M1.1: Motion in a Straight Line with Constant Acceleration
Welcome to Mechanics! This chapter, often called Kinematics, is all about describing how things move. Instead of worrying about why things move (that comes later with forces), we focus purely on the maths of position, speed, and acceleration.
You’ll learn a powerful set of equations (the SUVAT equations) that let you predict exactly where an object will be, how fast it will be travelling, and how long it will take to get there—as long as its acceleration remains constant. Mastering these concepts is fundamental to all future Mechanics topics!
1. Understanding the Vocabulary of Motion
In Mechanics, we must be very precise. We use two types of quantities:
Vectors vs. Scalars (Why Direction Matters)
Vector Quantities are those that have both magnitude (size) and direction.
Scalar Quantities have only magnitude.
It is essential to decide on a positive direction before solving any problem (e.g., "Up is positive," or "Right is positive").
- Displacement (\(s\)): Vector. This is the change in position measured in a straight line from the start point to the end point. Units: metres (\(m\)).
- Distance: Scalar. This is the total path length travelled.
- Velocity (\(v\) or \(u\)): Vector. This is the rate of change of displacement. Units: metres per second (\(ms^{-1}\)).
- Speed: Scalar. This is the rate of change of distance.
- Acceleration (\(a\)): Vector. This is the rate of change of velocity. Units: metres per second squared (\(ms^{-2}\)).
- Time (\(t\)): Scalar. Units: seconds (\(s\)).
Analogy: Imagine walking 5m East and then 5m West.
Your total distance travelled is 10m.
Your final displacement (\(s\)) is 0m (since you returned to your start point).
Quick Review: Average Speed vs. Average Velocity
The syllabus requires you to know the difference:
Average Speed \( = \frac{\text{Total Distance}}{\text{Total Time}} \)
Average Velocity \( = \frac{\text{Total Displacement}}{\text{Total Time}} \)
2. Visualizing Motion: Kinematics Graphs
Graphs are a powerful way to understand motion, especially because they link directly to calculus concepts (gradients and areas).
2.1 Displacement-Time (\(s-t\)) Graphs
- The gradient (slope) of an \(s-t\) graph represents the velocity.
A straight, sloping line means constant velocity (zero acceleration). - The position (y-axis value) represents the displacement from the starting point.
2.2 Velocity-Time (\(v-t\)) Graphs
These graphs are the most important for constant acceleration problems:
- The gradient (slope) of a \(v-t\) graph represents the acceleration (\(a\)).
For constant acceleration, this graph will be a straight line. - The area under the \(v-t\) graph represents the displacement (\(s\)).
Did you know? The fundamental definitions of kinematics come from these graphs:
\( \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}} \)
Key Takeaway: If you see a constant acceleration problem, you are dealing with straight lines on a \(v-t\) graph. Gradient is \(a\), Area is \(s\).
3. The SUVAT Equations (Constant Acceleration)
These are the five crucial formulae you must know. They only work if the acceleration (\(a\)) is constant.
The Five SUVAT Variables:
- \(s\) = Displacement
- \(u\) = Initial Velocity
- \(v\) = Final Velocity
- \(a\) = Constant Acceleration
- \(t\) = Time
The Equations:
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. \( v^2 = u^2 + 2as \)
How to use SUVAT successfully:
- List: Write down the SUVAT variables and fill in the values you know (and the one you want to find).
- Direction: Choose a consistent positive direction (e.g., if a car is slowing down, its acceleration might be negative).
- Select: Pick the equation that includes the three known variables and the one unknown variable you are looking for.
- Solve: Substitute the values and calculate the result.
Memory Trick: To solve any problem, you must know at least three of the five variables. The fifth equation (\(v^2 = u^2 + 2as\)) is super useful because it allows you to solve problems where you don't know (or don't care about) the time \(t\).
🚨 Common Mistakes to Avoid 🚨
- Inconsistent Units: Ensure all lengths are in \(m\) and all times are in \(s\). If a speed is given in \(kmh^{-1}\), convert it to \(ms^{-1}\)!
- Direction Errors: If you set motion rightwards as positive, and acceleration is leftwards, \(a\) must be negative.
- Mixing up \(u\) and \(v\): Remember \(u\) is the initial velocity (start) and \(v\) is the final velocity (end of the specific time period).
Key Takeaway: The SUVAT equations are your toolbox. Learn to identify which variable is missing in the question, and then pick the corresponding formula.
4. Vertical Motion Under Gravity
When an object is moving purely vertically (up or down), the constant acceleration comes from gravity.
Key Facts about Gravity:
- The acceleration due to gravity is denoted by \(g\).
- In OxfordAQA A-Level Mechanics, \(g\) is always taken as \(9.8 \text{ ms}^{-2}\).
- Gravity always acts downwards.
- Air resistance is almost always ignored in these basic models (we treat the object as a particle).
When solving vertical problems, you must stick to your positive direction:
- If you define Up as positive: Then acceleration \(a = -9.8 \text{ ms}^{-2}\).
- If you define Down as positive: Then acceleration \(a = +9.8 \text{ ms}^{-2}\).
Special Case: Throwing an Object Upwards
If you throw a ball straight up into the air:
- As it travels upwards, its velocity is positive, and its acceleration is \(-9.8 \text{ ms}^{-2}\) (slowing it down).
- At its maximum height, the ball momentarily stops. This means its final velocity \(v = 0\) at that peak moment.
- As it falls back down, its velocity becomes negative, but the acceleration remains \(-9.8 \text{ ms}^{-2}\).
Don't worry if this seems tricky at first. Setting up the SUVAT list and choosing a consistent direction are the only skills you need!
Example Scenario: A ball is thrown vertically upwards with an initial speed of \(15 \text{ ms}^{-1}\). Find the maximum height reached.
1. Define Direction: Let Up be positive.
2. List SUVAT:
\(s\) = ? (This is the max height)
\(u\) = \(+15 \text{ ms}^{-1}\)
\(v\) = \(0 \text{ ms}^{-1}\) (At maximum height)
\(a\) = \(-9.8 \text{ ms}^{-2}\)
\(t\) = (Not known, not needed)
3. Select Equation: We use Equation 5 (no \(t\)): \( v^2 = u^2 + 2as \)
\( 0^2 = (15)^2 + 2(-9.8)s \)
\( 0 = 225 - 19.6s \)
\( s = \frac{225}{19.6} \approx 11.48 \text{ m} \)
Chapter Summary: Key Takeaways
- This entire chapter assumes constant acceleration.
- Always define your positive direction and use vector quantities (Displacement, Velocity, Acceleration) consistently.
- On a velocity-time graph, gradient = acceleration and area = displacement.
- To solve a problem, find three known SUVAT values to calculate the fourth.
- For vertical motion, set acceleration \(a = \pm 9.8 \text{ ms}^{-2}\) (depending on your choice of positive direction).