Non uniform acceleration

Introduction: Moving Beyond "Steady" Motion

Welcome! So far in Mechanics, you have likely spent a lot of time working with SUVAT equations. Those formulas are fantastic, but they have one major limitation: they only work when acceleration is constant (uniform). In the real world, things rarely move with a perfectly steady acceleration. Think about a sprinter starting a race or a car pulling away from traffic lights—the acceleration changes every split second!

In this chapter, we are going to learn how to handle non-uniform acceleration. Don't worry if this seems tricky at first; we will be using Calculus (Differentiation and Integration) to bridge the gap between displacement, velocity, and acceleration. If you can differentiate and integrate a simple power of \( x \), you already have the tools you need to succeed here!

The Kinematics Hierarchy

To master this topic, you just need to remember the order of three key things: Displacement (\( s \)), Velocity (\( v \)), and Acceleration (\( a \)). We can move between these three using calculus.

1. Moving "Down" the Chain: Differentiation

When you are given an expression for displacement in terms of time (\( t \)), you can find the others by "moving down" the chain using differentiation.

From Displacement to Velocity:
Velocity is the rate of change of displacement. In math terms, that means:
\( v = \frac{ds}{dt} \)

From Velocity to Acceleration:
Acceleration is the rate of change of velocity. That means:
\( a = \frac{dv}{dt} \)

Did you know? Because acceleration is the derivative of velocity, and velocity is the derivative of displacement, acceleration is also the second derivative of displacement! We write this as:
\( a = \frac{d^2s}{dt^2} \)

Quick Review: The "Differentiation Slide"
Imagine displacement is at the top of a slide. To go down to velocity and then acceleration, you differentiate. Use this whenever the question gives you an equation like \( s = 3t^2 - 5t + 2 \).

2. Moving "Up" the Chain: Integration

What if you start with acceleration and want to find velocity? You do the opposite! You move "up" the chain using integration.

From Acceleration to Velocity:
\( v = \int a \, dt \)

From Velocity to Displacement:
\( s = \int v \, dt \)

Crucial Point: The Constant of Integration (\( + c \))
When you integrate, you must include a \( + c \). In Mechanics, we often use the information in the question (like "the particle starts from rest") to find out what that constant is. Failing to add \( + c \) is the most common mistake students make!

Memory Aid: "DVA" (The Diva)
Write the letters D, V, A in a column:
D (Displacement)
V (Velocity)
A (Acceleration)
Going Down? Differentiate.
Going Up? Integrate.

Step-by-Step: Solving a Non-Uniform Acceleration Problem

Let's look at how to tackle a typical problem where we move from an equation for acceleration back to displacement.

Example: A particle moves in a straight line with acceleration \( a = 6t - 4 \). At time \( t = 0 \), its velocity is \( 3 \, ms^{-1} \) and its displacement is \( 0 \). Find an expression for displacement \( s \).

Step 1: Integrate acceleration to find velocity.
\( v = \int (6t - 4) \, dt = 3t^2 - 4t + c \)

Step 2: Use the "initial conditions" to find \( c \).
The question says when \( t = 0 \), \( v = 3 \).
\( 3 = 3(0)^2 - 4(0) + c \), so \( c = 3 \).
Our velocity equation is: \( v = 3t^2 - 4t + 3 \).

Step 3: Integrate velocity to find displacement.
\( s = \int (3t^2 - 4t + 3) \, dt = t^3 - 2t^2 + 3t + k \)
(We use a different letter, like \( k \), for the second constant!)

Step 4: Use initial conditions for displacement.
The question says when \( t = 0 \), \( s = 0 \).
\( 0 = (0)^3 - 2(0)^2 + 3(0) + k \), so \( k = 0 \).
Final Answer: \( s = t^3 - 2t^2 + 3t \).

Common Mistakes to Avoid

1. Using SUVAT for everything:
If the equation for acceleration has a \( t \) in it (e.g., \( a = 4t \)), DO NOT use \( v = u + at \). SUVAT only works for constant numbers like \( a = 9.8 \) or \( a = 5 \). If it changes with time, you must use calculus.

2. Forgetting the \( + c \):
In Mechanics, the \( + c \) often represents the initial velocity or initial displacement. If you leave it out, your whole path to the answer will be wrong!

3. Misinterpreting "At Rest":
When a question says the particle is "at rest," it means velocity is zero (\( v = 0 \)). This is often the key to finding a time value or a constant.

Summary: Key Takeaways

Differentiation:
Use it to find gradients. Gradient of displacement-time graph = Velocity. Gradient of velocity-time graph = Acceleration.

Integration:
Use it to find areas. Area under a velocity-time graph = Displacement. Area under an acceleration-time graph = Change in Velocity.

The Core Formulas:
\( v = \frac{ds}{dt} \)
\( a = \frac{dv}{dt} = \frac{d^2s}{dt^2} \)
\( s = \int v \, dt \)
\( v = \int a \, dt \)

Keep these formulas handy, remember your \( + c \), and you'll be well on your way to mastering kinematics with non-uniform acceleration!