Non uniform acceleration

Welcome to Non-Uniform Acceleration!

In your previous studies, you probably spent a lot of time with SUVAT equations. Those are great, but they only work when acceleration is a constant number. In the real world, acceleration changes all the time—think about a car pulling away from a traffic light or a sprinter exploding out of the blocks. This chapter shows you how to handle motion when acceleration isn't constant, using the power of Calculus.

Don't worry if calculus feels a bit "pure math" right now; in Mechanics, it's just a tool to help us see how things move over time.

1. The Core Relationship: Displacement, Velocity, and Acceleration

To master this chapter, you need to see Displacement (\(s\)), Velocity (\(v\)), and Acceleration (\(a\)) as a chain. Depending on which way you want to move along the chain, you either differentiate or integrate with respect to time (\(t\)).

Moving "Down" the Chain: Differentiation

If you have an expression for displacement and want to find acceleration, you differentiate:

• To get Velocity from Displacement: \(v = \frac{ds}{dt}\)
• To get Acceleration from Velocity: \(a = \frac{dv}{dt}\)
• To get Acceleration directly from Displacement: \(a = \frac{d^2s}{dt^2}\)

Moving "Up" the Chain: Integration

If you have the acceleration and want to find where the object is, you integrate:

• To get Velocity from Acceleration: \(v = \int a \, dt\)
• To get Displacement from Velocity: \(s = \int v \, dt\)

Memory Aid: The "D-V-A" Rule
Think of the order Displacement → Velocity → Acceleration.
If you go Down the list, you Differentiate (Both start with D!).
If you go Up the list, you Integrate.

Quick Review:
- Differentiation = Finding the rate of change (the gradient).
- Integration = Accumulating the change (the area under the graph).
- Important: Never use SUVAT if the acceleration expression contains a \(t\)!

2. Dealing with the Constant of Integration (\(+c\))

When you integrate to find velocity or displacement, you must always add a constant of integration (usually written as \(+c\)). In Mechanics, we find the value of this constant using initial conditions.

Example: A particle has acceleration \(a = 6t\). At time \(t = 0\), the velocity \(v = 2\).
1. Integrate: \(v = \int 6t \, dt = 3t^2 + c\).
2. Use the condition (\(t=0, v=2\)): \(2 = 3(0)^2 + c\), so \(c = 2\).
3. Final formula: \(v = 3t^2 + 2\).

Common Mistake to Avoid:
Many students assume \(c\) is always the value at \(t = 0\). While often true for polynomials, if your expression involves \(cos(t)\) or \(e^t\), \(c\) might be something totally different. Always plug the numbers in to check!

3. Motion in Two Dimensions (Vectors)

In the OCR H240 syllabus, you are expected to extend these ideas to 2D using vectors. The good news? It's exactly the same math, just done twice—once for the horizontal component (\(\mathbf{i}\)) and once for the vertical component (\(\mathbf{j}\)).

Vector Notation

We usually use \(\mathbf{r}\) or \(\mathbf{x}\) for the position vector:
\(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\)

The derivatives look like this:
Velocity: \(\mathbf{v} = \frac{d\mathbf{r}}{dt} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}\)
Acceleration: \(\mathbf{a} = \frac{d\mathbf{v}}{dt} = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}\)

Did you know?
The little dots over the letters (\(\dot{x}\) and \(\ddot{x}\)) are called Newton’s notation. One dot means you've differentiated with respect to time once; two dots mean twice!

Step-by-Step: Solving a Vector Problem

Suppose you are given \(\mathbf{v} = (3t^2)\mathbf{i} + (4t)\mathbf{j}\) and asked for the position vector \(\mathbf{r}\) at \(t=2\), given \(\mathbf{r} = \mathbf{0}\) at \(t=0\).

Step 1: Integrate the \(\mathbf{i}\) component: \(\int 3t^2 \, dt = t^3 + c_1\).
Step 2: Integrate the \(\mathbf{j}\) component: \(\int 4t \, dt = 2t^2 + c_2\).
Step 3: Use initial conditions to find \(c_1\) and \(c_2\). Since \(\mathbf{r} = \mathbf{0}\) at \(t=0\), both constants are \(0\).
Step 4: Write the position vector: \(\mathbf{r} = t^3\mathbf{i} + 2t^2\mathbf{j}\).
Step 5: Plug in \(t=2\): \(\mathbf{r} = (2)^3\mathbf{i} + 2(2)^2\mathbf{j} = 8\mathbf{i} + 8\mathbf{j}\).

Key Takeaway: Treat the \(\mathbf{i}\) and \(\mathbf{j}\) directions as completely separate problems that happen to be written in the same bracket.

4. Displacement vs. Distance Travelled

This is a classic "trick" area in exams. There is a subtle difference between these two terms:

• Displacement: How far you are from your starting point (Vector). Found by \(\int_{t_1}^{t_2} v \, dt\).
• Distance Travelled: The total ground covered (Scalar).

Analogy: Imagine walking 10 meters forward and 10 meters back. Your displacement is 0 (you're back where you started), but your distance travelled is 20 meters.

To find the total distance when velocity changes direction, you must:
1. Find when the object stops (set \(v = 0\)).
2. Calculate the distance for each "leg" of the journey separately.
3. Add the absolute values (positive versions) of those distances together.

Don't worry if this seems tricky at first! Drawing a quick sketch of the velocity-time graph usually makes it clear whether the object has turned around.

Summary Checklist

Before you tackle the practice questions, make sure you've got these "Must-Knows" down:

• Can I differentiate? (To go from \(s \rightarrow v \rightarrow a\))
• Can I integrate? (To go from \(a \rightarrow v \rightarrow s\))
• Did I remember \(+c\)? (And did I use the info in the question to solve for it?)
• Am I using SUVAT by mistake? (Check: if acceleration has a \(t\) in it, SUVAT is banned!)
• For vectors: Am I keeping my \(\mathbf{i}\)'s and \(\mathbf{j}\)'s separate?

Keep practicing! Kinematics with calculus is one of those topics that suddenly "clicks" once you've done a few problems. You've got this!