1.2 – Motion

Table of Contents

1.2 — Motion

1.2 — Motion #

IGCSE Physics | Core & Supplement

1. Speed and Velocity #

Speed #

Definition Speed is the distance an object travels per unit of time.

The formula for speed is:

$$v = \frac{s}{t}$$

$v$ = speed (m/s)
$s$ = distance travelled (m)
$t$ = time taken (s)

Worked Example — A car travels 150 m in 5 s. Calculate its speed.

Formula: $$v = \frac{s}{t}$$
Given: $s = 150\ \text{m}$, $t = 5\ \text{s}$
Substitute: $$v = \frac{150}{5}$$
Answer: $$v = 30\ \text{m/s}$$

Velocity #

Definition Velocity is speed in a given direction.

Speed vs Velocity: Speed only tells you how fast an object moves. Velocity also tells you which direction. For example, 30 m/s is a speed; 30 m/s north is a velocity.

2. Average Speed #

When an object’s speed changes during a journey, use average speed:

$$\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}$$

Worked Example — A cyclist travels 600 m in 2 minutes. Find the average speed.

Formula: $$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$$
Given: distance = 600 m, time = 2 min = 120 s
Substitute: $$\text{average speed} = \frac{600}{120}$$
Answer: $$\text{average speed} = 5\ \text{m/s}$$

3. Acceleration Supplement #

Definition Acceleration is the change in velocity per unit of time.

$$a = \frac{\Delta v}{\Delta t}$$

$a$ = acceleration (m/s²)
$\Delta v$ = change in velocity (m/s) — final velocity minus initial velocity
$\Delta t$ = time taken (s)

Worked Example — A car speeds up from 10 m/s to 30 m/s in 5 s. Find the acceleration.

Formula: $$a = \frac{\Delta v}{\Delta t}$$
Given: initial velocity = 10 m/s, final velocity = 30 m/s, $\Delta t$ = 5 s
Change in velocity: $\Delta v = 30 – 10 = 20\ \text{m/s}$
Substitute: $$a = \frac{20}{5}$$
Answer: $$a = 4\ \text{m/s}^2$$

Deceleration #

Definition Deceleration is a negative acceleration. The velocity is decreasing, so $\Delta v$ is negative.

Worked Example — A car slows from 20 m/s to 5 m/s in 3 s. Find the acceleration.

Formula: $$a = \frac{\Delta v}{\Delta t}$$
$\Delta v = 5 – 20 = -15\ \text{m/s}$, $\Delta t = 3\ \text{s}$
Substitute: $$a = \frac{-15}{3}$$
Answer: $$a = -5\ \text{m/s}^2$$

The negative sign shows the car is decelerating (slowing down).

4. Distance–Time Graphs #

A distance–time graph shows how far an object has travelled over time. The shape of the line tells you how it is moving.

Shape of Line	What it Means
Horizontal line	At rest — distance is not changing
Straight line sloping upward	Constant speed — equal distance in equal time
Curve getting steeper	Accelerating — speed is increasing
Curve getting less steep	Decelerating — speed is decreasing

Calculating Speed from the Gradient #

The gradient (slope) of a straight-line section equals the speed of the object:

$$\text{speed} = \frac{\text{change in distance}}{\text{change in time}}$$

Worked Example — A straight line goes from (0 s, 0 m) to (4 s, 20 m). Calculate the speed.

Formula: $$\text{speed} = \frac{\Delta s}{\Delta t}$$
$\Delta s = 20 – 0 = 20\ \text{m}$, $\Delta t = 4 – 0 = 4\ \text{s}$
Substitute: $$\text{speed} = \frac{20}{4}$$
Answer: $$\text{speed} = 5\ \text{m/s}$$

Remember: Only calculate the gradient for a straight-line section. A curved section means the speed is changing.

5. Speed–Time Graphs #

A speed–time graph shows how the speed of an object changes over time.

Shape of Line	What it Means
Horizontal line at $v = 0$	At rest
Horizontal line above zero	Constant speed
Line sloping up	Accelerating — speed is increasing
Line sloping down	Decelerating — speed is decreasing

Supplement — Constant vs Changing Acceleration: If the sloping line is straight, the acceleration is constant. If the line is a curve, the acceleration is changing.

Calculating Distance from the Area #

The area under a speed–time graph equals the distance travelled.

Worked Example 1 — Constant Speed: An object moves at 6 m/s for 10 s. Find the distance.

Shape under the graph: rectangle
Area = width × height = $10 \times 6$
Distance = $60\ \text{m}$

Worked Example 2 — Constant Acceleration from Rest: Speed increases from 0 to 12 m/s in 6 s. Find the distance.

Shape under the graph: triangle (speed starts at 0)
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 12$$
Distance = $36\ \text{m}$

Note: If the object does not start from rest, the shape is a trapezium. Use: $$\text{Distance} = \frac{1}{2} \times (v_1 + v_2) \times t$$ where $v_1$ and $v_2$ are the starting and ending speeds, and $t$ is the time.

Calculating Acceleration from the Gradient Supplement #

The gradient of a straight line on a speed–time graph equals the acceleration:

$$a = \frac{\Delta v}{\Delta t}$$

Worked Example — A line goes from (0 s, 4 m/s) to (5 s, 14 m/s). Find the acceleration.

Formula: $$a = \frac{\Delta v}{\Delta t}$$
$\Delta v = 14 – 4 = 10\ \text{m/s}$, $\Delta t = 5 – 0 = 5\ \text{s}$
Substitute: $$a = \frac{10}{5}$$
Answer: $$a = 2\ \text{m/s}^2$$

6. Free Fall #

Definition Free fall is when an object falls under gravity only, with no other forces acting on it.

Near the surface of the Earth, the acceleration due to gravity is approximately constant:

$$g \approx 9.8\ \text{m/s}^2$$

This means a freely falling object speeds up by about 9.8 m/s every second.

Remember: All objects near Earth’s surface fall with the same acceleration of $9.8\ \text{m/s}^2$, regardless of their mass — as long as there is no air resistance.

7. Falling with Air Resistance — Terminal Velocity Supplement #

When a real object falls through air or a liquid, two forces act on it:

Weight — pulls the object downward (constant)
Air resistance (drag) — acts upward, opposing the motion

What Happens as the Object Falls #

At the start, weight is greater than air resistance, so the object accelerates.
As speed increases, air resistance increases.
The acceleration decreases as the two forces get closer in size.
Eventually, air resistance equals the weight. Net force = 0.
The object reaches a constant speed called terminal velocity. It no longer accelerates.

Terminal Velocity Terminal velocity is the maximum constant speed a falling object reaches when air resistance equals its weight. At this point there is no net force, so there is no acceleration.

Without air resistance: Constant acceleration of $g = 9.8\ \text{m/s}^2$ all the way down.

With air resistance: Acceleration decreases over time. The object eventually reaches terminal velocity and falls at constant speed.