Further Mathematics | Kinematics of a particle moving in a straight line or plane

Welcome to Kinematics: The Study of Motion!

Hello there! If you are tackling Unit M2, you've already mastered the basics of motion (like SUVAT). In Further Maths, we make things much more realistic by introducing variable acceleration and motion in two dimensions (the plane).

This chapter is fundamental because it connects calculus (differentiation and integration) directly to physics. Don't worry if this seems tricky at first; we will break down the concepts step-by-step. By the end of this unit, you will be able to describe exactly where a particle is, how fast it is moving, and where it is headed, even if its acceleration is constantly changing!

Part 1: Motion in a Straight Line (Variable Acceleration)

1.1 The Language of 1D Kinematics

When a particle moves in a straight line, we describe its position relative to a fixed origin (\(O\)). We use three key quantities, all of which are functions of time \(t\):

• Displacement (\(s\)): The particle’s position relative to the origin. (Units: meters, m)
• Velocity (\(v\)): The rate of change of displacement. (Units: m s\(^{-1}\))
• Acceleration (\(a\)): The rate of change of velocity. (Units: m s\(^{-2}\))

1.2 Going Forward: Differentiation

When acceleration is variable (i.e., it depends on time), we use differentiation to find the instantaneous rate of change. Think of differentiation as finding the gradient of a curve at a specific point.

The Differentiation Relationships

To move from displacement to acceleration, you differentiate with respect to time, \(t\).

• Velocity is the derivative of displacement: \[v = \frac{ds}{dt}\]
• Acceleration is the derivative of velocity: \[a = \frac{dv}{dt}\]
• Therefore, acceleration is the second derivative of displacement: \[a = \frac{d^2s}{dt^2}\]

Memory Aid: Differentiation goes DOWN the alphabet: S -> V -> A.

1.3 Going Backward: Integration

If you are given acceleration and need to find velocity or displacement, you must use integration. Integration is like summing up all the tiny changes over time.

The Integration Relationships

• Velocity is the integral of acceleration: \[v = \int a \, dt\]
• Displacement is the integral of velocity: \[s = \int v \, dt\]

CRITICAL STEP: When integrating, you must always include the constant of integration, \(C\).

To find the value of \(C\), you need the initial conditions (the position or velocity at a specific time, usually \(t=0\)).

Key Takeaway (Part 1 Summary)

Kinematics is about moving between \(s\), \(v\), and \(a\). Differentiation takes you one step down (S to V to A). Integration takes you one step up (A to V to S), but requires boundary conditions to find \(C\).

Part 2: Kinematics in a Plane (Vectors)

2.1 Introducing 2D Motion

In the real world, particles rarely move in a perfectly straight line. When motion happens across a surface (like a ball rolling on a field or a plane flying), we need vectors to describe the movement in two dimensions (usually represented by the \(\mathbf{i}\) and \(\mathbf{j}\) components).

The Vector Quantities

We now use bold letters or arrows over the symbol to denote vectors (though here we will stick to bold):

• Position Vector (\(\mathbf{r}\)): Defines the location of the particle relative to the origin. \[\mathbf{r} = x\mathbf{i} + y\mathbf{j}\]
• Velocity Vector (\(\mathbf{v}\)): Defines the rate of change of position. \[\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\]
• Acceleration Vector (\(\mathbf{a}\)): Defines the rate of change of velocity. \[\mathbf{a} = a_x\mathbf{i} + a_y\mathbf{j}\]

2.2 Vector Calculus: The Power of Independence

The greatest simplification in 2D kinematics is that the horizontal (\(\mathbf{i}\)) motion is completely independent of the vertical (\(\mathbf{j}\)) motion. You can treat the components separately!

Vector Differentiation

To find the velocity vector, you differentiate the position vector's components individually:

\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = \left(\frac{dx}{dt}\right)\mathbf{i} + \left(\frac{dy}{dt}\right)\mathbf{j}\]

Similarly, for acceleration:

\[\mathbf{a} = \frac{d\mathbf{v}}{dt} = \left(\frac{dv_x}{dt}\right)\mathbf{i} + \left(\frac{dv_y}{dt}\right)\mathbf{j}\]

Vector Integration

When integrating, remember you must introduce a constant vector, \(\mathbf{C}\), which has both \(\mathbf{i}\) and \(\mathbf{j}\) components: \[\mathbf{C} = C_1\mathbf{i} + C_2\mathbf{j}\]

You will need the initial conditions for both components to find \(C_1\) and \(C_2\).

2.3 Finding Magnitude and Direction

Once you have a vector (velocity \(\mathbf{v}\) or acceleration \(\mathbf{a}\)), you often need to find its speed (the magnitude of the velocity) and its direction.

Magnitude (Speed)

The magnitude of any vector \(\mathbf{A} = A_x\mathbf{i} + A_y\mathbf{j}\) is found using Pythagoras’ Theorem:

\[|\mathbf{A}| = \sqrt{A_x^2 + A_y^2}\]

The speed \(|\mathbf{v}|\) is the length of the velocity vector.

Direction

The direction is usually expressed as an angle (\(\theta\)) relative to the positive \(\mathbf{i}\) axis (the positive x-axis).

\[\tan \theta = \frac{\text{component in } \mathbf{j} \text{ direction}}{\text{component in } \mathbf{i} \text{ direction}} = \frac{A_y}{A_x}\]

Crucial Tip: Always sketch the vector components to ensure you place the angle in the correct quadrant (0-360 degrees). For example, if \(A_x\) is negative and \(A_y\) is positive, the vector is in the second quadrant.

Key Takeaway (Part 2 Summary)

In 2D kinematics, we use vectors. The crucial technique is separating the components. Treat the \(\mathbf{i}\) parts and the \(\mathbf{j}\) parts as two independent 1D problems, and then combine them or use Pythagoras to find the overall speed and direction.

Part 3: Advanced Concepts and Problem-Solving Techniques

3.1 Identifying Turning Points and Maximum/Minimum Velocity

A particle moving in 1D may momentarily stop or change direction. This happens when its velocity is zero.

• Momentarily at Rest: Set \(v = 0\) and solve for \(t\).
• Maximum/Minimum Velocity: This occurs when the acceleration is zero. Set \(a = \frac{dv}{dt} = 0\) and solve for \(t\). (You may need to use the second derivative test, \(\frac{d^2v}{dt^2}\), to confirm if it's a max or min, but often the context will be clear).

3.2 Using Vectors to Determine Collinear Motion

In 2D, we sometimes need to determine if two events happen on the same line or if three points are collinear.

If two vectors, \(\mathbf{A}\) and \(\mathbf{B}\), are parallel, they must be scalar multiples of each other.

Example: If a particle has velocity \(\mathbf{v} = (k-t)\mathbf{i} + (2t)\mathbf{j}\), and we are told the motion is parallel to the vector \(\mathbf{i} + 4\mathbf{j}\) at \(t=1\):

1. Find \(\mathbf{v}\) at \(t=1\): \(\mathbf{v} = (k-1)\mathbf{i} + 2\mathbf{j}\).
2. Since \(\mathbf{v}\) is parallel to \(\mathbf{i} + 4\mathbf{j}\), one must be a multiple (\(\lambda\)) of the other: \[(k-1)\mathbf{i} + 2\mathbf{j} = \lambda(\mathbf{i} + 4\mathbf{j})\]
3. Equate components:
   • \(\mathbf{j}\): \(2 = 4\lambda \implies \lambda = 0.5\)
   • \(\mathbf{i}\): \(k-1 = \lambda = 0.5 \implies k = 1.5\)

3.3 Intercepts and Collisions

Kinematics often involves two particles, \(P_1\) and \(P_2\).

• Interception: Particle \(P_1\) intercepts a fixed point or trajectory when its position vector \(\mathbf{r}_1\) is equal to that fixed position at some time \(t\).
• Collision: Two particles collide if and only if their position vectors are identical at the exact same time \(t\). \[\mathbf{r}_1(t) = \mathbf{r}_2(t)\] This means the \(\mathbf{i}\) components must be equal, AND the \(\mathbf{j}\) components must be equal, using the same value of \(t\).