Motion

CORE VS EXTENDED GUIDE

• Core students study only the unlabelled sections.
• Extended students must study everything, including Extended Extended points.
• Extended = Core + Supplement.

Speed and the Speed Equation

Key Equations

Speed:

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

Variables:

• $v$ = speed, in metres per second (m/s or m s⁻¹)
• $s$ = distance travelled, in metres (m)
• $t$ = time taken, in seconds (s)

SI unit of speed: m s⁻¹

Rearrangements:

$$s=v\times t$$

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

ProportionalitySpeed is directly proportional to distance when time is constant. Speed is inversely proportional to time when distance is constant — so doubling the time halves the speed.

Speed is the distance travelled per unit time.

Speed tells you how far an object moves in each second, without any reference to direction. Because speed has magnitude only and no direction, it is a scalar quantity. A car travelling at 15 m s⁻¹ covers 15 metres every second, regardless of which way it is heading.

Worked Example

A cyclist travels 840 m in 2 minutes. Calculate the cyclist's speed.

Converting time to SI units:

$$t=\frac{2\times 60}{1}=120\text{ s}$$

Equation used — Speed

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

Given

$$s=840\text{ m}$$

$$t=120\text{ s}$$

Substitution:

$$v=\frac{840}{120}$$

$$v=7.0{\text{ m s}}^{-1}$$

MisconceptionStudents often write "speed = distance over time" in definitions — this loses the mark. The locked definition requires the phrase "distance travelled per unit time." Exam cue: memorise "per unit time," not "divided by time."

Velocity as Speed with Direction

Velocity is speed in a stated direction.

Velocity includes both how fast an object moves and the direction it moves in. Because it has both magnitude and direction, velocity is a vector quantity. A car travelling at 30 m s⁻¹ northward has a different velocity from one travelling at 30 m s⁻¹ southward, even though both have the same speed.

MisconceptionSpeed and velocity are not interchangeable. An object moving in a circle at constant speed has a changing velocity because the direction changes continuously. Exam cue: if the question says "velocity," your answer must include a direction.

Average Speed

Key Equations

Average speed:

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

SI unit: m s⁻¹

The average speed of a journey accounts for the entire distance and the entire time, even when the object speeds up, slows down, or stops along the way. A car may travel at varying speeds throughout a journey, but the average speed gives one single value summarising the whole trip.

Average speed does not describe any particular instant during the journey — it smooths out all variations. Therefore, the actual speed at any moment could be higher or lower than the average.

Worked Example

A runner completes the first 200 m of a race in 25 s, then the next 200 m in 35 s. Calculate the runner's average speed for the whole race.

Finding total distance:

$$\text{total distance}=200+200=400\text{ m}$$

Finding total time:

$$\text{total time}=25+35=60\text{ s}$$

Equation used — Average speed

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

Substitution:

$$\text{average speed}=\frac{400}{60}$$

$$\text{average speed}=6.67{\text{ m s}}^{-1}\approx 6.7{\text{ m s}}^{-1}\text{ (2 s.f.)}$$

Examiner InsightA common error is to average two speeds (e.g. add two speeds and divide by two). This only works if the time spent at each speed is equal. CIE questions usually give different times for different sections, so you must always use total distance ÷ total time. Exam cue: always find total distance and total time separately first.

Distance–Time Graphs

Distance–time graphs display how far an object has travelled over a period of time. The horizontal axis (x-axis) represents time in seconds, and the vertical axis (y-axis) represents distance in metres. Every point on the line tells you the total distance the object has covered at that moment.

Reading a distance–time graph:

• The gradient (slope) of the line represents the speed of the object. A steeper gradient means a greater speed.
• A horizontal line (zero gradient) means the object is at rest — it is not moving.
• A straight line sloping upward means the object moves with constant speed — equal distances are covered in equal time intervals.
• A curve that gets steeper means the object is accelerating — the speed is increasing over time.
• A curve that gets less steep (flattening out) means the object is decelerating — the speed is decreasing over time.

Calculating speed from the gradient:

$$\text{gradient}=\frac{\Delta s}{\Delta t}=\text{speed}$$

Choose two points on the straight-line section, read off the distance and time values, and divide the change in distance by the change in time.

Exam TipA common error is drawing a line that slopes downward and labelling it "deceleration" — on a distance–time graph, the distance cannot decrease (unless the object returns), so a less steep upward curve indicates deceleration, not a downward slope.

MisconceptionStudents often confuse distance–time graphs with speed–time graphs. On a distance–time graph, a straight upward slope means constant speed (not acceleration). Acceleration appears as a curve on a distance–time graph. Exam cue: check the y-axis label before interpreting any graph.

Worked Example

From a distance–time graph, a straight-line section passes through (0 s, 0 m) and (8 s, 56 m). Calculate the speed during this section.

Equation used — Gradient of a distance–time graph

$$\text{speed}=\frac{\Delta s}{\Delta t}$$

Given

$$\Delta s=56-0=56\text{ m}$$

$$\Delta t=8-0=8\text{ s}$$

Substitution:

$$\text{speed}=\frac{56}{8}$$

$$\text{speed}=7.0{\text{ m s}}^{-1}$$

Speed–Time Graphs

Speed–time graphs display how the speed of an object changes over time. The horizontal axis represents time in seconds and the vertical axis represents speed in m s⁻¹.

Reading a speed–time graph:

• A horizontal line means the object moves with constant speed.
• A straight line sloping upward means the object has constant acceleration.
• A straight line sloping downward means the object has constant deceleration.
• A curve indicates changing acceleration. Extended A curve that becomes less steep over time indicates that the acceleration is decreasing.
• A point where the line meets the x-axis (speed = 0) means the object is at rest at that instant.

The area under a speed–time graph represents the distance travelled. For motion with constant speed, the area forms a rectangle. For motion with constant acceleration, the area forms a triangle or trapezium.

To calculate the area:

• Rectangle: area = base × height = time × speed
• Triangle: area = ½ × base × height = ½ × time × speed
• For combined shapes, split the area into rectangles and triangles, calculate each separately, and add them.
• 

Worked Example

A car accelerates uniformly from rest to 20 m s⁻¹ in 10 s, then travels at constant speed for a further 15 s. Calculate the total distance travelled.

Step 1: Finding the distance during acceleration (triangle area).

$${\text{distance}}_{1}=\frac{1}{2}\times \text{base}\times \text{height}$$

$${\text{distance}}_{1}=\frac{1}{2}\times 10\times 20$$

$${\text{distance}}_{1}=100\text{ m}$$

Step 2: Finding the distance during constant speed (rectangle area).

$${\text{distance}}_{2}=\text{base}\times \text{height}$$

$${\text{distance}}_{2}=15\times 20$$

$${\text{distance}}_{2}=300\text{ m}$$

Step 3: Finding total distance.

$$\text{total distance}=100+300=400\text{ m}$$

Examiner InsightCIE commonly gives a speed–time graph with several stages and asks for the total distance. Students must identify each section's shape (triangle, rectangle, trapezium) and calculate the area of each. Exam cue: always label each section clearly and add all areas together.

Qualitative Motion Analysis from Graphs

Given data or the shape of a graph, an object's motion can be identified qualitatively without calculation. The following table summarises the key features for both graph types:

The gradient of the distance–time graph gives the speed, so a changing gradient means changing speed — which is acceleration or deceleration. On a speed–time graph, acceleration and deceleration are visible directly from whether the line slopes upward or downward.

Free Fall and the Acceleration Due to Gravity

The acceleration of free fall ($g$) for an object near the surface of the Earth is approximately constant and is approximately 9.8 m/s² (often rounded to 10 m s⁻² in calculations unless told otherwise).

This means that any object falling freely (with no air resistance) increases its velocity by approximately 9.8 m s⁻¹ every second, regardless of its mass. A feather and a hammer dropped in a vacuum hit the ground at the same time because both experience the same gravitational acceleration.

The value of $g$ acts vertically downward toward the centre of the Earth.

MisconceptionStudents often believe that heavier objects fall faster. In the absence of air resistance, all objects near the Earth's surface accelerate at the same rate, approximately 9.8 m s⁻². Exam cue: state "in the absence of air resistance" when explaining equal acceleration.

Extended Acceleration and the Acceleration Equation

Key Equations

Acceleration:

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

Variables:

• $a$ = acceleration, in metres per second squared (m s⁻²)
• $\Delta v$ = change in velocity = final velocity ($v$) − initial velocity ($u$), in m s⁻¹
• $\Delta t$ = time taken for the change, in seconds (s)

The symbol $\Delta $ (delta) means "change in." So $\Delta v$ is read as "the change in velocity."

SI unit of acceleration: m s⁻²

Rearrangements:

$$\Delta v=a\times \Delta t$$

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

ProportionalityAcceleration is directly proportional to the change in velocity when time is constant — doubling $\Delta v$ doubles the acceleration. Acceleration is inversely proportional to time when $\Delta v$ is constant — doubling $\Delta t$ halves the acceleration.

Acceleration is the change in velocity per unit time.

Because velocity is a vector, acceleration is also a vector — it has both magnitude and direction. When an object speeds up in a given direction, the acceleration acts in that direction. When an object slows down, the acceleration acts in the opposite direction to the motion.

Worked Example

A car increases its velocity from 8.0 m s⁻¹ to 20 m s⁻¹ in 6.0 s. Calculate the acceleration.

Equation used — Acceleration

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

Given

$$\Delta v=20-8.0=12{\text{ m s}}^{-1}$$

$$\Delta t=6.0\text{ s}$$

Substitution:

$$a=\frac{12}{6.0}$$

$$a=2.0{\text{ m s}}^{-2}$$

Extended Acceleration from Speed–Time Graphs

On a speed–time graph, the gradient of the line equals the acceleration. A steeper gradient means a greater acceleration. A positive gradient represents acceleration, and a negative gradient represents deceleration.

For a straight-line section, the acceleration is constant because the gradient does not change. For a curved section, the acceleration is changing — if the curve becomes less steep, the acceleration is decreasing; if the curve becomes steeper, the acceleration is increasing.

To calculate the acceleration from a straight-line section:

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

Choose two points on the straight section, find the change in speed from the y-axis and the change in time from the x-axis, then divide.

Examiner InsightCIE examiners often ask students to find acceleration from a specific section of a speed–time graph. Students must pick two points on the correct section (not from different sections) and show the gradient calculation clearly. Exam cue: always show Δv and Δt explicitly — do not just state the answer.

Extended Deceleration as Negative Acceleration

A deceleration occurs when an object slows down — its speed decreases over time. Deceleration is a negative acceleration because the change in velocity is negative (the final velocity is less than the initial velocity).

Taking the direction of initial motion as positive: if an object is moving forward and slowing down, then $\Delta v$ is negative, which produces a negative value of $a$.

In calculations, a deceleration is handled by substituting the velocities correctly so that $\Delta v$ comes out negative. The negative sign in the answer indicates deceleration — it should not be discarded.

Worked Example

A bicycle travelling at 12 m s⁻¹ brakes and comes to rest in 8.0 s. Calculate the acceleration.

Taking the direction of initial motion as positive:

Equation used — Acceleration

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

Given

$$u=12{\text{ m s}}^{-1}$$

$$v=0{\text{ m s}}^{-1}$$

$$\Delta t=8.0\text{ s}$$

$$\Delta v=v-u=0-12=-12{\text{ m s}}^{-1}$$

Substitution:

$$a=\frac{-12}{8.0}$$

$$a=-1.5{\text{ m s}}^{-2}$$

The negative sign indicates a deceleration of 1.5 m s⁻².

MisconceptionStudents sometimes drop the negative sign or write deceleration as a positive number. A deceleration must be shown as negative acceleration in calculations. Writing a = 1.5 m s⁻² without acknowledging the negative sign would lose marks. Exam cue: always state the positive direction at the start, and keep the sign in your final answer.

Extended Falling with and without Air Resistance — Terminal Velocity

Objects falling near the Earth's surface experience a downward gravitational force (weight). In a vacuum (without air resistance), the only force acting is weight, so the object accelerates at $g$ ≈ 9.8 m s⁻² throughout the fall.

When air resistance is present, the motion is different because a second force acts. As an object begins to fall, its speed is low and air resistance is small, so the resultant downward force is large and the object accelerates quickly. As the speed increases, air resistance increases because the object collides with more air molecules per second and with greater force. This reduces the resultant downward force, so the acceleration decreases.

Eventually, the air resistance grows large enough to equal the weight. At this point the resultant force is zero, so the acceleration is zero and the object falls at a constant velocity. This constant maximum velocity is called terminal velocity.

The full causal chain for terminal velocity:

1. Object released — weight acts downward, air resistance is initially zero → resultant force is large → object accelerates.

2. Speed increases → air resistance increases → resultant force decreases → acceleration decreases.

3. Air resistance equals weight → resultant force is zero → acceleration is zero → object moves at constant (terminal) velocity.

The same principle applies to objects falling through liquids, except the resistive force is greater (due to the higher density of the liquid), so terminal velocity is reached sooner and at a lower speed.