Welcome to Kinematics! Your Guide to Motion
Hello! If you're studying M2 Mechanics, you're about to delve into Kinematics—the geometry of motion. Simply put, Kinematics is the mathematical description of how objects move, without worrying why they move (that's the job of Forces and Newton's Laws!).
This chapter connects many parts of your curriculum: basic algebra, problem-solving, graph sketching, and crucially, AS Pure Maths calculus (Differentiation and Integration).
Don't worry if this seems tricky at first. We will break down motion into two simple categories: constant acceleration and variable acceleration, using clear steps for both!
1. The Kinematics Foundation: Key Definitions
In mechanics, precise language is vital. Distance and displacement are not the same! Speed and velocity are not the same!
1.1 Scalars vs. Vectors
A scalar quantity only has magnitude (size).
A vector quantity has both magnitude and direction.
- Distance (Scalar): The total path length travelled. Example: I walked 5 km.
- Displacement (Vector, $s$): The shortest path length from the start point to the end point. Example: I am 2 km East of my starting position.
Analogy: If you jog 5 times around a 1 km track, your distance travelled is 5 km, but your final displacement is 0 km (you ended where you started!).
- Speed (Scalar): Rate of change of distance.
- Velocity (Vector, $v$): Rate of change of displacement.
- Acceleration (Vector, $a$): Rate of change of velocity. If acceleration is constant, it means the velocity changes by the same amount every second.
Quick Review: Fundamental Variables
\(s\): Displacement (m)
\(u\): Initial Velocity (\(\text{ms}^{-1}\))
\(v\): Final Velocity (\(\text{ms}^{-1}\))
\(a\): Constant Acceleration (\(\text{ms}^{-2}\))
\(t\): Time (s)
2. Motion with Constant Acceleration (SUVAT) (M1.1)
When an object is moving in a straight line with a constant acceleration, we use the five standard Kinematic (or SUVAT) equations. You only need three known values to find any of the other two unknowns.
2.1 The Constant Acceleration Equations
These formulae relate the five variables defined above. They are only valid when the acceleration, $a$, is constant.
- \(\boldsymbol{v = u + at}\)
- \(\boldsymbol{s = ut + \frac{1}{2}at^2}\)
- \(\boldsymbol{s = vt - \frac{1}{2}at^2}\)
- \(\boldsymbol{v^2 = u^2 + 2as}\)
- \(\boldsymbol{s = \frac{1}{2}(u + v)t}\)
Memory Aid: Choosing the Right Equation
To choose which formula to use, look for the variable that is NOT involved in your problem (the "missing" variable).
- Formula 1 (\(s\) missing): Use if you don't care about displacement.
- Formula 2 (\(v\) missing): Use if you don't know or don't need the final velocity.
- Formula 3 (\(u\) missing): Use if you don't know or don't need the initial velocity.
- Formula 4 (\(t\) missing): Use if you don't know or don't need the time.
- Formula 5 (\(a\) missing): Use if you don't know or don't need the acceleration.
2.2 Vertical Motion Under Gravity
Motion due to gravity is a special case of constant acceleration in a straight line (the vertical line). When an object is thrown or dropped, the only force acting on it (ignoring air resistance) is gravity.
The acceleration due to gravity is denoted by \(g\), and you must use the standard value: \(g = 9.8 \text{ ms}^{-2}\).
Crucial Step: Always decide which direction is positive!
- If you take UP as positive, then \(a = -g = -9.8 \text{ ms}^{-2}\).
- If you take DOWN as positive, then \(a = +g = +9.8 \text{ ms}^{-2}\).
Common Mistake to Avoid: When a particle reaches its maximum height, its velocity (\(v\)) is instantaneously zero, but its acceleration is still \(g\) (it's always accelerating downwards!).
Key Takeaway: Constant Acceleration
If the acceleration is constant, list your SUVAT variables, establish a positive direction, and pick the equation that skips the variable you don't need.
3. Kinematics Graphs (M1.1)
Graphs are essential for visualizing motion and for solving problems, especially those involving multiple stages of movement.
3.1 Displacement-Time (\(s-t\)) Graphs
- The gradient of an \(s-t\) graph represents the velocity.
- A straight, positive gradient means constant positive velocity.
- A curve means the velocity is changing (there is acceleration).
- If the graph is horizontal, the object is stationary (\(v=0\)).
3.2 Velocity-Time (\(v-t\)) Graphs
This is the most important graph in Kinematics!
- The gradient of a \(v-t\) graph represents the acceleration (\(a = \frac{dv}{dt}\)).
- The area under the curve of a \(v-t\) graph represents the displacement (\(s = \int v \, dt\)).
Tip: If the acceleration is constant (like in SUVAT), the \(v-t\) graph will be a straight line, making the area calculation easy (often using trapeziums or triangles).
3.3 Acceleration-Time (\(a-t\)) Graphs
- The area under the curve of an \(a-t\) graph represents the change in velocity (\(\Delta v = \int a \, dt\)).
Key Takeaway: Kinematic Graphs
Remember the hierarchy: Position/Displacement \(\rightarrow\) Velocity \(\rightarrow\) Acceleration.
Gradient moves you DOWN the hierarchy (e.g., \(s\) to \(v\)).
Area moves you UP the hierarchy (e.g., \(a\) to \(v\), or \(v\) to \(s\)).
4. Motion with Variable Acceleration (Calculus) (M1.2)
If the acceleration is not constant, you cannot use the SUVAT equations. Instead, the position, velocity, and acceleration are related functions of time, and we must use Differentiation and Integration. This requires knowledge from AS Unit P1.
4.1 Moving Down the Hierarchy (Differentiation)
If displacement \(s\) is a function of time \(t\), then:
- Velocity is the rate of change of displacement:
\(\boldsymbol{v = \frac{ds}{dt}}\) - Acceleration is the rate of change of velocity:
\(\boldsymbol{a = \frac{dv}{dt} = \frac{d^2s}{dt^2}}\)
Example: If the position of a particle is given by \(s(t) = 3t^2 - 5t\), then the velocity is \(v(t) = \frac{ds}{dt} = 6t - 5\), and the acceleration is \(a(t) = \frac{dv}{dt} = 6\) (constant acceleration in this case).
4.2 Moving Up the Hierarchy (Integration)
Integration is the reverse process of differentiation.
- Velocity is the integral of acceleration with respect to time:
\(\boldsymbol{v = \int a \, dt}\) - Displacement is the integral of velocity with respect to time:
\(\boldsymbol{s = \int v \, dt}\)
CRITICAL POINT: The Constant of Integration (\(+c\))
When you integrate, you MUST include the constant of integration, \(c\). This constant is found by using the initial conditions (the position or velocity when \(t=0\)).
Step-by-Step Integration:
1. Integrate the expression for acceleration to get the velocity function, including \((+c)\).
2. Use the given initial velocity (e.g., \(v=u\) when \(t=0\)) to find \(c\).
3. Integrate the completed velocity function to find the displacement function, including a new constant \(c'\).
4. Use the given initial displacement (e.g., \(s=0\) when \(t=0\)) to find \(c'\).
Key Takeaway: Variable Acceleration
Calculus is your tool when acceleration changes. If you are given acceleration and need displacement, you must integrate twice and use the initial conditions (at \(t=0\)) to find your constants of integration.
5. Vector Kinematics (Motion in 2D or 3D) (M2.2)
So far, we have only looked at motion along a straight line. Now we extend these ideas using vectors to handle motion in a plane (\(\mathbf{i}\) and \(\mathbf{j}\) components) or in space (\(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) components).
5.1 Position, Velocity, and Acceleration Vectors
The position of a particle is given by the position vector \(\mathbf{r}\), which is usually a function of time, \(t\).
Position: \(\boldsymbol{r(t) = f(t)i + g(t)j + h(t)k}\)
The core principle remains: differentiation and integration work exactly the same way, but applied component by component.
Differentiation (Finding velocity and acceleration)
To find velocity \(\mathbf{v}\), differentiate each component of \(\mathbf{r}\):
\(\boldsymbol{v = \frac{dr}{dt} = f'(t)i + g'(t)j + h'(t)k}\)
To find acceleration \(\mathbf{a}\), differentiate each component of \(\mathbf{v}\):
\(\boldsymbol{a = \frac{dv}{dt} = f''(t)i + g''(t)j + h''(t)k}\)
Integration (Finding velocity and position)
To find velocity from acceleration, integrate each component of \(\mathbf{a}\):
\(\boldsymbol{v = \int a \, dt}\) (Remember a constant vector of integration, e.g., \(\mathbf{C} = c_1\mathbf{i} + c_2\mathbf{j} + c_3\mathbf{k}\))
To find position from velocity, integrate each component of \(\mathbf{v}\):
\(\boldsymbol{r = \int v \, dt}\)
Step Tip: When solving vector integration problems, you essentially solve three separate 1D problems simultaneously (one for \(\mathbf{i}\), one for \(\mathbf{j}\), and one for \(\mathbf{k}\)), using the initial conditions (e.g., initial position or velocity vector) to find your constants.
5.2 Finding Speed (Magnitude of Velocity)
When you have the velocity vector \(\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\) (in 2D), the speed is the magnitude of the velocity vector, found using Pythagoras' Theorem:
\(\boldsymbol{\text{Speed } = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2}}\)
For 3D motion, the formula extends to:
\(\boldsymbol{|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}}\)
Did you know? This principle is crucial in more advanced mechanics topics, such as Projectile Motion, where the horizontal motion is usually constant velocity (\(a_x=0\)) and the vertical motion is constant acceleration (\(a_y = -g\)). All vector kinematics problems build on this framework!
Key Takeaway: Vector Kinematics
Vector problems are just 1D problems separated by components. Apply differentiation or integration to the \(\mathbf{i}\) terms, the \(\mathbf{j}\) terms, and the \(\mathbf{k}\) terms individually. Speed is the magnitude of the velocity vector.
Chapter Summary
You now have the tools to describe any motion, whether simple or complex:
- Constant Acceleration: Use the powerful SUVAT equations.
- Variable Acceleration (1D): Use Pure Maths calculus. Differentiate to go from displacement to velocity to acceleration. Integrate (remembering \(+c\)) to go backward.
- Variable Acceleration (2D/3D): Use vectors. Apply calculus component-wise, and use Pythagoras to find the speed.
Practice identifying whether acceleration is constant or variable before you start solving. Good luck! You've got this!