Introduction: Moving Beyond SUVAT

Welcome to Further Kinematics! Up until now, you’ve likely spent a lot of time with the "SUVAT" equations. Those are great, but they only work when acceleration is constant. In the real world, things are rarely that simple. Think of a rocket blasting off (where mass changes) or a car facing increasing air resistance as it speeds up. In these cases, acceleration changes.

In this chapter, we use the power of calculus (differentiation and integration) to solve problems where acceleration depends on time, velocity, or displacement. Don't worry if this seems tricky at first—once you learn which "tool" to pick from your mathematical toolbox, it becomes much easier!

1. The Fundamentals: The Calculus Bridge

Before we dive into the new stuff, let’s quickly review the bridge that connects our three main variables. If you can remember this flow, you’re halfway there!

The Hierarchy:
1. Displacement (\(x\))
2. Velocity (\(v = \frac{dx}{dt}\))
3. Acceleration (\(a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\))

  • To go DOWN the list (e.g., Displacement to Velocity), we Differentiate.
  • To go UP the list (e.g., Acceleration to Velocity), we Integrate.

Quick Review:
\( v = \int a \, dt \)
\( x = \int v \, dt \)
Always remember to add your constant of integration (\(+ C\))! You usually find this value using "initial conditions" (like "at \(t=0, v=2\)").

2. Case 1: Acceleration as a Function of Time \(a = f(t)\)

This is the most straightforward scenario. If your acceleration is given in terms of \(t\) (e.g., \(a = 3t^2 + 2\)), you simply integrate with respect to \(t\).

The Process:
1. Write down the relationship: \(\frac{dv}{dt} = f(t)\).
2. Integrate both sides with respect to \(t\) to find velocity: \(v = \int f(t) \, dt\).
3. Integrate again to find displacement: \(x = \int v \, dt\).

Example: A particle has acceleration \(a = 6t\). Find \(v\) if \(v=2\) when \(t=0\).
\(\frac{dv}{dt} = 6t \)
\(v = \int 6t \, dt = 3t^2 + C\)
Using \(t=0, v=2\): \(2 = 3(0)^2 + C \Rightarrow C = 2\).
So, \(v = 3t^2 + 2\).

Key Takeaway: If you see \(t\) in the acceleration formula, just integrate directly.

3. Case 2: Acceleration as a Function of Velocity \(a = f(v)\)

This happens when the acceleration depends on how fast the object is already going—like a parachute opening. Since acceleration is \(\frac{dv}{dt}\), we get an equation like \(\frac{dv}{dt} = f(v)\).

The "Separation" Trick:
To solve this, you need to use a technique from Pure Maths called Separation of Variables. You want all the \(v\) terms on one side and the \(t\) terms on the other.

Step-by-Step:
1. Start with \(\frac{dv}{dt} = f(v)\).
2. Rearrange to: \(\frac{1}{f(v)} \, dv = 1 \, dt\).
3. Integrate both sides: \(\int \frac{1}{f(v)} \, dv = \int 1 \, dt\).
4. This will give you an equation linking \(v\) and \(t\).

Did you know?

Terminal velocity occurs when acceleration becomes zero. If you have \(a = f(v)\), you can find the terminal velocity by setting \(f(v) = 0\) and solving for \(v\)!

Key Takeaway: If you see \(v\) in the acceleration formula, flip it and integrate to find \(t\).

4. Case 3: Acceleration as a Function of Displacement \(a = f(x)\)

This is the classic "Further Mechanics" scenario. Here, the acceleration depends on where the particle is (like a mass on a spring). We need to link acceleration (\(a\)) and displacement (\(x\)), but the standard \(\frac{dv}{dt}\) involves time.

The Essential Identity:
In these problems, we use a special form of acceleration:
\( a = v\frac{dv}{dx} \)

Why does this work? It's just the Chain Rule! \(\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}\). Since \(\frac{dx}{dt}\) is just velocity (\(v\)), we get \(v\frac{dv}{dx}\).

The Process:
1. Set up the equation: \(v\frac{dv}{dx} = f(x)\).
2. Separate the variables: \(v \, dv = f(x) \, dx\).
3. Integrate both sides: \(\int v \, dv = \int f(x) \, dx\).
4. This gives: \(\frac{1}{2}v^2 = \int f(x) \, dx + C\).

Memory Aid: Think of "V-D-V". When acceleration depends on \(x\), use \(v \frac{dv}{dx}\). It rhymes with "Vroom" if you say it fast enough!

Key Takeaway: If you see \(x\) in the acceleration formula, use \(a = v\frac{dv}{dx}\) to skip over time and link speed to position directly.

5. Common Mistakes to Avoid

  • The Missing +C: This is the #1 reason students lose marks. Always include the constant of integration immediately after integrating.
  • Confusion of Variables: If the question asks for velocity in terms of time, but you used \(v\frac{dv}{dx}\), you’ll get velocity in terms of displacement. Read the question carefully to see which variable is needed!
  • Algebraic Slips: When separating variables, like in \(\int \frac{1}{v} \, dv\), remember your log rules (\(\ln|v|\)).
  • Limits vs. Constants: You can use definite integrals (with limits) instead of \(+ C\) if you prefer. Just make sure the upper and lower limits match up (e.g., \(t=0\) at the bottom, \(t=T\) at the top).

Summary Checklist

Before you tackle a problem, ask yourself: "What does the acceleration depend on?"

  • Depends on Time (\(t\))? Use \(a = \frac{dv}{dt}\) and integrate.
  • Depends on Velocity (\(v\))? Use \(a = \frac{dv}{dt}\), separate variables, and integrate \(\frac{1}{f(v)}\).
  • Depends on Displacement (\(x\))? Use \(a = v\frac{dv}{dx}\), separate variables, and integrate \(v\).

Encouraging Note: Don't worry if the integration looks scary! In Paper 4C, the focus is on setting up the mechanics correctly. The actual calculus will be similar to what you’ve practiced in your Pure Core modules. You've got this!