Motion in a straight line with constant acceleration - Mathematics (9660) - Oxford AQA International AS Level

Welcome to Mechanics: Motion in a Straight Line!

Welcome to the world of Mechanics (M1)! In this chapter, we are going to look at how objects move when they travel in a straight line with a constant (steady) acceleration. This is the foundation of almost everything you will study in Mechanics. Whether it is a car speeding up away from a traffic light or a ball falling through the air, the rules we learn here will help you predict exactly where an object will be and how fast it will be going.

Don’t worry if this seems a bit "physics-heavy" at first! We will break every concept down into small, manageable steps with plenty of tricks to help you remember the formulas.

1. The Fundamentals: Distance, Displacement, Speed, and Velocity

Before we can calculate motion, we need to be very clear about the words we use. In everyday life, we use "distance" and "displacement" interchangeably, but in Mechanics, they have specific meanings.

Scalar vs. Vector

In Mechanics, we divide measurements into two groups:

Scalars: These only have a size (magnitude). Example: 5 meters.
Vectors: These have a size AND a direction. Example: 5 meters to the Right.

Distance vs. Displacement

Distance (Scalar): The total ground covered. If you walk 10m forward and 10m back, your distance is 20m.

Displacement, \( s \) (Vector): Your position relative to where you started. If you walk 10m forward and 10m back, your displacement is 0m because you are back where you started!

Speed vs. Velocity

Speed (Scalar): How fast an object is moving. \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \).

Velocity, \( v \) or \( u \) (Vector): Speed in a given direction. \( \text{Velocity} = \frac{\text{Displacement}}{\text{Time}} \).

Acceleration, \( a \)

Acceleration is the rate at which velocity changes. If a car goes from 0 to 20 m/s, it is accelerating. In this chapter, we only deal with constant acceleration, meaning the velocity changes by the same amount every second.

Key Takeaway: Always check your signs! In Mechanics, positive (+) and negative (-) usually just tell you the direction (e.g., + is Up, - is Down).

2. Kinematics Graphs

Sometimes, a picture is worth a thousand equations. You need to be able to sketch and interpret two types of graphs.

Displacement-Time Graphs

The gradient (slope) of the line represents the velocity.
A straight diagonal line means constant velocity.
A horizontal line means the object is stationary (displacement isn't changing).

Velocity-Time Graphs

These are the most common graphs in the M1 exam!

The gradient represents the acceleration.
The area under the graph represents the displacement (distance travelled).

Step-by-Step: Finding Displacement from a Graph
1. Identify the shapes under the velocity line (usually triangles or rectangles).
2. Calculate the area of each shape: \( \text{Area of Rectangle} = \text{base} \times \text{height} \); \( \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} \).
3. Add them together to find the total displacement.

Quick Review:
Gradient of Displacement-Time = Velocity
Gradient of Velocity-Time = Acceleration
Area under Velocity-Time = Displacement

3. The SUVAT Equations

When an object moves with constant acceleration in a straight line, we use five famous equations. We call them the SUVAT equations because of the five variables involved:

\( s \): displacement (m)
\( u \): initial (starting) velocity (m/s)
\( v \): final velocity (m/s)
\( a \): constant acceleration (\( \text{m/s}^2 \))
\( t \): time (s)

The 5 Big Formulas

1. \( v = u + at \)

2. \( s = \frac{1}{2}(u + v)t \)

3. \( s = ut + \frac{1}{2}at^2 \)

4. \( s = vt - \frac{1}{2}at^2 \)

5. \( v^2 = u^2 + 2as \)

Memory Aid: To remember the letters, think of the phrase: "Silly Uncle Victor Ate Toast".

How to Solve SUVAT Problems:

Don't panic if a question looks long! Just follow these steps:
1. List your variables: Write down \( s, u, v, a, t \) and fill in the values you know from the question.
2. Identify what you need: Put a question mark next to the variable you are trying to find.
3. Choose the equation: Find the equation that uses your known values and your target variable.
4. Substitute and solve: Plug in the numbers and calculate the answer.

Did you know? These equations only work if acceleration is constant. If the acceleration changes (e.g., \( a = 3t \)), you cannot use SUVAT!

4. Vertical Motion Under Gravity

When an object is thrown up or dropped, it is in "free fall." In the Oxford AQA syllabus, we assume there is no air resistance. This means the object accelerates downwards due to gravity at a constant rate.

The Magic Number: \( g \)

The acceleration due to gravity is denoted by \( g \). For your exams, always use:
\( g = 9.8 \, \text{ms}^{-2} \)

Setting Your Directions

This is where most students make mistakes. You must decide which direction is positive at the start of the problem.

If you choose UP as positive: then \( a = -9.8 \) (because gravity pulls down).
If you choose DOWN as positive: then \( a = +9.8 \).

Key Facts for Vertical Motion:

If an object is dropped from rest, \( u = 0 \).
At the highest point of a throw, the velocity \( v = 0 \) for a split second.
The time it takes to go up to the highest point is the same as the time it takes to come back down to the same level.

Common Mistake to Avoid: Forgetting that \( a \) is negative if you've decided that "up" is the positive direction. If you throw a ball up at 10 m/s, your \( u = 10 \) but your \( a = -9.8 \)!

5. Average Speed

Sometimes a question will ask for the Average Speed of a journey. This is different from average velocity.

\( \text{Average Speed} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}} \)

Example: If a particle moves 10m right in 2 seconds, then 4m left in 1 second:
Total distance = \( 10 + 4 = 14\text{m} \).
Total time = \( 2 + 1 = 3\text{s} \).
Average speed = \( 14 / 3 = 4.67 \, \text{m/s} \).
(Note: The average velocity would be \( (10 - 4) / 3 = 2 \, \text{m/s} \) because displacement is only 6m).

Key Takeaway: For average speed, ignore the direction and just add up the total distance!

Summary Checklist

Do I know the difference between displacement and distance?
Can I find the area under a Velocity-Time graph to find displacement?
Have I memorized the 5 SUVAT equations?
Am I using \( a = 9.8 \) for gravity problems?
Have I checked if my signs (+/-) are consistent for direction?

You've got this! Mechanics is all about practice. Try listing your SUVAT variables for every problem, and the rest will fall into place.

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