Kinematics

Welcome to Kinematics (A2 Applied Mathematics)

Kinematics is the study of how objects move—describing their position, velocity, and acceleration without worrying about the forces that cause the motion. In AS Mathematics, you worked with constant acceleration using the standard SUVAT equations. In A2 Unit A2 2 (Mechanics), we take this to the next level in two major ways:

1. Variable acceleration: What happens when acceleration changes over time? We use calculus (differentiation and integration) to connect displacement, velocity, and acceleration.
2. Motion in two dimensions: We describe motion across a plane using vectors (\(\mathbf{i}\) and \(\mathbf{j}\)) and model projectiles moving under gravity.

Don't worry if this seems like a big step up from AS—once you understand the connections between displacement, velocity, and acceleration, the steps become very logical and repeatable!

Standard CCEA Exam Conventions to Keep in Mind:
• Acceleration due to gravity is taken as \(g = 9.8\text{ m s}^{-2}\) unless stated otherwise.
• Final numerical answers must be rounded to 3 significant figures unless exact forms are requested.
• Always include appropriate SI units (e.g., \(\text{m}\), \(\text{m s}^{-1}\), \(\text{m s}^{-2}\), \(\text{s}\)).

Key Takeaway: If acceleration is constant, you can use SUVAT; if acceleration changes with time \(t\), you must use calculus!

1. Variable Acceleration in 1D: The Calculus Toolkit

When an object's acceleration is not constant, the standard SUVAT equations fail. Instead, we use calculus to transition between position, velocity, and acceleration.

Key Definitions & Relationships

• Displacement (\(s\)): Position relative to a fixed origin. (Vector quantity)
• Velocity (\(v\)): Rate of change of displacement with respect to time: \(v = \frac{ds}{dt}\).
• Speed: The scalar magnitude of velocity: \(\text{Speed} = |v|\).
• Acceleration (\(a\)): Rate of change of velocity with respect to time: \(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\).

Moving Between Quantities

To go from Displacement \(\rightarrow\) Velocity \(\rightarrow\) Acceleration, we differentiate with respect to time \(t\):
\(v = \frac{ds}{dt}\)
\(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\)

To go backwards from Acceleration \(\rightarrow\) Velocity \(\rightarrow\) Displacement, we integrate with respect to time \(t\):
\(v = \int a \, dt + c\)
\(s = \int v \, dt + d\)

Memory Trick: Think of the ladder D-V-A (Displacement, Velocity, Acceleration).
• Going Down the ladder (D to V to A) \(\rightarrow\) Differentiate.
• Going Up the ladder (A to V to D) \(\rightarrow\) Integrate.

Don't Forget the Constant of Integration!

Whenever you integrate, you must add an arbitrary constant of integration (\(+c\)). You will usually be given initial conditions (e.g., "the particle starts from rest when \(t = 0\)" means \(v = 0\) when \(t = 0\)) to calculate the exact numerical value of \(c\).

Worked Example: Variable Acceleration in 1D

A particle moves in a straight line such that its acceleration at time \(t\) seconds is given by \(a = 6t - 4\text{ m s}^{-2}\) for \(t \ge 0\). At \(t = 0\), the particle has a velocity of \(3\text{ m s}^{-1}\) and is located at the origin \(s = 0\). Find its velocity and displacement when \(t = 2\text{ s}\).

Step 1: Find velocity by integrating acceleration:
\(v = \int (6t - 4) \, dt = 3t^2 - 4t + c\)
Using initial conditions: when \(t = 0\), \(v = 3\):
\(3 = 3(0)^2 - 4(0) + c \implies c = 3\)
So, \(v(t) = 3t^2 - 4t + 3\text{ m s}^{-1}\).

When \(t = 2\text{ s}\):
\(v(2) = 3(2)^2 - 4(2) + 3 = 12 - 8 + 3 = 7\text{ m s}^{-1}\).

Step 2: Find displacement by integrating velocity:
\(s = \int (3t^2 - 4t + 3) \, dt = t^3 - 2t^2 + 3t + d\)
Using initial conditions: when \(t = 0\), \(s = 0\):
\(0 = (0)^3 - 2(0)^2 + 3(0) + d \implies d = 0\)
So, \(s(t) = t^3 - 2t^2 + 3t\text{ m}\).

When \(t = 2\text{ s}\):
\(s(2) = (2)^3 - 2(2)^2 + 3(2) = 8 - 8 + 6 = 6\text{ m}\).

Key Takeaway: Differentiate down to find acceleration; integrate up and find the constant \(+ c\) to find velocity and displacement.

2. Displacement vs. Total Distance Travelled

A classic examiner trap in CCEA papers is asking for total distance travelled over a time interval \([t_1, t_2]\) instead of just displacement.

• Displacement: Simply the difference in position: \(s(t_2) - s(t_1)\).
• Distance Travelled: The actual total path length covered by the particle.

If a particle moves forward and then reverses direction, its displacement will decrease, but the total distance travelled keeps increasing! Think of walking 5 metres forward and 3 metres backward: your displacement is \(+2\text{ m}\), but your distance travelled is \(8\text{ m}\).

Method for Total Distance:

1. Find when the particle changes direction by setting velocity \(v = 0\) and solving for \(t\).
2. Check if any turning points lie inside your given time interval.
3. Calculate the position \(s\) at the start time, at every turning point, and at the end time.
4. Sum the absolute distances between each consecutive position.

Common Pitfall Alert:

Never just calculate \(s(t_{\text{end}}) - s(t_{\text{start}})\) when asked for total distance! Always check for turning points where \(v = 0\).

Key Takeaway: Turning points occur when \(v = 0\). Split the motion into separate intervals around these points to calculate total distance.

3. Kinematics with Vectors (2D Motion)

In two dimensions, motion is expressed using unit vectors \(\mathbf{i}\) (horizontal / East) and \(\mathbf{j}\) (vertical / North), or as column vectors.

Vector Calculus Relations

• Position vector: \(\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\)
• Velocity vector: \(\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \dot{x}(t)\mathbf{i} + \dot{y}(t)\mathbf{j} = \begin{pmatrix} \dot{x}(t) \\ \dot{y}(t) \end{pmatrix}\)
• Acceleration vector: \(\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2} = \ddot{x}(t)\mathbf{i} + \ddot{y}(t)\mathbf{j} = \begin{pmatrix} \ddot{x}(t) \\ \ddot{y}(t) \end{pmatrix}\)

When integrating with vectors, remember that the constant of integration is also a vector constant: \(\mathbf{c} = c_1\mathbf{i} + c_2\mathbf{j}\).

Magnitude and Direction (Speed & Bearings)

Calculus gives you the vector components \(\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\). Exam questions often ask for the speed and direction of motion:

• Speed: The magnitude of velocity:
\(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\)

• Direction / Angle (\(\theta\)): The angle with the horizontal or \(\mathbf{i}\) direction is given by:
\(\tan\theta = \left|\frac{v_y}{v_x}\right|\)

• Bearings: Measured clockwise from North (\(000^\circ\) to \(360^\circ\)). Sketch a quick diagram showing the \(\mathbf{i}\) (East) and \(\mathbf{j}\) (North) components to find the three-figure bearing accurately.

Worked Example: Vector Kinematics

A particle moves in a horizontal plane such that its velocity at time \(t\) seconds is given by \(\mathbf{v} = (3t^2 - 6)\mathbf{i} + (4t)\mathbf{j}\text{ m s}^{-1}\).
(a) Find the acceleration vector when \(t = 3\text{ s}\).
(b) Find the speed of the particle when \(t = 2\text{ s}\).

Solution (a):
Differentiate \(\mathbf{v}\) with respect to \(t\):
\(\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3t^2 - 6)\mathbf{i} + \frac{d}{dt}(4t)\mathbf{j} = (6t)\mathbf{i} + 4\mathbf{j}\text{ m s}^{-2}\)
When \(t = 3\text{ s}\):
\(\mathbf{a} = 6(3)\mathbf{i} + 4\mathbf{j} = 18\mathbf{i} + 4\mathbf{j}\text{ m s}^{-2}\).

Solution (b):
Substitute \(t = 2\text{ s}\) into \(\mathbf{v}\):
\(\mathbf{v} = (3(2)^2 - 6)\mathbf{i} + 4(2)\mathbf{j} = (12 - 6)\mathbf{i} + 8\mathbf{j} = 6\mathbf{i} + 8\mathbf{j}\text{ m s}^{-1}\)
Calculate magnitude (speed):
\(\text{Speed} = |\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ m s}^{-1}\).

Key Takeaway: Treat the \(\mathbf{i}\) and \(\mathbf{j}\) components independently when differentiating and integrating, then use Pythagoras to find magnitudes.

4. Projectiles: 2D Motion Under Gravity

A projectile is an object launched into the air that moves under the sole influence of gravity (assuming air resistance is negligible).

The Fundamental Principle of Projectiles

Horizontal motion and vertical motion are completely independent of each other:
• Horizontal Motion: Zero acceleration (\(a_x = 0\)). Velocity is constant throughout flight.
• Vertical Motion: Constant acceleration downwards due to gravity (\(a_y = -g = -9.8\text{ m s}^{-2}\)).

Initial Velocity Components

If a particle is projected with initial speed \(U\) at an angle \(\theta\) above the horizontal:
• Horizontal initial velocity: \(u_x = U\cos\theta\)
• Vertical initial velocity: \(u_y = U\sin\theta\)

Equations of Motion at Time \(t\)

Horizontal (Constant Speed):
• Velocity: \(v_x = U\cos\theta\)
• Displacement: \(x = (U\cos\theta)t\)

Vertical (Constant Acceleration \(a_y = -g\)):
• Velocity: \(v_y = U\sin\theta - gt\)
• Displacement: \(y = (U\sin\theta)t - \frac{1}{2}gt^2\)
• Velocity-displacement relation: \(v_y^2 = (U\sin\theta)^2 - 2gy\)

5. Key Projectile Landmarks & Trajectory Equations

Many exam questions focus on specific points during a projectile's flight.

1. Maximum Height (\(H_{\text{max}}\))

At the highest point, the vertical component of velocity is momentarily zero: \(v_y = 0\).
• Time to reach maximum height:
\(0 = U\sin\theta - gt \implies t_{\text{top}} = \frac{U\sin\theta}{g}\)
• Maximum height above projection point:
\(v_y^2 = u_y^2 - 2gy \implies 0 = (U\sin\theta)^2 - 2gH \implies H_{\text{max}} = \frac{U^2\sin^2\theta}{2g}\)

2. Time of Flight (\(T\)) and Horizontal Range (\(R\))

For a projectile returning to its original launch height (where \(y = 0\)):
• Setting \(y = (U\sin\theta)t - \frac{1}{2}gt^2 = 0 \implies t(U\sin\theta - \frac{1}{2}gt) = 0\):
Total time of flight: \(T = \frac{2U\sin\theta}{g}\)
• Horizontal range \(R\) (substituting \(T\) into the horizontal distance equation):
\(R = (U\cos\theta)T = (U\cos\theta)\left(\frac{2U\sin\theta}{g}\right) = \frac{U^2(2\sin\theta\cos\theta)}{g} = \frac{U^2\sin(2\theta)}{g}\)

3. The Cartesian Equation of the Trajectory

By eliminating the parameter \(t\) between the horizontal and vertical equations, we obtain the path (parabola) in terms of \(x\) and \(y\):
From \(x = (U\cos\theta)t\), we get \(t = \frac{x}{U\cos\theta}\).
Substitute \(t\) into \(y = (U\sin\theta)t - \frac{1}{2}gt^2\):
\(y = (U\sin\theta)\left(\frac{x}{U\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{U\cos\theta}\right)^2\)
Using \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and \(\frac{1}{\cos^2\theta} = 1 + \tan^2\theta\):
\(y = x\tan\theta - \frac{gx^2}{2U^2\cos^2\theta} = x\tan\theta - \frac{gx^2(1 + \tan^2\theta)}{2U^2}\)

This formula is particularly powerful when solving for the projection angle \(\theta\) given a target coordinate \((x, y)\), as it forms a quadratic equation in \(\tan\theta\)!

Worked Example: Projectile Calculation

A ball is projected from ground level with speed \(28\text{ m s}^{-1}\) at an angle of \(30^\circ\) above the horizontal. Taking \(g = 9.8\text{ m s}^{-2}\), find:
(a) The time taken to reach the maximum height.
(b) The horizontal range of the ball.

Solution:
Initial components:
\(u_x = 28\cos(30^\circ) = 28 \times \frac{\sqrt{3}}{2} = 14\sqrt{3}\text{ m s}^{-1}\)
\(u_y = 28\sin(30^\circ) = 28 \times \frac{1}{2} = 14\text{ m s}^{-1}\)

(a) Maximum height occurs when \(v_y = 0\):
\(v_y = u_y - gt \implies 0 = 14 - 9.8t\)
\(t = \frac{14}{9.8} = \frac{10}{7} \approx 1.43\text{ s}\) (to 3 s.f.).

(b) Total time of flight \(T = 2 \times \frac{10}{7} = \frac{20}{7}\text{ s}\):
\(R = u_x \times T = (14\sqrt{3}) \times \left(\frac{20}{7}\right) = 40\sqrt{3} \approx 69.3\text{ m}\) (to 3 s.f.).

Key Takeaway: Always separate the horizontal (constant speed) and vertical (constant acceleration under gravity) parts of the motion first!

6. Summary & Quick Review Checklist

Before sitting your CCEA Unit A2 2 exam, make sure you are confident with each of the following points:

• Calculus in 1D:
- Differentiate: \(s \rightarrow v \rightarrow a\)
- Integrate: \(a \rightarrow v \rightarrow s\) (always find \(+ c\) using boundary conditions)
- Total distance requires finding turning points (\(v = 0\)).

• 2D Vectors:
- \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), \(\mathbf{v} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}\), \(\mathbf{a} = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}\)
- \(\text{Speed} = \sqrt{v_x^2 + v_y^2}\)

• Projectiles:
- Horizontally: \(a_x = 0 \implies x = (U\cos\theta)t\)
- Vertically: \(a_y = -9.8\text{ m s}^{-2} \implies y = (U\sin\theta)t - 4.9t^2\)
- At max height: \(v_y = 0\)
- Landing below launch level: \(y\) is negative!

• Exam Standards: Use \(g = 9.8\text{ m s}^{-2}\), provide answers to 3 significant figures, and always include units.