Motion in a straight line with variable acceleration

Welcome to Motion with Variable Acceleration!

Hi there! So far in Mechanics, you’ve likely spent a lot of time working with constant acceleration using the SUVAT equations. But in the real world, acceleration rarely stays the same. Think about a car pulling away from a traffic light or a sprinter starting a race—they don't just click into one acceleration and stay there; it changes every second!
In this chapter, we are going to learn how to describe motion when the acceleration is variable (changing). Since we can't use SUVAT anymore, we will use our "mathematical power tools": Calculus (Differentiation and Integration). Don't worry if you find calculus a bit intimidating; we’ll break it down step-by-step!

Section 1: The "Big Three" – s, v, and a

To master this topic, you need to understand the relationship between three key things:

1. Displacement (\(s\)): Where the object is relative to a starting point.
2. Velocity (\(v\)): How fast the displacement is changing.
3. Acceleration (\(a\)): How fast the velocity is changing.

The Calculus Ladder

Think of these three as being on a ladder. To move between them, you either climb up or climb down:

Going DOWN the ladder (Differentiation):
If you have Displacement, you differentiate to get Velocity.
If you have Velocity, you differentiate to get Acceleration.

\(v = \frac{ds}{dt}\)
\(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\)

Going UP the ladder (Integration):
If you have Acceleration, you integrate to get Velocity.
If you have Velocity, you integrate to get Displacement.

\(v = \int a \, dt\)
\(s = \int v \, dt\)

Memory Aid: Just remember Displacement, Velocity, Acceleration (**DVA**). Moving from D to A requires Differentiation (Both start with D!).

Key Takeaway: SUVAT is for constant numbers; Calculus is for functions of time (\(t\)). If you see a \(t\) in your acceleration formula, put the SUVAT equations away!

Section 2: Finding Velocity and Acceleration (Differentiation)

When you are given displacement as a function of time, like \(s = 2t^3 - 4t\), you are looking at how the position changes as time ticks by. To find the velocity at any specific moment, we look at the gradient of the displacement-time graph.

Step-by-Step: Moving Down the Ladder

1. Start with Displacement: \(s = 4t^3 + 2t^2\)
2. Differentiate once for Velocity: Multiply by the power, then subtract one from the power.
\(v = \frac{ds}{dt} = 12t^2 + 4t\)
3. Differentiate again for Acceleration:
\(a = \frac{dv}{dt} = 24t + 4\)

Encouraging Phrase: Don't worry if this seems tricky at first! Just remember the rule from your Pure Maths (P1) unit: for \(at^n\), the derivative is \(ant^{n-1}\). It's the same rule here, just using \(t\) instead of \(x\)!

Common Mistake to Avoid: Students often forget that "at rest" means \(v = 0\). If a question asks for the time when the particle is momentarily at rest, set your velocity equation to zero and solve for \(t\).

Section 3: Finding Velocity and Displacement (Integration)

This is the reverse process. If you know how the acceleration is changing, you can work backward to find the velocity and displacement.

The Mystery of "+ c" (The Constant of Integration)

When you integrate, you must add a constant (usually called \(c\)). In Mechanics, this constant is vital because it represents the initial conditions (what was happening at \(t = 0\)).

Analogy: Imagine you are told a car is accelerating at \(2 ms^{-2}\). You can't know how fast it's going later unless you know its starting speed. That starting speed is your \(+ c\)!

Step-by-Step: Moving Up the Ladder

1. Start with Acceleration: \(a = 6t - 2\)
2. Integrate for Velocity:
\(v = \int (6t - 2) \, dt = 3t^2 - 2t + c\)
3. Find \(c\): Use the information in the question. If it says "the particle starts with velocity \(5 ms^{-1}\)", then when \(t = 0, v = 5\).
\(5 = 3(0)^2 - 2(0) + c \rightarrow c = 5\).
So, \(v = 3t^2 - 2t + 5\).
4. Integrate again for Displacement:
\(s = \int (3t^2 - 2t + 5) \, dt = t^3 - t^2 + 5t + d\) (Use a different letter like \(d\) for the second constant!).

Quick Review Box:
- Differentiation: Power down, multiply, reduce power. No \(+c\).
- Integration: Increase power, divide by new power. Always add \(+c\)!

Section 4: Important Definitions and Turning Points

Questions will often use specific phrases. Here is a "translation guide" for you:

"Initially" or "At the start": This always means \(t = 0\).
"At the origin": This means displacement \(s = 0\).
"Momentarily at rest": This means velocity \(v = 0\).
"Constant velocity": This means acceleration \(a = 0\).

Did you know?

The total distance traveled isn't always the same as the displacement! If a particle moves forward and then comes back, the displacement might be zero, but the distance traveled is the total length of the path. To find total distance, you need to check if the particle changes direction (where \(v = 0\)) during the time interval.

Section 5: Summary Checklist

To succeed in variable acceleration questions, always ask yourself:

1. Is acceleration constant? If yes, use SUVAT. If there is a \(t\) in the equation, use Calculus.
2. Which way am I moving on the ladder?
- \(s \rightarrow v \rightarrow a\) : Differentiate.
- \(a \rightarrow v \rightarrow s\) : Integrate.
3. Did I include \(+ c\)? (When integrating).
4. Have I used the initial conditions? (To find the value of \(c\)).

Final Encouragement: You've got this! Practice a few problems where you differentiate a polynomial and a few where you integrate one. Once you get the "ladder" concept down, the rest is just simple algebra!