How long does it take to catch the speeding car, and how far does it travel?
Let t be the time to catch up. Distance of speeding car = 20t. Distance of police car = ½ × 4 × t² = 2t². Setting them equal: 20t = 2t² → t² = 10t → t = 10 s. Distance = 20 × 10 = 200 m.
Explain why a car turning a corner at constant speed is accelerating, even though its speed is not changing.
Acceleration is the rate of change of velocity. Velocity is a vector — it has both size and direction. When a car turns a corner, its direction changes, so its velocity changes, even if its speed remains constant. Therefore, it is accelerating. This is called centripetal acceleration.
Two objects are dropped from the same height, but one is heavier than the other. Which hits the ground first? Explain.
In the absence of air resistance, both hit the ground at the same time. This is because the acceleration due to gravity is the same for all objects, regardless of mass. In the presence of air resistance, the heavier object may hit first because it is less affected by air resistance relative to its weight.
Motion is the poetry of the universe — and with the language of linear motion, you can read its verses.
