Linear Motion

PHYSICS • Form 4 • Chapter 2: Force and Motion I

2.1 Linear Motion

Linear motion is motion in a straight line. Learn how distance, displacement, speed, velocity, and acceleration describe how objects move — whether at rest, at constant speed, or changing speed.

Learning Objectives

Define distance, displacement, speed, velocity, and acceleration.
Differentiate between scalar (distance, speed) and vector (displacement, velocity) quantities.
Calculate distance, displacement, speed, velocity, and acceleration using formulas.
Use ticker timer and photogate systems to investigate linear motion.
Solve problems involving linear motion using the four linear motion equations.

Distance vs Displacement in Linear Motion

See how distance follows the route while displacement is the shortest straight-line path with direction.

Short Explanation

Distance and Displacement

Distance is the length of the route covered by an object (scalar). Displacement is the shortest straight-line distance between initial and final positions, with direction (vector).

Speed and Velocity

Speed is the rate of distance traveled (scalar). Velocity is the rate of displacement (vector, includes direction).

Acceleration

Acceleration is the rate of change of velocity. A negative acceleration means the object is decelerating.

Measurement Tools

A ticker timer works at 50 Hz (1 tick = 0.02 s). A photogate system is more accurate (measures time to 0.001 s, no friction from tape).

Key Formulas

\(\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \quad v = \dfrac{d}{t}\) \(\text{Velocity} = \dfrac{\text{Displacement}}{\text{Time}} \quad v = \dfrac{s}{t}\) \(\text{Acceleration} = \dfrac{v - u}{t} \quad a = \dfrac{v - u}{t}\)

where \(v\) = final velocity, \(u\) = initial velocity, \(t\) = time, \(s\) = displacement, \(d\) = distance

Distance	Displacement
Length of route covered	Shortest distance between initial and final positions (with direction)
Scalar quantity	Vector quantity
Magnitude depends on route	Magnitude is straight-line distance

Worked Example 1: Radzi’s Run

Radzi runs 100 m to the right, then 20 m back to the left, in 20 s.

Distance: \(100\,\text{m} + 20\,\text{m} = 120\,\text{m}\)

Displacement: \(100\,\text{m} + (-20\,\text{m}) = 80\,\text{m}\) (right)

Speed: \(\dfrac{120\,\text{m}}{20\,\text{s}} = 6\,\text{m/s}\)

Velocity: \(\dfrac{80\,\text{m}}{20\,\text{s}} = 4\,\text{m/s}\) (right)

Worked Example 2: Plane Deceleration

A plane slows from \(u = 75\,\text{m/s}\) to \(v = 5\,\text{m/s}\) in \(t = 20\,\text{s}\).

Acceleration: \(a = \dfrac{5 - 75}{20} = -3.5\,\text{m/s}^2\) (negative = deceleration)

Worked Example 3: School Bus Acceleration

A bus starts from rest (\(u = 0\,\text{m/s}\)), accelerates at \(a = 2\,\text{m/s}^2\) for \(t = 5\,\text{s}\).

Final velocity: \(v = 0 + (2)(5) = 10\,\text{m/s}\)

Linear Motion Equations (Uniform Acceleration)

\(v = u + at\) Equation 1

\(s = \tfrac{1}{2}(u + v)t\) Equation 2

\(s = ut + \tfrac{1}{2}at^2\) Equation 3

\(v^2 = u^2 + 2as\) Equation 4

Worked Example 4: Sports Car Displacement

A car accelerates from \(u = 40\,\text{m/s}\) to \(v = 50\,\text{m/s}\) in \(t = 3\,\text{s}\).

Displacement: \(s = \tfrac{1}{2}(40 + 50)(3) = 135\,\text{m}\)

Worked Example 5: Athlete Acceleration

An athlete starts from rest (\(u = 0\,\text{m/s}\)), runs \(s = 40\,\text{m}\) in \(t = 8.0\,\text{s}\).

Acceleration: \(40 = 0 + \tfrac{1}{2}a(8)^2 \Rightarrow a = 1.25\,\text{m/s}^2\)

Try to Answer First

Answer in your mind, then press “Check Answer”.

Explain the difference between distance and displacement.

Check Answer

Answer: Distance is the length of the route covered (scalar), while displacement is the shortest straight-line distance between initial and final positions (with direction, vector).

A car moving at 30 m/s stops in 5 s. What is its acceleration?

Check Answer

Answer: \(a = \dfrac{v - u}{t}\), so \(a = \dfrac{0 - 30}{5} = -6\ \mathrm{m/s^2}\). The negative sign means deceleration.

A transporter accelerates from 1 m/s to 5 m/s in 0.5 min (30 s). Calculate its displacement.

Check Answer

Answer: First, find acceleration: \(a = \dfrac{5 - 1}{30} = 0.133\ \mathrm{m/s^2}\). Then, \(s = (1)(30) + \tfrac{1}{2}(0.133)(30)^2 = 90\ \mathrm{m}\).

Common Mistakes

!Confusing distance (route length) with displacement (shortest path with direction).
!Mixing up speed (scalar) and velocity (vector, includes direction).
!Forgetting to include direction in vector quantities such as velocity and displacement.
!Sign errors in acceleration — not recognizing negative acceleration as deceleration.

Concept Misunderstandings

Misunderstanding

Deceleration is a separate physical quantity from acceleration.

Correct Concept

Deceleration is not a separate quantity. It is acceleration in the opposite direction of motion (negative acceleration).

Misunderstanding

An object moving at constant speed always has acceleration.

Correct Concept

Constant velocity (constant speed in a straight line) means zero acceleration. There is no change in velocity.

Misunderstanding

A ticker timer at 50 Hz means 1 tick = 1 s.

Correct Concept

50 Hz means 50 ticks per second, so 1 tick = 0.02 s. This is used to calculate time intervals on ticker tape.

Summary

Linear motion is motion in a straight line, described by distance, displacement, speed, velocity, and acceleration.
Distance (scalar) is the route length; displacement (vector) is the shortest straight-line distance with direction.
Speed (scalar) = distance ÷ time; velocity (vector) = displacement ÷ time.
Acceleration (vector) is the rate of change of velocity: \(a = \dfrac{v - u}{t}\).
A ticker timer operates at 50 Hz (1 tick = 0.02 s); a photogate system is more accurate (no friction from tape).
The four linear motion equations relate \(u\), \(v\), \(a\), \(t\), and \(s\): \(v = u + at\), \(s = \tfrac{1}{2}(u+v)t\), \(s = ut + \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\).

Short Activity

Fill in the blanks with the correct terms. Type your answers and press “Check Answer”.

Fill in the Blanks

1 Linear motion is motion in a line.

2 Distance is a quantity, while displacement is a quantity.

3 Speed is calculated as divided by time, and velocity is divided by time.

4 Acceleration is the rate of change of .

Keywords

Distance Displacement Speed Velocity Acceleration Ticker timer Photogate system Linear motion equations

Linear Motion

PHYSICS • Form 4 • Chapter 2: Force and Motion I

2.1 Linear Motion

Linear motion is motion in a straight line. Learn how distance, displacement, speed, velocity, and acceleration describe how objects move — whether at rest, at constant speed, or changing speed.

Learning Objectives

Define distance, displacement, speed, velocity, and acceleration.
Differentiate between scalar (distance, speed) and vector (displacement, velocity) quantities.
Calculate distance, displacement, speed, velocity, and acceleration using formulas.
Use ticker timer and photogate systems to investigate linear motion.
Solve problems involving linear motion using the four linear motion equations.

Distance vs Displacement in Linear Motion

See how distance follows the route while displacement is the shortest straight-line path with direction.

Short Explanation

Distance and Displacement

Distance is the length of the route covered by an object (scalar). Displacement is the shortest straight-line distance between initial and final positions, with direction (vector).

Speed and Velocity

Speed is the rate of distance traveled (scalar). Velocity is the rate of displacement (vector, includes direction).

Acceleration

Acceleration is the rate of change of velocity. A negative acceleration means the object is decelerating.

Measurement Tools

A ticker timer works at 50 Hz (1 tick = 0.02 s). A photogate system is more accurate (measures time to 0.001 s, no friction from tape).

Key Formulas

where \(v\) = final velocity, \(u\) = initial velocity, \(t\) = time, \(s\) = displacement, \(d\) = distance

Distance	Displacement
Length of route covered	Shortest distance between initial and final positions (with direction)
Scalar quantity	Vector quantity
Magnitude depends on route	Magnitude is straight-line distance

Worked Example 1: Radzi’s Run

Radzi runs 100 m to the right, then 20 m back to the left, in 20 s.

Distance: \(100\,\text{m} + 20\,\text{m} = 120\,\text{m}\)

Displacement: \(100\,\text{m} + (-20\,\text{m}) = 80\,\text{m}\) (right)

Speed: \(\dfrac{120\,\text{m}}{20\,\text{s}} = 6\,\text{m/s}\)

Velocity: \(\dfrac{80\,\text{m}}{20\,\text{s}} = 4\,\text{m/s}\) (right)

Worked Example 2: Plane Deceleration

A plane slows from \(u = 75\,\text{m/s}\) to \(v = 5\,\text{m/s}\) in \(t = 20\,\text{s}\).

Acceleration: \(a = \dfrac{5 - 75}{20} = -3.5\,\text{m/s}^2\) (negative = deceleration)

Worked Example 3: School Bus Acceleration

A bus starts from rest (\(u = 0\,\text{m/s}\)), accelerates at \(a = 2\,\text{m/s}^2\) for \(t = 5\,\text{s}\).

Final velocity: \(v = 0 + (2)(5) = 10\,\text{m/s}\)

Linear Motion Equations (Uniform Acceleration)

\(v = u + at\) Equation 1

\(s = \tfrac{1}{2}(u + v)t\) Equation 2

\(s = ut + \tfrac{1}{2}at^2\) Equation 3

\(v^2 = u^2 + 2as\) Equation 4

Worked Example 4: Sports Car Displacement

A car accelerates from \(u = 40\,\text{m/s}\) to \(v = 50\,\text{m/s}\) in \(t = 3\,\text{s}\).

Displacement: \(s = \tfrac{1}{2}(40 + 50)(3) = 135\,\text{m}\)

Worked Example 5: Athlete Acceleration

An athlete starts from rest (\(u = 0\,\text{m/s}\)), runs \(s = 40\,\text{m}\) in \(t = 8.0\,\text{s}\).

Acceleration: \(40 = 0 + \tfrac{1}{2}a(8)^2 \Rightarrow a = 1.25\,\text{m/s}^2\)

Try to Answer First

Answer in your mind, then press “Check Answer”.

Explain the difference between distance and displacement.

Check Answer

Answer: Distance is the length of the route covered (scalar), while displacement is the shortest straight-line distance between initial and final positions (with direction, vector).

A car moving at 30 m/s stops in 5 s. What is its acceleration?

Check Answer

Answer: \(a = \dfrac{v - u}{t}\), so \(a = \dfrac{0 - 30}{5} = -6\ \mathrm{m/s^2}\). The negative sign means deceleration.

A transporter accelerates from 1 m/s to 5 m/s in 0.5 min (30 s). Calculate its displacement.

Check Answer

Answer: First, find acceleration: \(a = \dfrac{5 - 1}{30} = 0.133\ \mathrm{m/s^2}\). Then, \(s = (1)(30) + \tfrac{1}{2}(0.133)(30)^2 = 90\ \mathrm{m}\).

Common Mistakes

!Confusing distance (route length) with displacement (shortest path with direction).
!Mixing up speed (scalar) and velocity (vector, includes direction).
!Forgetting to include direction in vector quantities such as velocity and displacement.
!Sign errors in acceleration — not recognizing negative acceleration as deceleration.

Concept Misunderstandings

Misunderstanding

Deceleration is a separate physical quantity from acceleration.

Correct Concept

Deceleration is not a separate quantity. It is acceleration in the opposite direction of motion (negative acceleration).

Misunderstanding

An object moving at constant speed always has acceleration.

Correct Concept

Constant velocity (constant speed in a straight line) means zero acceleration. There is no change in velocity.

Misunderstanding

A ticker timer at 50 Hz means 1 tick = 1 s.

Correct Concept

50 Hz means 50 ticks per second, so 1 tick = 0.02 s. This is used to calculate time intervals on ticker tape.

Summary

Linear motion is motion in a straight line, described by distance, displacement, speed, velocity, and acceleration.
Distance (scalar) is the route length; displacement (vector) is the shortest straight-line distance with direction.
Speed (scalar) = distance ÷ time; velocity (vector) = displacement ÷ time.
Acceleration (vector) is the rate of change of velocity: \(a = \dfrac{v - u}{t}\).
A ticker timer operates at 50 Hz (1 tick = 0.02 s); a photogate system is more accurate (no friction from tape).
The four linear motion equations relate \(u\), \(v\), \(a\), \(t\), and \(s\): \(v = u + at\), \(s = \tfrac{1}{2}(u+v)t\), \(s = ut + \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\).

Short Activity

Fill in the blanks with the correct terms. Type your answers and press “Check Answer”.

Fill in the Blanks

1 Linear motion is motion in a line.

2 Distance is a quantity, while displacement is a quantity.

3 Speed is calculated as divided by time, and velocity is divided by time.

4 Acceleration is the rate of change of .

Keywords

Distance Displacement Speed Velocity Acceleration Ticker timer Photogate system Linear motion equations

Linear Motion

2.1 Linear Motion

Learning Objectives

Distance vs Displacement in Linear Motion

Short Explanation

Distance and Displacement

Speed and Velocity

Acceleration

Measurement Tools

Worked Example 1: Radzi’s Run

Worked Example 2: Plane Deceleration

Worked Example 3: School Bus Acceleration

Linear Motion Equations (Uniform Acceleration)

Worked Example 4: Sports Car Displacement

Worked Example 5: Athlete Acceleration

Try to Answer First

Common Mistakes

Concept Misunderstandings

Summary

Short Activity

Fill in the Blanks

Keywords

Linear Motion

2.1 Linear Motion

Learning Objectives

Distance vs Displacement in Linear Motion

Short Explanation

Distance and Displacement

Speed and Velocity

Acceleration

Measurement Tools

Worked Example 1: Radzi’s Run

Worked Example 2: Plane Deceleration

Worked Example 3: School Bus Acceleration

Linear Motion Equations (Uniform Acceleration)

Worked Example 4: Sports Car Displacement

Worked Example 5: Athlete Acceleration

Try to Answer First

Common Mistakes

Concept Misunderstandings

Summary

Short Activity

Fill in the Blanks

Keywords

Chapter : Force and Motion I

Topic : Linear Motion

Related notes