Kinematics of a particle moving in a straight line - Further Mathematics - Pearson A-Level

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🚀 Kinematics of a Particle Moving in a Straight Line

Welcome to the fascinating world of Kinematics! Don't worry if 'Mechanics' sounds intimidating; Kinematics is essentially the maths of motion. We study how things move, without worrying why they move (that comes later with Forces!).

This chapter is fundamental to Unit M1. It teaches you the essential tools needed to describe the movement of objects (which we model as particles) travelling along a single, straight path. Mastering these concepts will unlock many future topics, so let's dive in!

What You Will Learn:

The definitions of displacement, velocity, and acceleration.
How to use the famous constant acceleration formulas (SUVAT).
How to use differentiation and integration when acceleration is changing (variable).

1. Defining the Fundamentals

In kinematics, we deal with three key measurements that describe the motion of a particle:

1.1 Displacement ($s$)

Displacement ($s$) is the shortest straight-line distance from a fixed reference point (usually called the origin, $O$).

It is a vector quantity: direction matters!
If you start at $O$, move 5m right ($s = +5$), and then move 5m left ($s = -5$), your final displacement is $s = 0$.
Key Difference: Displacement is NOT the same as distance travelled. Distance travelled is the total path length (a scalar quantity).

Analogy: If you walk around a circular running track and finish where you started, your distance travelled is the track length, but your displacement is zero.

1.2 Velocity ($v$)

Velocity ($v$) is the rate of change of displacement with respect to time.

It is also a vector quantity (speed in a specific direction).
If the particle is speeding up in the positive direction, $v$ is positive.
If the particle is moving in the negative direction, $v$ is negative, even if the particle is speeding up!
Speed is the magnitude of the velocity ($|v|$).

1.3 Acceleration ($a$)

Acceleration ($a$) is the rate of change of velocity with respect to time.

If a particle is moving right and accelerating right, its speed increases.
If a particle is moving right but accelerating left (decelerating), its speed decreases.
The standard unit for acceleration is $\text{m/s}^2$.

🔑 Quick Review: Kinematic Bridge

Displacement, velocity, and acceleration are all linked through time.

Displacement $\xrightarrow{\text{Rate of Change}}$ Velocity $\xrightarrow{\text{Rate of Change}}$ Acceleration

2. Motion Under CONSTANT Acceleration (SUVAT)

The majority of M1 kinematics problems involve movement where the acceleration ($a$) never changes. This is fantastic because it allows us to use a set of five incredibly powerful formulas, often known as the SUVAT equations.

Don't panic! You do not need to memorise the derivations, but you must know how to use them.

2.1 The Five SUVAT Variables

When tackling a constant acceleration problem, always list these five variables:

$s$: Displacement (m)
$u$: Initial Velocity (m/s)
$v$: Final Velocity (m/s)
$a$: Constant Acceleration ($\text{m/s}^2$)
$t$: Time (s)

💡 Top Tip: Always choose one direction (e.g., upwards or right) as the positive direction and stick to it. All vectors pointing that way are positive; all vectors pointing the opposite way are negative.

2.2 The SUVAT Equations

To use these formulas, you usually need to know three of the variables to find a fourth.

$$ v = u + at \quad \text{ (Missing } s \text{)} $$ $$ s = ut + \frac{1}{2}at^2 \quad \text{ (Missing } v \text{)} $$ $$ v^2 = u^2 + 2as \quad \text{ (Missing } t \text{)} $$ $$ s = \frac{1}{2}(u+v)t \quad \text{ (Missing } a \text{)} $$ $$ s = vt - \frac{1}{2}at^2 \quad \text{ (Missing } u \text{)} $$

Step-by-Step Guide to Solving SUVAT Problems

Draw a Diagram: Even a simple line drawing helps define your starting point and direction.
List SUVAT: Write down $s, u, v, a, t$.
Input Knowns: Fill in the values you know (remembering units and signs!).
Identify Unknown: What are you trying to find?
Choose the Formula: Select the equation that uses your three knowns and the one unknown, ignoring the fifth variable.
Solve: Substitute the values and rearrange the formula.

2.3 Special Case: Vertical Motion under Gravity

When a particle is moving freely under gravity (e.g., a ball thrown up), the acceleration is constant:

Acceleration: $a = g$ (the acceleration due to gravity).
Value: Unless otherwise stated, use $g = 9.8 \text{ m/s}^2$.

When solving vertical problems, you must be very careful with signs:

If you define UPWARDS as positive: $a = -9.8 \text{ m/s}^2$.
If you define DOWNWARDS as positive: $a = +9.8 \text{ m/s}^2$.

⚠️ Common Mistake Alert: At the highest point of a projectile’s path, the particle is momentarily stationary. Therefore, $v = 0$ at the maximum height (but acceleration $a$ is still $-9.8 \text{ m/s}^2$!).

Key Takeaway: SUVAT is your best friend when acceleration is a fixed number. Always define your positive direction first!

3. Motion Under VARIABLE Acceleration (Calculus)

What if the acceleration is not constant? For instance, what if $a$ changes based on the particle's position or the time elapsed? This is where the power of Calculus comes in.

Don't worry if this seems tricky at first—it just means we are using instantaneous rates of change instead of average rates of change.

3.1 The Calculus Definitions

Because velocity is the rate of change of displacement, and acceleration is the rate of change of velocity, we can use differentiation:

Going Down the Kinematic Bridge (Differentiating)

1. Displacement to Velocity:

$$ v = \frac{ds}{dt} $$

2. Velocity to Acceleration:

$$ a = \frac{dv}{dt} $$

Going Up the Kinematic Bridge (Integrating)

If we want to go the other way, we use integration. Remember that when you integrate, you must include the constant of integration ($C$). You find $C$ by using initial conditions (usually conditions when $t=0$).

1. Acceleration to Velocity:

$$ v = \int a \, dt $$

2. Velocity to Displacement:

$$ s = \int v \, dt $$

3.2 Practical Applications

Typically, in a variable acceleration problem, you will be given an expression for $s$, $v$, or $a$ in terms of time $t$.

Example: A particle’s displacement is given by $s = t^3 - 6t^2 + 5t$.

Step 1: Find Velocity ($v$)

$$ v = \frac{ds}{dt} = 3t^2 - 12t + 5 $$

Step 2: Find Acceleration ($a$)

$$ a = \frac{dv}{dt} = 6t - 12 $$

Notice that $a$ depends on $t$; therefore, this is variable acceleration, and SUVAT cannot be used!

🛑 When the Particle is at Rest (Crucial Concept)

A particle is momentarily at rest when its velocity is zero ($v=0$). If you are asked to find the time when the particle is at rest, you must set the velocity expression to zero and solve the resulting equation (often a quadratic).

🧠 Memory Aid: The SDA Triangle

Imagine the letters S, V, A in a line. To move from S to V, or V to A, you Differentiate. To move from A to V, or V to S, you Integrate.

Key Takeaway: If the problem involves an expression for $s, v,$ or $a$ that depends on time, you MUST use calculus (differentiation or integration).

4. Graphical Interpretation of Motion

Graphs are a visual way to understand kinematics. While displacement-time graphs and acceleration-time graphs are useful, the Velocity-Time (V-T) Graph is the most critical tool in M1.

4.1 Velocity-Time Graphs (V-T)

A V-T graph plots velocity ($v$) on the vertical axis against time ($t$) on the horizontal axis.

Feature 1: The Gradient ($\text{Slope}$)

The gradient of the V-T graph represents the acceleration.

A straight, positive slope means constant, positive acceleration (speeding up).
A zero slope (horizontal line) means zero acceleration (constant velocity/SUVAT).
A curved slope means variable acceleration (must use calculus methods to find instantaneous acceleration).

$$ \text{Acceleration} = \frac{\Delta v}{\Delta t} = \text{Gradient} $$

Feature 2: The Area Under the Curve

The area between the curve and the time axis represents the displacement.

Area above the axis (positive velocity) = Positive displacement.
Area below the axis (negative velocity) = Negative displacement.
Total Displacement: The signed area (Area Above - Area Below).
Total Distance Travelled: The sum of the magnitudes of all areas (Area Above + Area Below).

Example: If the area above the axis is 10m and the area below is 3m:
Displacement = $10 - 3 = 7\text{ m}$
Distance Travelled = $10 + 3 = 13\text{ m}$

⚠️ Common Mistake Alert: Students often confuse displacement and distance travelled when the velocity goes negative (i.e., when the particle turns around). Always calculate the areas separately!

4.2 Sketching V-T Graphs

For constant acceleration, the V-T graph is always a straight line segment, because the gradient (acceleration) is constant. For variable acceleration, the graph will be a curve.

Did you know?

The concepts of displacement and velocity were formalised by Isaac Newton in the late 17th century when he developed calculus, which he called "fluxions," specifically to understand mechanics and orbital motion!

Summary and Next Steps

You have now grasped the core concepts of straight-line kinematics. Remember these three pillars:

SUVAT: Use only when acceleration ($a$) is constant.
Calculus: Use differentiation ($\frac{d}{dt}$) to find rates of change, and integration ($\int dt$) to go backwards, when motion is variable.
Graphs: Use the gradient for acceleration and the area for displacement/distance.

Practice converting between the three methods. Success in this chapter comes from rigorous practice and precise use of signs (positive/negative directions)! You’ve got this!