Study Notes: Kinematics of a Particle (M2)
Hello future Mathematician! Welcome to Kinematics in M2. If you enjoyed the basics of motion in M1, get ready for the upgrade! In M1, acceleration was always constant. Here in M2, we explore the exciting world of variable acceleration, meaning acceleration changes over time.
This is where Calculus truly shines in Mechanics. Don't worry if differentiation and integration seem challenging; we’ll break down exactly how these tools allow us to predict the precise position, velocity, and acceleration of a moving particle at any moment.
Key Skill: You will be using differentiation and integration extensively to move between displacement, velocity, and acceleration.
1. Kinematics of a Particle Moving in a Straight Line
This section deals with motion along a single axis (like a car driving straight along a road). The crucial difference from M1 is that the acceleration, \(a\), is now typically a function of time, \(t\).
1.1 Defining the Core Variables
For a particle moving in a straight line, we define three related quantities:
- Displacement (\(x\) or \(s\)): The position of the particle relative to a fixed origin.
- Velocity (\(v\)): The rate of change of displacement. It has magnitude (speed) and direction.
- Acceleration (\(a\)): The rate of change of velocity.
The Calculus Relationships (The Kinematic Bridge)
This is the most important concept in M2 Kinematics.
Memory Aid: Think D-V-A. Differentiation moves you Down (Displacement to Velocity to Acceleration). Integration moves you Up.
1. Moving Down (Differentiation)
If you know the displacement \(x\) as a function of time \(t\), you can find the velocity \(v\) by differentiating:
$$v = \frac{\mathrm{d}x}{\mathrm{d}t}$$
And you can find the acceleration \(a\) by differentiating the velocity \(v\):
$$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2}$$
Example: If the displacement is given by \(x = 2t^3 - 5t\), then the velocity is \(v = 6t^2 - 5\) and the acceleration is \(a = 12t\).
2. Moving Up (Integration)
If you know the acceleration \(a\), you can find the velocity \(v\) by integrating:
$$v = \int a \, \mathrm{d}t$$
And you can find the displacement \(x\) by integrating the velocity \(v\):
$$x = \int v \, \mathrm{d}t$$
1.2 Finding the Constant of Integration (\(C\))
Attention: Integration always introduces an unknown Constant of Integration. In mechanics, this constant represents the initial conditions of the particle (its starting velocity, or starting position).
You must use the information given in the problem (e.g., "The particle starts from rest," or "When \(t=0\), \(x=4\)") to find this constant immediately after integrating.
Step-by-Step: Acceleration to Displacement
- Start with \(a = f(t)\).
- Integrate \(a\) to find \(v\). Include \(+ C_1\).
- Use the initial velocity condition (usually at \(t=0\)) to find \(C_1\).
- Integrate the resulting expression for \(v\) to find \(x\). Include \(+ C_2\).
- Use the initial displacement condition (usually at \(t=0\)) to find \(C_2\).
Quick Review: You need one piece of initial information for every integration you perform.
1.3 Finding Maximum Speed and Turning Points
In straight-line motion, a particle reverses its direction when its velocity is zero. These points are crucial for finding distance traveled.
-
To Find Times When Direction Changes:
Set the velocity function to zero: \(\mathbf{v = 0}\). Solve for \(t\). -
To Find Maximum or Minimum Displacement/Velocity:
Use the rules of maxima/minima from Pure Mathematics: set the derivative to zero.- To find maximum/minimum velocity, set acceleration to zero (\(a = \frac{\mathrm{d}v}{\mathrm{d}t} = 0\)).
- To find maximum/minimum displacement, set velocity to zero (\(v = \frac{\mathrm{d}x}{\mathrm{d}t} = 0\)).
1.4 Displacement vs. Distance Traveled
This is a common trap for students!
- Displacement is the straight-line difference between the start point and the end point. (It can be negative.)
- Distance Traveled is the total length of the path taken. (It must be positive.)
If a particle changes direction during the time interval (i.e., \(v=0\) at some point \(t\) within the interval), the distance traveled is NOT the same as the final displacement.
Process for Finding Distance Traveled:
- Find the times \(t\) when \(v=0\).
- If these turning points fall within the required time interval, calculate the displacement \(x\) at each turning point, the start time, and the end time.
- Calculate the absolute displacement for each segment of the journey, and add the magnitudes together.
Analogy: If you start at 0, walk to 5, and then walk back to 2. Your final displacement is 2. Your distance traveled is \(5 + 3 = 8\).
Key Takeaway for Straight Line Motion: The relationship between \(x\), \(v\), and \(a\) is governed entirely by differentiation and integration. Always check initial conditions to find the constants.
2. Vector Kinematics (Motion in a Plane)
Sometimes, particles don't just move in a straight line; they move in a 2D plane (like a fly buzzing around a room). We describe this motion using vectors, specifically the unit vectors \(\mathbf{i}\) (horizontal component) and \(\mathbf{j}\) (vertical component).
2.1 Position, Velocity, and Acceleration Vectors
The principles of differentiation and integration remain exactly the same, but they must be applied independently to the \(\mathbf{i}\) component and the \(\mathbf{j}\) component.
Position Vector (\(\mathbf{r}\))
The position \(\mathbf{r}\) of a particle is given relative to an origin \(O\):
$$\mathbf{r} = x\mathbf{i} + y\mathbf{j}$$
Velocity Vector (\(\mathbf{v}\))
Velocity is the rate of change of the position vector.
$$\mathbf{v} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)\mathbf{i} + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)\mathbf{j} = v_x\mathbf{i} + v_y\mathbf{j}$$
Acceleration Vector (\(\mathbf{a}\))
Acceleration is the rate of change of the velocity vector.
$$\mathbf{a} = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = \left(\frac{\mathrm{d}v_x}{\mathrm{d}t}\right)\mathbf{i} + \left(\frac{\mathrm{d}v_y}{\mathrm{d}t}\right)\mathbf{j}$$
2.2 Vector Integration
To go from acceleration to velocity, or velocity to position, you integrate each component separately.
Crucial Point: When integrating vector functions, the constant of integration (\(\mathbf{C}\)) is also a vector!
Example: If \(\mathbf{a} = (6t)\mathbf{i} + (2)\mathbf{j}\), then:
$$\mathbf{v} = \int \mathbf{a} \, \mathrm{d}t = (3t^2)\mathbf{i} + (2t)\mathbf{j} + \mathbf{C}$$
If you are told that at \(t=0\), \(\mathbf{v} = 4\mathbf{i} - 1\mathbf{j}\), substitute \(t=0\):
\(4\mathbf{i} - 1\mathbf{j} = (0)\mathbf{i} + (0)\mathbf{j} + \mathbf{C}\). Therefore, \(\mathbf{C} = 4\mathbf{i} - 1\mathbf{j}\).
Common Mistake to Avoid: Forgetting to apply the initial vector conditions to both the \(\mathbf{i}\) and \(\mathbf{j}\) components of the constant vector \(\mathbf{C}\).
2.3 Magnitude and Direction
While \(\mathbf{v}\) is the velocity vector, the speed is the magnitude of the velocity vector. This is found using Pythagoras’ theorem.
If \(\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\), then:
Speed \(|\mathbf{v}|\):
$$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$$
The magnitude of the acceleration is found similarly: \(|\mathbf{a}| = \sqrt{a_x^2 + a_y^2}\).
Finding the Direction of Motion
The direction of motion is always the direction of the velocity vector \(\mathbf{v}\). We usually define this direction as the angle (\(\theta\)) the vector makes with the positive \(\mathbf{i}\) direction (the positive x-axis).
If \(\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\), then:
$$\tan \theta = \frac{v_y}{v_x}$$
Did you know? Speed is a scalar quantity (just a number, like 50 km/h), but velocity is a vector quantity (like 50 km/h due North-East).
2.4 Solving Simultaneous Vector Problems
Vector problems often involve finding when a particle is moving parallel to \(\mathbf{i}\) or \(\mathbf{j}\), or when two particles collide.
- Parallel to \(\mathbf{i}\) (Horizontal): This means the vertical component of the required vector is zero. Set the \(\mathbf{j}\) component to zero. (e.g., \(v_y = 0\)).
- Parallel to \(\mathbf{j}\) (Vertical): This means the horizontal component of the required vector is zero. Set the \(\mathbf{i}\) component to zero. (e.g., \(v_x = 0\)).
- Collision: Two particles, A and B, collide if their position vectors are identical at the same time \(t\). Set \(\mathbf{r}_A = \mathbf{r}_B\). This gives two equations (one for \(\mathbf{i}\) and one for \(\mathbf{j}\)) that must be satisfied by the same value of \(t\).
Key Takeaway for Vector Kinematics: Treat the horizontal (\(\mathbf{i}\)) and vertical (\(\mathbf{j}\)) components as two separate, independent straight-line kinematics problems. Remember that the constant of integration is a vector.
Final Review and Encouragement
You’ve covered the entire M2 Kinematics syllabus! The key to mastering this chapter is practice with differentiation and integration.
Quick Checklist:
- \(x \xrightarrow{\text{d/d}t} v \xrightarrow{\text{d/d}t} a\)
- \(a \xrightarrow{\int \mathrm{d}t} v \xrightarrow{\int \mathrm{d}t} x\)
- Always find the Constant of Integration using initial conditions (\(t=0\)).
- Distance traveled requires checking if \(v=0\) (turning points).
- In 2D (Vector Kinematics), apply all calculus rules separately to the \(\mathbf{i}\) and \(\mathbf{j}\) components.
- Speed is the magnitude of velocity: \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\).
Don't worry if integrating to find position seems daunting at first; the more you practice finding those vector constants, the more natural it will become. Keep up the excellent work!