Welcome to the World of Motion!
In this chapter, we are going to explore Constant Acceleration. Have you ever been in a car that pulls away smoothly from a green light, or watched a ball drop from a window? These are real-life examples of objects moving where their speed changes at a steady, predictable rate. By the end of these notes, you’ll be able to predict exactly where an object will be and how fast it will be going at any moment. Don’t worry if Mechanics feels a bit "physics-heavy" at first—we’ll break it down step-by-step!
1. Meeting the "SUVAT" Family
To solve problems in this chapter, we use five key variables. We often call these the SUVAT variables. Think of them as the "ingredients" for our motion recipes.
\(s\) = Displacement: How far the object is from its starting point (in metres, \(m\)). Note: This is different from distance because it has a direction!
\(u\) = Initial Velocity: How fast the object was going at the very start (in \(m s^{-1}\)).
\(v\) = Final Velocity: Final How fast it is going at the end of the time we are looking at (in \(m s^{-1}\)).
\(a\) = Acceleration: The steady rate at which the velocity changes (in \(m s^{-2}\)).
\(t\) = Time: How long the motion lasted (in seconds, \(s\)).
Quick Tip: Sign Matters!
Because \(s, u, v,\) and \(a\) are vectors, direction is everything. If you decide that "up" is positive, then anything moving "down" must be given a negative value. If you forget this, your math might tell you a ball is flying into space when it’s actually hitting the ground!
Key Takeaway: Always list your "SUVAT" values before starting a calculation. Usually, if you know any 3 of them, you can find the other 2!
2. The Big Five Equations
These are the core tools in your mechanics toolbox. You’ll use these when acceleration is constant. If the acceleration changes (like a car jerking around), these won't work!
1. \( v = u + at \) (Use this if you don't know \(s\))
2. \( s = ut + \frac{1}{2}at^2 \) (Use this if you don't know \(v\))
3. \( s = \frac{1}{2}(u + v)t \) (Use this if you don't know \(a\))
4. \( v^2 = u^2 + 2as \) (Use this if you don't know \(t\))
5. \( s = vt - \frac{1}{2}at^2 \) (Use this if you don't know \(u\))
Analogy: The Toolbox
Imagine you are fixing a bike. You don't use a hammer for every job. If you don't have "Time" (\(t\)), you reach for Equation 4. If you don't have "Final Velocity" (\(v\)), you reach for Equation 2. Always pick the equation that uses the information you have to find the information you want.
Key Takeaway: You don't need to memorize which is which "number," but you should be able to recognize which variable is missing in each one.
3. Where do these equations come from? (Derivations)
The OCR syllabus requires you to understand how we find these equations. We can do this in three main ways:
A. Using Velocity-Time Graphs
If you draw a graph of velocity (\(v\)) against time (\(t\)), a constant acceleration looks like a straight diagonal line.
- The gradient (steepness) of the line is the acceleration (\(a\)).
- The area under the line represents the displacement (\(s\)).
Example: To get \( s = \frac{1}{2}(u + v)t \), we simply calculate the area of the trapezium formed under the graph line!
B. Using Calculus (Integration)
Acceleration is the rate of change of velocity. In math terms: \( a = \frac{dv}{dt} \).
If we integrate \(a\) with respect to \(t\), we get: \( v = \int a \, dt = at + c \).
Since \( v = u \) when \( t = 0 \), our constant \(c\) must be \(u\).
This gives us our first equation: \( v = u + at \)!
C. Using Substitution
You can create new equations by plugging one into another. For example, if you take \( v = u + at \) and rearrange it to find \(t\), then plug that into the area formula, you will eventually discover \( v^2 = u^2 + 2as \).
Key Takeaway: Displacement is the area under a Velocity-Time graph; Acceleration is the gradient.
4. Vertical Motion Under Gravity
One of the most common ways to test constant acceleration is by dropping or throwing things. On Earth, if we ignore air resistance, everything falls with the same constant acceleration.
The Magic Number: \( g = 9.8 \, m s^{-2} \)
Unless the question tells you otherwise, always use \(9.8\) for gravity.
Step-by-Step: The "Ball Thrown Upwards" Problem
1. Pick a direction: Let's say "Up" is positive (\(+\)).
2. Set your acceleration: Since gravity pulls things down, your \( a = -9.8 \).
3. At the highest point: Remember that when an object reaches its peak, its velocity is momentarily zero (\( v = 0 \)).
4. Total flight: If a ball is caught at the same height it was thrown, its total displacement is zero (\( s = 0 \)).
Did you know? Even though a ball stops for a tiny fraction of a second at the very top of its path, its acceleration is still \(9.8 \, m s^{-2}\) downwards! If the acceleration were zero, the ball would just float there forever.
Key Takeaway: Gravity always acts downwards. Be very careful with your \(+\) and \(-\) signs here!
5. How to Solve Any Constant Acceleration Problem
Don't worry if these problems seem tricky at first. Follow this "Recipe for Success" every time:
Step 1: Draw a simple diagram. Even a stick figure and a box helps! Mark which way is positive.
Step 2: Write out "S, U, V, A, T" in a list. Fill in the numbers you know from the question.
Step 3: Identify what you need to find. Put a question mark next to it.
Step 4: Choose your equation. Pick the one that has your "knowns" and your "unknown."
Step 5: Substitute and solve. Plug in the numbers and do the algebra.
Common Mistakes to Avoid:
- Mixing Units: Ensure everything is in metres and seconds. If a speed is in \(km/h\), convert it to \(m/s\) first!
- Square Root Blues: When using \( v^2 = u^2 + 2as \), remember that taking a square root gives a positive and a negative answer. Think about which direction the object is moving to pick the right one.
- Deceleration: If an object is slowing down, its acceleration should usually be negative compared to its velocity.
Key Takeaway: Slow down and list your variables. Most mistakes happen by rushing the setup, not the calculation!
Quick Review Box
Is the acceleration constant? Yes? Use SUVAT. No? Use Calculus (Differentiation/Integration).
What is \(s\)? It is displacement (change in position), not total distance.
What is \(g\)? It is \(9.8 \, m s^{-2}\), acting downwards.
Graph Area: Area under Velocity-Time = Displacement.
Graph Gradient: Gradient of Velocity-Time = Acceleration.