Welcome to Kinematics!
Welcome to one of the most exciting and fundamental topics in Applied Mathematics! Kinematics is the study of how objects move without worrying about the forces that cause the motion. Whether it is a sprinter blasting off the starting blocks, a football soaring through the air, or a car braking to avoid a collision, kinematics provides the mathematical tools to describe their position, velocity, and acceleration at any moment.
Don't worry if mechanics feels daunting at first. We will break every concept down into clear, bite-sized steps with everyday examples to build your confidence.
1. The Language of Motion: Vectors vs. Scalars
Before jumping into formulas, we must understand the language used to describe movement. In mechanics, quantities are divided into two categories:
• Scalar quantities: These have a magnitude (size) only. Examples include time (\(t\)), distance, and speed.
• Vector quantities: These have both a magnitude and a direction. Examples include displacement (\(s\)), initial velocity (\(u\)), final velocity (\(v\)), and acceleration (\(a\)).
Distance vs. Displacement
• Distance is the total length of the path travelled, regardless of direction. It is always positive.
• Displacement (\(s\)) is the straight-line distance from your starting point to your finishing point in a specific direction. It can be positive, negative, or zero.
Everyday Analogy: If you walk \(400\text{ m}\) all the way around a running track back to the exact spot you started, your distance travelled is \(400\text{ m}\), but your displacement is \(0\text{ m}\)!
Speed vs. Velocity
• Speed is the rate at which distance changes over time: \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\).
• Velocity (\(v\)) is the rate of change of displacement over time: \(\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}\). Velocity tells you how fast something moves and in what direction (e.g., \(+15\text{ ms}^{-1}\) means forwards, \(-15\text{ ms}^{-1}\) means backwards).
Acceleration
• Acceleration (\(a\)) is the rate at which velocity changes over time: \(\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}}\). The standard unit of acceleration is \(\text{ms}^{-2}\).
• A negative acceleration is often called deceleration or retardation, meaning the object is slowing down in the positive direction (or speeding up in the negative direction).
Key Takeaway: Always keep track of direction! In 1D kinematics, we define one direction as positive (usually upwards or to the right). Anything pointing the opposite way must carry a negative sign.
2. Motion Graphs
Graphs are a fantastic visual way to understand motion. In CCEA AS Mathematics, you need to work with two main types of motion graphs: Displacement–Time graphs and Velocity–Time graphs.
Displacement–Time (\(s\)–\(t\)) Graphs
• Gradient: The gradient (slope) of a displacement–time graph gives the velocity.
• A horizontal line means the object is stationary (\(v = 0\)).
• A straight, sloped line represents constant velocity.
• A curved line represents changing velocity, which means the object is accelerating.
Velocity–Time (\(v\)–\(t\)) Graphs
• Gradient: The gradient of a velocity–time graph gives the acceleration (\(a = \frac{\Delta v}{\Delta t}\)).
• Area under the graph: The area bounded by the graph line and the time axis gives the displacement.
• Total Distance: To find the total distance from a velocity–time graph, find the sum of all areas treated as positive values (even parts below the time axis).
Finding Average Values from Graphs
• \(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\)
• \(\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}\)
Memory Trick: Remember the acronym "GAV" for \(v\)–\(t\) graphs: Gradient = Acceleration, Area = Volume of movement (Displacement).
Key Takeaway: For \(s\)–\(t\) graphs, gradient = velocity. For \(v\)–\(t\) graphs, gradient = acceleration and area = displacement.
3. Constant Acceleration: The SUVAT Equations
When an object moves in a straight line with constant (uniform) acceleration, we can use a set of five powerful formulas known as the SUVAT equations.
The Five SUVAT Variables
• \(s\) = Displacement (\(\text{m}\))
• \(u\) = Initial velocity (\(\text{ms}^{-1}\))
• \(v\) = Final velocity (\(\text{ms}^{-1}\))
• \(a\) = Constant acceleration (\(\text{ms}^{-2}\))
• \(t\) = Time taken (\(\text{s}\))
The SUVAT Formulae
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. \(v^2 = u^2 + 2as\)
Step-by-Step Strategy to Solve SUVAT Problems
Step 1: Choose and write down your positive direction (e.g., "Take right as positive").
Step 2: List the letters \(s, u, v, a, t\) down the margin.
Step 3: Fill in the values you know (you need at least 3 known values) and identify what you want to find.
Step 4: Select the equation containing your 3 knowns and your 1 unknown variable.
Step 5: Substitute the numbers carefully and solve for the unknown.
Common Hidden Clues in Exam Questions
• "Starts from rest" \(\implies u = 0\)
• "Comes to rest" / "Brakes to a halt" \(\implies v = 0\)
• "Returns to its original position" \(\implies s = 0\)
Key Takeaway: SUVAT equations only apply when acceleration is constant. If acceleration depends on time (e.g., \(a = 3t\)), you MUST use calculus instead!
4. Vertical Motion Under Gravity
When an object moves vertically in the air without engine power or significant air resistance, it moves under the influence of gravity alone. This is called free fall.
Key Rules for Vertical Motion
• Gravity always acts straight downwards.
• The acceleration due to gravity is denoted by \(g\). In the CCEA specification, \(g = 9.8\text{ ms}^{-2}\) unless specified otherwise.
• Choose a positive direction:
– If you choose upwards as positive: \(a = -9.8\text{ ms}^{-2}\).
– If you choose downwards as positive: \(a = +9.8\text{ ms}^{-2}\).
Special Properties of Projectile Flight
• At the maximum height: The object momentarily stops rising, so its vertical velocity is zero (\(v = 0\)).
• Symmetry of vertical flight: For an object projected vertically upwards from ground level that returns to the same level:
– Time to reach the top = Time to fall back down.
– Landing speed = Initial launch speed (\(v = -u\)).
Example Walkthrough: A ball is thrown vertically upwards from ground level with speed \(19.6\text{ ms}^{-1}\). How long does it take to reach maximum height?
• Take upwards as positive.
• \(u = 19.6\text{ ms}^{-1}\), \(v = 0\text{ ms}^{-1}\) (at max height), \(a = -9.8\text{ ms}^{-2}\), \(t = ?\)
• Using \(v = u + at\):
\(0 = 19.6 + (-9.8)t\)
\(9.8t = 19.6 \implies t = 2\text{ s}\).
Key Takeaway: Always declare your positive direction at the start of a gravity question. If upward is positive, \(a = -9.8\text{ ms}^{-2}\).
5. Variable Acceleration (Calculus in Kinematics)
In the real world, acceleration is rarely constant. A rocket burns fuel and gets lighter, speeding up at a changing rate. When acceleration, velocity, or displacement are expressed as functions of time \(t\), we use differentiation and integration.
The Calculus Relationship
Displacement (\(s\)) \(\xrightarrow{\text{Differentiate}}\) Velocity (\(v\)) \(\xrightarrow{\text{Differentiate}}\) Acceleration (\(a\))
Acceleration (\(a\)) \(\xrightarrow{\text{Integrate}}\) Velocity (\(v\)) \(\xrightarrow{\text{Integrate}}\) Displacement (\(s\))
1. Moving Down (Differentiation)
• Velocity is the derivative of displacement with respect to time:
\(v = \frac{ds}{dt}\)
• Acceleration is the derivative of velocity with respect to time:
\(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\)
2. Moving Up (Integration)
• Velocity from acceleration:
\(v = \int a \, dt\)
• Displacement from velocity:
\(s = \int v \, dt\)
Don't Forget the Constant of Integration (\(+ c\))!
When performing indefinite integration, you will obtain a constant of integration (e.g., \(+ c\)). You must use given initial conditions or boundary conditions (such as "when \(t = 0, v = 4\)") to calculate the numerical value of \(c\).
Finding Maxima and Minima in Kinematics
• Maximum displacement / height occurs when \(\frac{ds}{dt} = v = 0\).
• Maximum velocity / speed occurs when \(\frac{dv}{dt} = a = 0\).
Key Takeaway: Differentiate to go from \(s \to v \to a\). Integrate to go from \(a \to v \to s\), and always solve for \(c\) using the given initial values!
6. Summary of Common Mistakes to Avoid
• Mixing up distance and displacement: Remember that displacement can be negative, but distance is always positive.
• Using SUVAT when acceleration varies: Always check if acceleration is constant! If \(a\) contains \(t\) (like \(a = 6t - 2\)), do not use SUVAT—use integration instead.
• Inconsistent signs in gravity problems: If you decide "up" is positive, both displacement below the release point and gravity must have negative values.
• Forgetting \(+ c\) when integrating: Always write \(+ c\) immediately after integrating, and substitute the given values of \(t\) and \(v\) (or \(s\)) to find it.