Linear motion

Overview

Linear motion is motion along a straight line. A learner studies it in Physics because many everyday movements can first be understood in one direction before more complicated paths are considered: a cyclist moving along a straight road, a stone falling vertically, a cart moving along a track, or a student walking from one point to another.

This chapter introduces the language used to describe such motion. Distance tells how much ground is covered. Displacement tells the change in position in a stated direction. Speed tells how fast distance changes. Velocity tells how fast displacement changes. Acceleration tells how fast velocity changes.

The main skill is to connect words, units, formulas, and graphs. A table of readings, a distance-time graph, and a velocity-time graph can all describe the same movement if their meanings are interpreted carefully.

Prerequisites

• Physical quantities and SI units - Learners should know that a physical quantity has a magnitude and a unit.
• Measuring instruments in Physics - Length and time measurements support motion experiments.
• Measuring instruments and physical quantities - Matching a measured quantity to a suitable instrument helps learners collect motion data.
• Mathematical relationships among physical quantities - Formula substitution and rearrangement are needed for motion calculations.
• Graphical presentation of experimental results - Graph axes, scales, gradients, and areas support interpretation of motion graphs.

Learning Scope

This chapter covers linear motion in one dimension, distance, displacement, scalar and vector quantities, speed, velocity, acceleration, SI units, simple calculations, and basic interpretation of distance-time and velocity-time graphs.

This page does not teach projectile motion, circular motion, Newton's laws of motion, momentum, friction, or advanced equations of uniformly accelerated motion. Those ideas belong to later or related mechanics pages.

Subtopics

Motion Along A Straight Line

An object is in motion when its position changes with time. In linear motion, the change of position happens along a straight line. The straight line may be horizontal, vertical, or sloping; the important idea is that the motion can be described along one direction.

To describe linear motion clearly, choose:

• a reference point, such as the starting point
• a positive direction, such as east, upward, or forward
• a time interval, such as from $t = 0 \ \text{s}$ to $t = 10 \ \text{s}$

Key insight: Motion is not described fully by saying "it moved". A Physics description should tell how far, in what direction when needed, and over what time.

For example, a trolley moving $8 \ \text{m}$ to the right in $4 \ \text{s}$ has motion that can be described with displacement, velocity, and time because the direction is known.

Distance And Displacement

Distance is the total length of the path travelled. It does not depend on direction. Distance is a scalar quantity.

Displacement is the change in position from the starting point to the final point in a stated direction. It depends on direction. Displacement is a vector quantity.

If a learner walks $30 \ \text{m}$ east and then $10 \ \text{m}$ west, the total distance is:

$$ 30 \ \text{m} + 10 \ \text{m} = 40 \ \text{m} $$

The displacement from the starting point is:

$$ 30 \ \text{m east} - 10 \ \text{m east} = 20 \ \text{m east} $$

Key insight: Distance can only increase or stay the same as an object moves. Displacement can increase, decrease, become zero, or be negative depending on the chosen direction.

The SI unit for both distance and displacement is the metre, written as $\text{m}$.

Scalars And Vectors

A scalar quantity has magnitude only. Magnitude means size or amount. Distance, speed, time, and mass are scalar quantities.

A vector quantity has magnitude and direction. Displacement, velocity, acceleration, and force are vector quantities.

For example:

• $5 \ \text{m}$ is a distance because no direction is stated.
• $5 \ \text{m east}$ is a displacement because a direction is included.
• $12 \ \text{m/s}$ is a speed.
• $12 \ \text{m/s east}$ is a velocity.

Key insight: Direction is not decoration. It changes the type of quantity and can change the answer in a motion problem.

Speed

Speed is the rate of change of distance with time. It tells how fast an object covers distance.

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

In symbols:

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

where $v$ is speed, $s$ is distance, and $t$ is time. Some books use $u$ or other letters in later motion work, so learners should focus first on the meaning of the formula.

The SI unit of speed is metre per second, written as $\text{m/s}$ or $\text{m s}^{-1}$.

If a student runs $60 \ \text{m}$ in $12 \ \text{s}$, the speed is:

$$ \begin{aligned} v &= \frac{s}{t} \\ &= \frac{60 \ \text{m}}{12 \ \text{s}} \\ &= 5 \ \text{m/s} \end{aligned} $$

Key insight: Average speed uses total distance divided by total time. It does not say the speed was the same at every moment.

Velocity

Velocity is the rate of change of displacement with time. It tells how fast position changes in a stated direction.

$$ \text{velocity} = \frac{\text{displacement}}{\text{time}} $$

In symbols:

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

where $\Delta x$ means change in position or displacement. The SI unit of velocity is also $\text{m/s}$, but velocity must include direction when the direction matters.

If a cart moves $20 \ \text{m}$ east in $5 \ \text{s}$, its average velocity is:

$$ \begin{aligned} v &= \frac{20 \ \text{m east}}{5 \ \text{s}} \\ &= 4 \ \text{m/s east} \end{aligned} $$

Key insight: Speed and velocity can have the same numerical value in straight-line motion when the object keeps moving in one direction. They differ when direction changes.

Acceleration

Acceleration is the rate of change of velocity with time. It describes how quickly velocity changes.

$$ \text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}} $$

In symbols:

$$ a = \frac{v - u}{t} $$

where:

• $a$ is acceleration
• $u$ is initial velocity
• $v$ is final velocity
• $t$ is time taken

The SI unit of acceleration is metre per second squared, written as $\text{m/s}^2$ or $\text{m s}^{-2}$.

If the velocity of a bicycle changes from $2 \ \text{m/s}$ to $8 \ \text{m/s}$ in $3 \ \text{s}$, then:

$$ \begin{aligned} a &= \frac{v-u}{t} \\ &= \frac{8 \ \text{m/s} - 2 \ \text{m/s}}{3 \ \text{s}} \\ &= \frac{6 \ \text{m/s}}{3 \ \text{s}} \\ &= 2 \ \text{m/s}^2 \end{aligned} $$

Key insight: Acceleration does not always mean "speeding up" in ordinary language. In Physics, it means velocity is changing. A change in speed, a change in direction, or both can involve acceleration. In this linear-motion chapter, the focus is on change of speed along a straight line.

If final velocity is less than initial velocity, the acceleration may be negative relative to the chosen positive direction. Negative acceleration in a straight line often means the object is slowing down, but the sign must be interpreted using the chosen direction.

Uniform And Non-Uniform Motion

Uniform motion means equal distances are covered in equal time intervals in the same direction. Its speed or velocity is constant.

Non-uniform motion means unequal distances are covered in equal time intervals, or the velocity changes. Accelerated motion is non-uniform because velocity changes with time.

For example, if a trolley covers $2 \ \text{m}$ every second along the same straight line, its speed is constant. If it covers $1 \ \text{m}$ in the first second, $3 \ \text{m}$ in the second second, and $5 \ \text{m}$ in the third second, its speed is changing.

Key insight: A table of distance and time can show whether motion is uniform before a graph is drawn.

Distance-Time Graphs

A distance-time graph has time on the horizontal axis and distance on the vertical axis. It shows how distance from the starting point changes with time.

For a distance-time graph:

• a horizontal line means the object is stationary because distance is not changing
• a straight sloping line means constant speed
• a steeper line means greater speed
• a curved line means changing speed

The gradient of a distance-time graph gives speed:

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

Using two points on a straight section:

$$ \text{speed} = \frac{d_2 - d_1}{t_2 - t_1} $$

For example, if a straight section passes through $(2 \ \text{s}, 4 \ \text{m})$ and $(6 \ \text{s}, 12 \ \text{m})$, then:

$$ \begin{aligned} \text{speed} &= \frac{12 \ \text{m} - 4 \ \text{m}}{6 \ \text{s} - 2 \ \text{s}} \\ &= \frac{8 \ \text{m}}{4 \ \text{s}} \\ &= 2 \ \text{m/s} \end{aligned} $$

Key insight: The graph's steepness is not the distance itself. Steepness tells the rate of change of distance, which is speed.

In this basic chapter, a distance-time graph is normally not used to show motion backwards because distance travelled does not decrease. If position or displacement is graphed instead, a downward slope can show motion in the negative direction.

Velocity-Time Graphs

A velocity-time graph has time on the horizontal axis and velocity on the vertical axis. It shows how velocity changes with time.

For a velocity-time graph:

• a horizontal line above zero means constant positive velocity
• a horizontal line at zero means the object is at rest
• an upward straight sloping line means uniform positive acceleration
• a downward straight sloping line means velocity is decreasing
• a line below zero means motion in the negative direction

The gradient of a velocity-time graph gives acceleration:

$$ \text{acceleration} = \frac{\text{change in velocity}}{\text{change in time}} $$

Using two points on a straight section:

$$ a = \frac{v_2 - v_1}{t_2 - t_1} $$

The area under a velocity-time graph gives displacement for motion along a straight line:

$$ \text{displacement} = \text{area under the velocity-time graph} $$

For constant velocity, the area is a rectangle:

$$ \text{displacement} = v \times t $$

For velocity increasing uniformly from zero, the area is a triangle:

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

Key insight: On a velocity-time graph, height means velocity, gradient means acceleration, and area means displacement. Confusing these three meanings is one of the most common graph errors.

Units And Formula Discipline

Motion calculations are safest when all quantities are written with SI units before substitution:

• distance and displacement: metre, $\text{m}$
• time: second, $\text{s}$
• speed and velocity: metre per second, $\text{m/s}$
• acceleration: metre per second squared, $\text{m/s}^2$

If distance is given in kilometres or time in hours, convert before using SI answers unless the question clearly asks for another unit.

Useful conversions include:

$$ 1 \ \text{km} = 1000 \ \text{m} $$

$$ 1 \ \text{h} = 3600 \ \text{s} $$

Key insight: A numerical answer without a unit is incomplete in Physics. The unit helps show whether the calculation used the correct relationship.

Key Terms

• Acceleration: rate of change of velocity with time; SI unit $\text{m/s}^2$.
• Distance: total length of path travelled; a scalar quantity; SI unit $\text{m}$.
• Displacement: change in position in a stated direction; a vector quantity; SI unit $\text{m}$.
• Linear motion: motion along a straight line.
• Scalar: a quantity with magnitude only.
• SI unit: an internationally agreed standard unit used for measurement in science.
• Speed: rate of change of distance with time; a scalar quantity; SI unit $\text{m/s}$.
• Vector: a quantity with magnitude and direction.
• Velocity: rate of change of displacement with time; a vector quantity; SI unit $\text{m/s}$.

Worked Examples

Example 1: Average Speed

A bus travels $150 \ \text{m}$ in $10 \ \text{s}$. Find its average speed.

Use:

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

Then substitute:

$$ \begin{aligned} v &= \frac{s}{t} \\ &= \frac{150 \ \text{m}}{10 \ \text{s}} \\ &= 15 \ \text{m/s} \end{aligned} $$

The average speed is $15 \ \text{m/s}$.

Example 2: Distance And Displacement

A learner walks $40 \ \text{m}$ north and then $15 \ \text{m}$ south along the same straight path. Find the total distance and the displacement.

Distance is the total path length:

$$ \begin{aligned} \text{distance} &= 40 \ \text{m} + 15 \ \text{m} \\ &= 55 \ \text{m} \end{aligned} $$

For displacement, take north as positive:

$$ \begin{aligned} \text{displacement} &= 40 \ \text{m north} - 15 \ \text{m north} \\ &= 25 \ \text{m north} \end{aligned} $$

The total distance is $55 \ \text{m}$, while the displacement is $25 \ \text{m north}$.

Example 3: Average Velocity

A cart moves $18 \ \text{m}$ east in $6 \ \text{s}$. Find its average velocity.

Velocity uses displacement, not total path length:

$$ \begin{aligned} v &= \frac{\Delta x}{t} \\ &= \frac{18 \ \text{m east}}{6 \ \text{s}} \\ &= 3 \ \text{m/s east} \end{aligned} $$

The average velocity is $3 \ \text{m/s east}$.

Example 4: Acceleration

A motorcycle increases its velocity from $5 \ \text{m/s}$ to $17 \ \text{m/s}$ in $4 \ \text{s}$. Find its acceleration.

Use:

$$ a = \frac{v-u}{t} $$

Then substitute:

$$ \begin{aligned} a &= \frac{17 \ \text{m/s} - 5 \ \text{m/s}}{4 \ \text{s}} \\ &= \frac{12 \ \text{m/s}}{4 \ \text{s}} \\ &= 3 \ \text{m/s}^2 \end{aligned} $$

The acceleration is $3 \ \text{m/s}^2$.

Example 5: Gradient Of A Distance-Time Graph

A straight section of a distance-time graph passes through $(0 \ \text{s}, 0 \ \text{m})$ and $(8 \ \text{s}, 24 \ \text{m})$. Find the speed represented by this section.

For a distance-time graph:

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

Therefore:

$$ \begin{aligned} \text{speed} &= \frac{24 \ \text{m} - 0 \ \text{m}}{8 \ \text{s} - 0 \ \text{s}} \\ &= \frac{24 \ \text{m}}{8 \ \text{s}} \\ &= 3 \ \text{m/s} \end{aligned} $$

The graph section represents a constant speed of $3 \ \text{m/s}$.

Example 6: Area Under A Velocity-Time Graph

An object moves with a constant velocity of $6 \ \text{m/s}$ for $5 \ \text{s}$. Find the displacement from the velocity-time graph idea.

The area under the velocity-time graph is a rectangle:

$$ \begin{aligned} \text{displacement} &= \text{velocity} \times \text{time} \\ &= 6 \ \text{m/s} \times 5 \ \text{s} \\ &= 30 \ \text{m} \end{aligned} $$

The displacement is $30 \ \text{m}$ in the direction of the positive velocity.

Common Mistakes

• Confusing distance and displacement. Distance is total path length; displacement is change in position with direction.
• Giving velocity without direction when direction is needed.
• Treating speed and acceleration as the same idea. Speed tells how fast distance changes; acceleration tells how fast velocity changes.
• Forgetting SI units in final answers.
• Using kilometres with seconds or metres with hours without converting units.
• Reading the height of a distance-time graph as speed. The height is distance; the gradient is speed.
• Reading the height of a velocity-time graph as acceleration. The height is velocity; the gradient is acceleration.
• Saying that zero velocity always means zero acceleration. An object can be momentarily at rest while its velocity is about to change, but this needs careful interpretation beyond the simplest cases.
• Assuming negative acceleration always means "moving backwards". It means acceleration is in the negative direction relative to the chosen positive direction.

Practice Tasks

1. Define distance, displacement, speed, velocity, and acceleration.
2. State the SI unit of distance, time, speed, velocity, and acceleration.
3. Classify each quantity as scalar or vector: distance, displacement, speed, velocity, acceleration, time.
4. A runner covers $80 \ \text{m}$ in $10 \ \text{s}$. Find the average speed.
5. A bicycle moves $120 \ \text{m}$ east in $20 \ \text{s}$. Find the average velocity.
6. A car travels $300 \ \text{m}$ in $15 \ \text{s}$. Find its average speed.
7. A student walks $25 \ \text{m}$ east and then $10 \ \text{m}$ west. Find the total distance and displacement.
8. A cart's velocity changes from $4 \ \text{m/s}$ to $16 \ \text{m/s}$ in $6 \ \text{s}$. Find its acceleration.
9. A ball moving along a straight line slows from $20 \ \text{m/s}$ to $8 \ \text{m/s}$ in $3 \ \text{s}$. Calculate the acceleration and explain the sign if the original direction is positive.
10. A distance-time graph has a straight section from $(2 \ \text{s}, 5 \ \text{m})$ to $(7 \ \text{s}, 20 \ \text{m})$. Find the speed.
11. A velocity-time graph changes from $2 \ \text{m/s}$ at $1 \ \text{s}$ to $14 \ \text{m/s}$ at $5 \ \text{s}$. Find the acceleration.
12. A velocity-time graph is horizontal at $9 \ \text{m/s}$ from $0 \ \text{s}$ to $6 \ \text{s}$. Find the displacement.
13. Explain why a person can have a total distance of $100 \ \text{m}$ but a displacement of $0 \ \text{m}$.
14. Draw a simple distance-time graph for an object that is stationary for $3 \ \text{s}$ and then moves at constant speed.
15. Draw a simple velocity-time graph for an object that starts from rest and accelerates uniformly.

Generated Question Layer

• Direct recall: definitions of distance, displacement, speed, velocity, acceleration, scalar, vector, and SI unit.
• Unit fluency: identifying and converting $\text{m}$, $\text{s}$, $\text{m/s}$, and $\text{m/s}^2$.
• Formula substitution: calculating speed, velocity, acceleration, distance, displacement, and time from simple data.
• Direction reasoning: comparing total distance with displacement in one-dimensional journeys.
• Graph interpretation: reading stationary motion, constant speed, changing speed, constant velocity, and uniform acceleration from graph shape.
• Graph calculation: finding speed from distance-time gradient, acceleration from velocity-time gradient, and displacement from velocity-time area.
• Misconception checks: distinguishing graph height, graph gradient, and graph area.

Learner Aid Opportunities

• diagram: Add straight-line journey sketches showing starting point, final point, distance, and displacement.
• chart: Add a comparison table for distance/displacement, speed/velocity, and scalar/vector.
• graph: Add labelled distance-time and velocity-time graph examples with gradients and areas marked.
• animation: Show an object moving while distance-time and velocity-time graphs update.
• interactive: Let learners change speed or acceleration and observe graph shape changes.
• LLM tutor: Use adaptive prompts to check whether a learner is confusing distance with displacement or graph height with gradient.

Exam-Derived Signals

• No past-paper or examination-format mappings have been reviewed for this Physics topic yet.
• The 2022 CSEE examination format may provide future assessment signals, but it does not replace the official 2023 syllabus placement or define the scope used here.

Source And Review Notes

• Official syllabus status: extracted from the 2023 Physics syllabus as the Form I topic Linear motion with scope covering speed, velocity, acceleration, distance, and displacement.
• Curriculum authority: the official syllabus and existing structured curriculum record define topic identity, form, competence, and hub placement.
• Learner expansion status: original explanatory notes written from the official topic summary and existing repo context.
• Exam signal status: not mapped or reviewed in this milestone.
• External enrichment status: not used.
• Textbook status: not used.
• Review risk: worked examples and graph explanations should be checked by a Physics reviewer for Form I level fit and local classroom vocabulary.