Mechanics (Paper 4): Kinematics of Motion in a Straight Line

Hello! Welcome to Kinematics! This chapter is the absolute foundation of Mechanics. It’s all about describing how objects move—how far they travel, how fast they go, and how quickly they change speed. We are dealing only with motion in a single straight line (one dimension). Mastering these concepts is crucial because they appear frequently in Paper 4, often combined with forces and Newton's Laws later on.

Don't worry if you sometimes confuse the terms. We will break down the precise definitions and provide clear tools (graphs, formulas, and calculus) to handle any problem!

1. Defining Motion: Scalars vs. Vectors

In mechanics, we must distinguish between quantities that only have magnitude (scalars) and quantities that have both magnitude and direction (vectors). Since we are moving in a straight line, direction is simply positive (forward) or negative (backward).

Key Definitions: Scalars (Magnitude Only)
  • Distance: The total length of the path travelled.
    Example: If you walk 5m forward and then 2m backward, the distance is 7m.
  • Speed: The rate of change of distance (how fast you are moving, regardless of direction).
Key Definitions: Vectors (Magnitude and Direction)
  • Displacement (\(s\)): The shortest distance from the starting point (origin) to the final position.
    Example: If you walk 5m forward and then 2m backward, your displacement is +3m.
  • Velocity (\(v\)): The rate of change of displacement. It tells you both the speed AND the direction.
    (A velocity of -5 m/s means you are moving at a speed of 5 m/s in the negative direction.)
  • Acceleration (\(a\)): The rate of change of velocity. It measures how quickly the velocity vector is changing.
    • A positive acceleration means velocity is increasing (or becoming less negative).
    • The term Deceleration means the object's speed is decreasing, which occurs when the velocity and acceleration vectors point in opposite directions.
Quick Review: Sign Convention is Vital!

You must choose one direction as positive (e.g., up or right) and stick to it throughout the problem. Displacement, velocity, and acceleration in the opposite direction must be entered as negative values.

2. Graphical Interpretation of Motion

Graphs are powerful tools for visualizing motion. The syllabus focuses on two main types:

2.1. Displacement-Time (\(s\)-\(t\)) Graphs

The graph plots the object's position (\(s\)) against time (\(t\)).

  • Gradient: The gradient of the \(s\)-\(t\) graph represents the velocity.
    • A horizontal line (zero gradient) means zero velocity (the object is stationary).
    • A constant positive gradient means constant positive velocity.
    • A curved line means variable velocity (the object is accelerating or decelerating).

Analogy: Think of the slope of a hill. A steep uphill slope means you are moving fast (high positive velocity). A steep downhill slope means you are moving fast in the negative direction (high negative velocity).

2.2. Velocity-Time (\(v\)-\(t\)) Graphs

These graphs are arguably the most important in kinematics.

  • Gradient: The gradient of the \(v\)-\(t\) graph represents the acceleration (\(a\)).
    • A constant, non-zero gradient means constant acceleration (this is where SUVAT equations apply).
    • A zero gradient means constant velocity.
  • Area: The area under the \(v\)-\(t\) graph represents the displacement (\(s\)).
    • Areas above the \(t\)-axis are positive displacements.
    • Areas below the \(t\)-axis are negative displacements.
    • To find the total distance travelled, you must calculate the absolute area (ignore the negative sign for areas below the axis) and sum them up.

3. Motion with Constant Acceleration (SUVAT)

The most common type of kinematics problem involves constant acceleration (such as motion under gravity). For these problems, we use the five standard SUVAT equations.

We rely on five variables:

  • \(s\): Displacement (m)
  • \(u\): Initial velocity (m/s)
  • \(v\): Final velocity (m/s)
  • \(a\): Constant acceleration (\(m/s^2\))
  • \(t\): Time (s)
The Five SUVAT Equations

To solve a SUVAT problem, you usually need to know three of these variables to find the fourth.

  1. \(v = u + at\) (Missing: s)
  2. \(s = \frac{1}{2}(u + v)t\) (Missing: a)
  3. \(s = ut + \frac{1}{2}at^2\) (Missing: v)
  4. \(s = vt - \frac{1}{2}at^2\) (Missing: u)
  5. \(v^2 = u^2 + 2as\) (Missing: t)
Tip for Struggling Students: Choosing the Right Formula

List the five letters (S, U, V, A, T). Identify the three values you are given and the one value you need to find. The variable that is missing/irrelevant tells you exactly which equation to use.

Example: If a problem doesn't mention the acceleration (\(a\)) and asks for displacement (\(s\)), use formula 2.

Real-World Application: Motion under Gravity

In problems where a particle is moving freely vertically (like throwing a ball up or dropping an object), the acceleration is constant:
\(a = g\) (The acceleration due to gravity).
The syllabus expects you to use the approximate numerical value: \(g = 10 \text{ m/s}^2\).

Crucial Convention for Gravity: Decide if 'Up' or 'Down' is positive.
If you choose Up (+): then \(a = -10 \text{ m/s}^2\).
If you choose Down (+): then \(a = +10 \text{ m/s}^2\).

4. Motion with Variable Acceleration (Calculus)

When the acceleration is not constant, it means \(a\) is usually expressed as a function of time (\(t\)). In this case, you must use techniques from Pure Mathematics 1 (Differentiation and Integration) to solve the problem.

4.1. The Chain of Differentiation

Differentiation tells us the instantaneous rate of change.

  • Displacement (\(s\)) to Velocity (\(v\)):

    \(v = \frac{ds}{dt}\)

  • Velocity (\(v\)) to Acceleration (\(a\)):

    \(a = \frac{dv}{dt}\) or \(a = \frac{d^2s}{dt^2}\)

Did you know? If \(s\) is a function of \(t\), you are simply differentiating that function with respect to \(t\). This is why prior knowledge of differentiation (P1) is assumed!

4.2. The Chain of Integration

Integration is the reverse process of differentiation and is used to find total change or displacement.

  • Acceleration (\(a\)) to Velocity (\(v\)):

    \(v = \int a \ dt\)

  • Velocity (\(v\)) to Displacement (\(s\)):

    \(s = \int v \ dt\)

Common Mistake: The Constant of Integration (\(C\))

When you integrate, you must always include the constant of integration, \(C\). You determine the value of \(C\) by using the initial conditions (the values of \(s\) or \(v\) when \(t = 0\)).

Example: If \(v = 2t + C\), and you know that the initial velocity is \(5 \text{ m/s}\) (i.e., \(v=5\) when \(t=0\)), then \(5 = 2(0) + C\), so \(C=5\).

5. Connecting Different Stages of Motion

Many complex kinematics questions involve motion broken into several stages (e.g., accelerating, then moving at constant velocity, then decelerating).

The key to solving these is recognizing what quantities are continuous between stages:

  • The final velocity (\(v\)) of Stage 1 becomes the initial velocity (\(u\)) of Stage 2.
  • The displacement (\(s\)) is cumulative (if you start measuring displacement from the beginning).

You solve each stage separately using SUVAT or calculus, then link them using the velocity or time at the transition points.

Chapter Key Takeaways

1. Vectors vs. Scalars: Use positive/negative signs consistently for displacement, velocity, and acceleration.

2. Graphs: The gradient relates \(s \to v \to a\). The area under the \(v\)-\(t\) graph is displacement.

3. Constant Acceleration: Use the five SUVAT formulas. Select the formula that excludes the unknown or irrelevant variable.

4. Variable Acceleration: Use calculus. Differentiate to move from \(s \to v \to a\). Integrate to move from \(a \to v \to s\), remembering to find the constant of integration, \(C\).