Linear motion

Q: What is the key difference between distance and displacement?

Distance is a scalar quantity representing the total path length traveled, whereas displacement is a vector quantity representing the straight-line distance in a specified direction from a reference point.

Q: How are average velocity and instantaneous velocity defined?

Average velocity is total displacement divided by total time (\(v_{\text{avg}} = \frac{\Delta s}{\Delta t}\)), whereas instantaneous velocity is the rate of change of displacement at a specific single instant in time.

Q: What is acceleration and what are its standard units?

Acceleration is the rate of change of velocity with respect to time (\(a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}\)), and its unit is \(\text{ms}^{-2}\).

Q: What physical quantity does the gradient of a displacement–time (\(s\text{--}t\)) graph represent?

The gradient of a displacement–time graph represents instantaneous velocity (\(v = \frac{\Delta s}{\Delta t}\)).

Q: What do the gradient and the area under the line on a velocity–time (\(v\text{--}t\)) graph represent?

The gradient represents acceleration (\(a = \frac{\Delta v}{\Delta t}\)), and the area between the line and the time axis represents displacement (\(s\)).

Q: What does the area between the line and the time axis on an acceleration–time (\(a\text{--}t\)) graph represent?

The area represents the change in velocity (\(\Delta v\)) of the object over that time interval.

Q: Under what condition can the standard \(suvat\) kinematic equations be applied?

The \(suvat\) equations can only be applied when acceleration is uniform (constant) along a straight line.

Q: How is the equation \(s = ut + \frac{1}{2}at^2\) derived from first principles?

Substitute the velocity definition \(v = u + at\) into the average velocity displacement formula \(s = \frac{1}{2}(u + v)t\) and expand the brackets.

Q: How is the equation \(v^2 = u^2 + 2as\) derived from first principles?

Rearrange \(v = u + at\) for time \(t = \frac{v - u}{a}\) and substitute this into \(s = \frac{1}{2}(u + v)t\), then simplify using the difference of two squares.

Q: What is the acceleration of a vertically projected object at the highest point of its flight?

The acceleration is \(9.81\text{ ms}^{-2}\) downwards (the instantaneous velocity is zero, but acceleration remains constant).

Physics 1210 · CCEA A Level · 8 min read · Updated: 31 Aug 2026

Welcome to Linear Motion

Welcome to one of the most fundamental topics in AS Unit 1: Forces, Energy and Electricity! Kinematics and linear motion describe how objects move in a straight line. Whether you are analyzing a sprinter racing toward a finish line or an apple falling from a tree, the exact same physical rules apply.

Don't worry if this topic seems daunting or heavily mathematical at first. By breaking motion down into distinct measurable quantities, mastering graphs, and practicing structured problem-solving, you will build full confidence for your CCEA AS 1 exam.

1. Core Definitions and Fundamentals

Before jumping into calculations, we must be completely clear about our fundamental physical quantities. A common place where marks are lost in CCEA exams is confusing scalars (magnitude only) with vectors (magnitude and direction).

Displacement vs. Distance

• Distance: The total length of the path traveled by an object. It is a scalar quantity (unit: \(\text{m}\)).
• Displacement (\(s\)): The straight-line distance moved in a specified direction from a fixed reference point. It is a vector quantity (unit: \(\text{m}\)).

Analogy: If you walk \(400\text{ m}\) around a standard running track and finish exactly where you started, your total distance is \(400\text{ m}\), but your displacement is \(0\text{ m}\) because your final position has not changed relative to your start.

Speed and Velocity

• Average Speed: The total distance traveled divided by the total time taken (scalar quantity, unit: \(\text{ms}^{-1}\)).
• Average Velocity (\(v_{\text{avg}}\)): The total displacement divided by the total time taken (vector quantity, unit: \(\text{ms}^{-1}\)):
\(v_{\text{avg}} = \frac{\Delta s}{\Delta t}\)
• Instantaneous Velocity (\(v\)): The rate of change of displacement at a specific instant in time (unit: \(\text{ms}^{-1}\)).

Acceleration and Deceleration

• Acceleration (\(a\)): The rate of change of velocity with respect to time. It is a vector quantity (unit: \(\text{ms}^{-2}\)):
\(a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}\)
where \(u\) is initial velocity, \(v\) is final velocity, and \(t\) is time.
• Deceleration / Retardation: A reduction in speed over time, represented mathematically as a negative acceleration in the direction of motion.

Quick Review: Key Takeaway
Always specify or account for direction when stating displacement, velocity, or acceleration. Displacement is straight-line change in position, not total path length!

2. Graphical Analysis of Linear Motion

CCEA examiners love testing your ability to interpret and extract numerical data from motion graphs. Always remember the two key mathematical tools: gradient (slope) and area under the line.

Displacement–Time Graphs (\(s\text{--}t\))

• Gradient: Represents instantaneous velocity (\(\frac{\Delta s}{\Delta t}\)).
• Straight line: Constant velocity (zero acceleration).
• Curved line: Changing velocity (the object is accelerating or decelerating).
• Horizontal line: The object is stationary (\(v = 0\text{ ms}^{-1}\)).

Examiner Tip: If an \(s\text{--}t\) graph is curved and you are asked for instantaneous velocity at a specific time, you must draw a tangent to the curve at that exact point and calculate its gradient (\(\frac{\Delta y}{\Delta x}\)). Never just divide the \(y\)-value by the \(x\)-value of that point!

Velocity–Time Graphs (\(v\text{--}t\))

• Gradient: Represents acceleration (\(a = \frac{\Delta v}{\Delta t}\)).
• Area between the line and the time axis: Represents the change in displacement (\(s\)).
• Regions above the time axis: Represent positive displacement (moving forward).
• Regions below the time axis: Represent negative displacement (moving backward in the opposite direction).

Important Note: To find the total displacement from a \(v\text{--}t\) graph, subtract the area below the time axis from the area above the time axis. To find the total distance, add all the areas together regardless of whether they are above or below the axis.

Acceleration–Time Graphs (\(a\text{--}t\))

• Line shape: For uniformly accelerated motion at AS level, \(a\text{--}t\) graphs are horizontal straight lines (\(a = \text{constant}\)).
• Area between the line and the time axis: Represents the change in velocity (\(\Delta v\)).

Quick Summary Table:
• \(s\text{--}t\) graph \(\rightarrow\) Gradient = Velocity
• \(v\text{--}t\) graph \(\rightarrow\) Gradient = Acceleration | Area = Displacement
• \(a\text{--}t\) graph \(\rightarrow\) Area = Change in Velocity

3. The Equations of Uniformly Accelerated Motion (SUVAT)

When an object moves with constant (uniform) acceleration along a straight line, we can use the five \(suvat\) equations. Each symbol represents a specific physical quantity:

• \(s\): Displacement (\(\text{m}\))
• \(u\): Initial velocity (\(\text{ms}^{-1}\))
• \(v\): Final velocity (\(\text{ms}^{-1}\))
• \(a\): Acceleration (\(\text{ms}^{-2}\))
• \(t\): Time taken (\(\text{s}\))

The Standard Equations

1. \(v = u + at\)
2. \(s = \frac{1}{2}(u + v)t\)
3. \(s = ut + \frac{1}{2}at^2\)
4. \(v^2 = u^2 + 2as\)
5. \(s = vt - \frac{1}{2}at^2\)

Crucial Rule: These equations apply only when acceleration is uniform (constant). Never apply \(suvat\) formulas to situations where acceleration changes, such as an object reaching terminal velocity in a fluid!

Derivations Required for CCEA Examinations

CCEA exam papers frequently ask candidates to derive these equations step-by-step from first principles.

Derivation 1: Deriving \(v = u + at\)
Start with the fundamental definition of acceleration:
\(a = \frac{v - u}{t}\)
Multiply both sides by \(t\):
\(at = v - u\)
Rearrange to solve for \(v\):
\(v = u + at\)

Derivation 2: Deriving \(s = ut + \frac{1}{2}at^2\)
Start with the average velocity displacement formula:
\(s = \frac{1}{2}(u + v)t\)
Substitute \(v = u + at\) into the equation:
\(s = \frac{1}{2}(u + [u + at])t\)
Combine terms inside the brackets:
\(s = \frac{1}{2}(2u + at)t\)
Expand the brackets:
\(s = ut + \frac{1}{2}at^2\)

Derivation 3: Deriving \(v^2 = u^2 + 2as\)
Rearrange \(v = u + at\) to express time \(t\) in terms of velocities and acceleration:
\(t = \frac{v - u}{a}\)
Substitute this expression for \(t\) into \(s = \frac{1}{2}(u + v)t\):
\(s = \left(\frac{u + v}{2}\right)\left(\frac{v - u}{a}\right)\)
Multiply the numerators (difference of two squares) and denominators:
\(s = \frac{v^2 - u^2}{2a}\)
Rearrange to solve for \(v^2\):
\(2as = v^2 - u^2\)
\(v^2 = u^2 + 2as\)

Key Takeaway:
To solve any \(suvat\) calculation: list the 5 variables, write down the 3 you know, identify the 1 you need, and pick the single equation that contains those 4 variables.

4. Motion Under Gravity (Free Fall)

When an object moves vertically solely under the influence of gravity (ignoring air resistance), it undergoes uniform acceleration.

• Acceleration due to gravity: \(g = 9.81\text{ ms}^{-2}\) downwards (the standard value used in CCEA AS Level Physics).

Setting Up a Consistent Sign Convention

Because displacement, velocity, and acceleration are vectors, you must choose a positive direction before starting a vertical motion calculation:

If you choose UPWARDS as POSITIVE (+):
• Initial upward velocity is positive: \(u > 0\)
• Acceleration is directed downward, so \(a = -g = -9.81\text{ ms}^{-2}\)
• Displacements above the launch level are positive (\(s > 0\)); displacements below the launch level are negative (\(s < 0\)).

Vertical Projection (Throwing an Object Upward)

When an object is thrown vertically upwards:
1. At the maximum height (apex): The instantaneous vertical velocity is zero (\(v = 0\text{ ms}^{-1}\)).
2. Acceleration at the apex: The acceleration is not zero; it remains \(9.81\text{ ms}^{-2}\) downwards throughout the entire flight.
3. Symmetry of motion: If the object returns to the same height from which it was launched, the time taken to rise equals the time taken to fall, and the landing speed equals the launch speed (in the opposite direction).

Experimental Determination of \(g\)

CCEA practical and written units frequently assess experiments used to measure \(g\):

1. Electromagnet and Trapdoor / Drop Sensor:
• A small steel ball bearing is held by an electromagnet.
• Switching off the current releases the ball and simultaneously starts an electronic timer.
• The ball drops through a measured height \(h\) and strikes a pressure plate/trapdoor, which stops the timer.
• Using \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(s = h\), and \(a = g\):
\(h = \frac{1}{2}gt^2\)
• By measuring \(t\) for several heights \(h\), a graph of \(h\) against \(t^2\) is plotted. The gradient equals \(\frac{1}{2}g\), so \(g = 2 \times \text{gradient}\).

2. Light Gates and Interrupt Card:
• A card of known length is dropped through a single light gate connected to a data logger to measure instantaneous velocity, or dropped through two light gates separated by a known vertical distance.
• Using \(v^2 = u^2 + 2gs\), \(g\) is calculated from the measured velocities and distance.

3. Video Analysis:
• High-speed digital video recorded alongside a calibrated vertical scale captures position against time frames to generate accurate \(s\text{--}t\) and \(v\text{--}t\) plots.

Key Takeaway:
In free fall near Earth's surface, all objects accelerate downward at \(9.81\text{ ms}^{-2}\) regardless of mass. Assign your sign convention at the start and maintain it rigorously throughout the problem.

5. Common Pitfalls and Examiner Tips

Avoid these common mistakes highlighted in CCEA Chief Examiner reports:

• Mixing up Distance and Displacement: Displacement is a vector. Always define it as straight-line distance in a specified direction from a start point.
• Sign Errors in Projectile / Gravity Calculations: If you throw an object up with \(u = +15\text{ ms}^{-1}\), you must use \(a = -9.81\text{ ms}^{-2}\). Making both positive means the object accelerates upward into space!
• Incorrect Graph Evaluation: On an \(s\text{--}t\) curve, do not divide coordinates \(\frac{y}{x}\) to find velocity. Always construct a large tangent triangle and calculate \(\frac{\Delta y}{\Delta x}\).
• Forgetting Negative Areas: When calculating total displacement on a \(v\text{--}t\) graph, subtract the area under the time axis from the area above it.
• Premature Rounding: Keep intermediate values in your calculator memory and round to appropriate significant figures (typically 2 or 3 s.f. matching the question data) only at the final answer.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the key difference between distance and displacement?

Distance is a scalar quantity representing the total path length traveled, whereas displacement is a vector quantity representing the straight-line distance in a specified direction from a reference point.

How are average velocity and instantaneous velocity defined?

Average velocity is total displacement divided by total time (\(v_{\text{avg}} = \frac{\Delta s}{\Delta t}\)), whereas instantaneous velocity is the rate of change of displacement at a specific single instant in time.

What is acceleration and what are its standard units?

Acceleration is the rate of change of velocity with respect to time (\(a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}\)), and its unit is \(\text{ms}^{-2}\).

What physical quantity does the gradient of a displacement–time (\(s\text{--}t\)) graph represent?

The gradient of a displacement–time graph represents instantaneous velocity (\(v = \frac{\Delta s}{\Delta t}\)).

What do the gradient and the area under the line on a velocity–time (\(v\text{--}t\)) graph represent?

The gradient represents acceleration (\(a = \frac{\Delta v}{\Delta t}\)), and the area between the line and the time axis represents displacement (\(s\)).

What does the area between the line and the time axis on an acceleration–time (\(a\text{--}t\)) graph represent?

The area represents the change in velocity (\(\Delta v\)) of the object over that time interval.

Under what condition can the standard \(suvat\) kinematic equations be applied?

The \(suvat\) equations can only be applied when acceleration is uniform (constant) along a straight line.

How is the equation \(s = ut + \frac{1}{2}at^2\) derived from first principles?

Substitute the velocity definition \(v = u + at\) into the average velocity displacement formula \(s = \frac{1}{2}(u + v)t\) and expand the brackets.

How is the equation \(v^2 = u^2 + 2as\) derived from first principles?

Rearrange \(v = u + at\) for time \(t = \frac{v - u}{a}\) and substitute this into \(s = \frac{1}{2}(u + v)t\), then simplify using the difference of two squares.

What is the acceleration of a vertically projected object at the highest point of its flight?

The acceleration is \(9.81\text{ ms}^{-2}\) downwards (the instantaneous velocity is zero, but acceleration remains constant).

How can the acceleration of free fall \(g\) be determined using an electromagnet and trapdoor experiment?

By measuring the fall time \(t\) over various heights \(h\) and plotting a graph of \(h\) against \(t^2\); the gradient equals \(\frac{1}{2}g\), so \(g = 2 \times \text{gradient}\).

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