Welcome to Further Kinematics!
Welcome to one of the most exciting and dynamic areas of Mechanics 2. In single-variable A-Level maths, you explored motion under constant acceleration using the familiar constant acceleration formulae (the classic 'suvat' equations). But what happens in the real world when acceleration is constantly shifting, changing with time, velocity, or position?
In Further Kinematics, we unlock the full toolkit of calculus and vectors. We will learn how to describe motion in one and two dimensions when forces vary, how to solve differential equations for velocity and displacement, and how to analyze the relative motion of two moving objects (such as two ships tracking each other across the open ocean). Don't worry if this seems a bit daunting at first—we will break down every single concept into bite-sized, friendly steps!
1. The Foundations: Calculus with Variable Acceleration
When acceleration is not constant, the standard constant acceleration formulae cannot be used. Instead, we return to the fundamental definitions of velocity and acceleration through differentiation and integration.
The Direction of Calculus: Differentiation vs Integration
Think of your motion variables as a three-story building:
Top floor: Displacement \(x\) (or position vector \(\mathbf{r}\))
Middle floor: Velocity \(v\) (or velocity vector \(\mathbf{v}\))
Bottom floor: Acceleration \(a\) (or acceleration vector \(\mathbf{a}\))
Moving Down (Differentiating with respect to time \(t\)):
\(v = \frac{\mathrm{d}x}{\mathrm{d}t}\)
\(a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\)
Moving Up (Integrating with respect to time \(t\)):
\(v = \int a \, \mathrm{d}t\)
\(x = \int v \, \mathrm{d}t\)
Memory Trick: Remember Down = Differentiate, In = Integrate (going up/inside). Always remember to include the constant of integration \(+ \, c\) and use initial conditions (boundary conditions like \(t = 0, v = u\)) to find its value!
Key Takeaway
Whenever acceleration depends on time \(t\), differentiate to find rates of change, and integrate to find accumulated quantities. Never use constant acceleration formulae unless you are 100% sure the acceleration is a fixed constant number!
2. Acceleration as a Function of Velocity or Displacement
In Mechanics 2, acceleration is often not given as a function of time \(t\). For example, air resistance depends on speed \(v\), or a spring's restoring force depends on displacement \(x\). How do we handle these?
Form 1: Acceleration as a function of velocity, \(a = f(v)\)
Since \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\), we can set up a first-order separable differential equation:
\(\frac{\mathrm{d}v}{\mathrm{d}t} = f(v)\)
Separating variables gives:
\(\int \frac{1}{f(v)} \, \mathrm{d}v = \int 1 \, \mathrm{d}t = t + c\)
If you need to connect velocity \(v\) and displacement \(x\) when \(a = f(v)\), use the chain rule variation described below.
Form 2: Acceleration as a function of displacement, \(a = f(x)\)
By the chain rule of calculus:
\(a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t} = v \frac{\mathrm{d}v}{\mathrm{d}x}\)
This gives us the crucial identity:
\(a = v \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v^2\right)\)
Setting this equal to \(f(x)\):
\(v \frac{\mathrm{d}v}{\mathrm{d}x} = f(x) \implies \int v \, \mathrm{d}v = \int f(x) \, \mathrm{d}x\)
\(\frac{1}{2}v^2 = \int f(x) \, \mathrm{d}x + c\)
Did you know? The expression \(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}m v^2\right) = m a = F\) is directly linked to the Work-Energy Principle! Integrating both sides gives Work Done \(=\) Change in Kinetic Energy.
Step-by-Step Decision Rule:
1. If \(a\) is given in terms of \(t\): Use \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\) and integrate with respect to \(t\).
2. If \(a\) is given in terms of \(v\) and you want \(t\): Use \(\frac{\mathrm{d}v}{\mathrm{d}t} = f(v)\) and separate variables.
3. If \(a\) is given in terms of \(x\) (or you want to connect \(v\) and \(x\)): Use \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\) and separate variables.
Key Takeaway
Memorize the identity \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\). It is the golden key for solving any problem where acceleration depends on position or where time \(t\) is not directly involved.
3. Vectors in 2D Kinematics
In two dimensions, motion is expressed using Cartesian unit vectors \(\mathbf{i}\) (horizontal) and \(\mathbf{j}\) (vertical), or column vectors.
Vector Relationships
Let position be \(\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}\).
Velocity vector: \(\mathbf{v}(t) = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}\)
Acceleration vector: \(\mathbf{a}(t) = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}\)
Magnitudes and Directions
Students often mix up velocity (a vector) with speed (a scalar), or displacement with distance. Keep these distinctions clear:
- Speed is the magnitude of the velocity vector: \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\)
- Distance from the origin is the magnitude of the position vector: \(|\mathbf{r}| = \sqrt{x^2 + y^2}\)
- Direction of motion is the angle that the velocity vector \(\mathbf{v}\) makes with a given axis or reference line.
Angle \(\theta\) above the positive \(x\)-axis: \(\tan \theta = \frac{v_y}{v_x}\)
Bearing (measured clockwise from North/\(\mathbf{j}\)): Draw a quick vector diagram to identify the quadrant correctly!
Key Takeaway
When dealing with 2D vectors, you can differentiate and integrate the \(\mathbf{i}\) and \(\mathbf{j}\) components separately. Always calculate the final magnitude using Pythagoras' Theorem if a question asks for speed or distance.
4. Relative Velocity
Imagine you are sitting in a train traveling at \(60\text{ km/h}\), and another train passes you in the same direction at \(70\text{ km/h}\). To you, the other train appears to be moving forward slowly at just \(10\text{ km/h}\). This is the intuitive core of relative velocity.
The Relative Velocity Equation
The velocity of an object \(A\) relative to an object \(B\) is denoted by \(_{A}\mathbf{v}_{B}\) (or \(\mathbf{v}_{A/B}\)):
\(_{A}\mathbf{v}_{B} = \mathbf{v}_A - \mathbf{v}_B\)
Similarly, the position of \(A\) relative to \(B\) is:
\(_{A}\mathbf{r}_{B} = \mathbf{r}_A - \mathbf{r}_B\)
Helpful Translation: "Velocity of \(A\) relative to \(B\)" simply means: What velocity does \(A\) appear to have if you pretend that \(B\) is completely stationary?
Key Vector Form for Constant Velocity:
If objects \(A\) and \(B\) move with constant velocities \(\mathbf{v}_A\) and \(\mathbf{v}_B\), and start with initial positions \(\mathbf{r}_{A0}\) and \(\mathbf{r}_{B0}\) at time \(t = 0\):
\(\mathbf{r}_A(t) = \mathbf{r}_{A0} + \mathbf{v}_A t\)
\(\mathbf{r}_B(t) = \mathbf{r}_{B0} + \mathbf{v}_B t\)
The relative position vector at time \(t\) is:
\(_{A}\mathbf{r}_{B}(t) = \mathbf{r}_A(t) - \mathbf{r}_B(t) = (\mathbf{r}_{A0} - \mathbf{r}_{B0}) + (\mathbf{v}_A - \mathbf{v}_B)t\)
Key Takeaway
To find the relative motion vector "of \(A\) relative to \(B\)", always subtract: \(A \text{ minus } B\). Keep this order consistent across all terms.
5. Interception and Closest Approach
Condition for Collision (Interception)
Two objects \(A\) and \(B\) will collide if there is some time \(T > 0\) such that their positions are identical:
\(\mathbf{r}_A(T) = \mathbf{r}_B(T) \implies \mathbf{r}_A(T) - \mathbf{r}_B(T) = \mathbf{0}\)
In terms of relative motion, this means the relative velocity \(_{A}\mathbf{v}_{B}\) must point directly along the line from \(B\)'s initial position towards \(A\)'s initial position (in the opposite direction of \(_{A}\mathbf{r}_{B}(0)\)).
Finding the Closest Approach (Shortest Distance)
If two objects do not collide, they will reach a point where the distance between them is at a minimum. There are two standard approaches to find this minimum distance and the time it occurs:
Method 1: Algebraic / Calculus Method (Recommended & Reliable)
1. Write down the relative position vector: \(\mathbf{d}(t) = \mathbf{r}_A(t) - \mathbf{r}_B(t) = X(t)\mathbf{i} + Y(t)\mathbf{j}\).
2. Find the square of the distance: \(D^2 = |\mathbf{d}(t)|^2 = [X(t)]^2 + [Y(t)]^2\).
3. To minimize \(D\), we minimize \(D^2\) (this avoids messy square roots!). Differentiate with respect to \(t\):
\(\frac{\mathrm{d}(D^2)}{\mathrm{d}t} = 0\)
4. Solve for \(t\). This gives the exact time of closest approach.
5. Substitute \(t\) back into \(D = \sqrt{X(t)^2 + Y(t)^2}\) to find the shortest distance.
Method 2: Geometric / Vector Projection Method
1. Treat object \(B\) as stationary at the origin of a relative frame.
2. Object \(A\) starts at relative position \(\mathbf{d}_0 = \mathbf{r}_{A0} - \mathbf{r}_{B0}\) and travels in the straight-line direction of relative velocity \(\mathbf{v}_{\text{rel}} = \mathbf{v}_A - \mathbf{v}_B\).
3. The shortest distance is the perpendicular distance from \(B\) to this straight path line of \(A\).
4. Use trigonometry or the cross product / dot product to find the perpendicular length: \(d_{\text{min}} = \frac{|\mathbf{d}_0 \times \mathbf{v}_{\text{rel}}|}{|\mathbf{v}_{\text{rel}}|}\).
Key Takeaway
For closest approach questions, minimizing \(D^2\) using differentiation is often the cleanest and most error-free method under exam conditions.
6. Common Pitfalls to Avoid
Mistake 1: Using constant acceleration equations when acceleration is variable.
Correction: Always check if \(a\) contains \(t, x,\) or \(v\). If it does, you must use calculus.
Mistake 2: Forgetting the constant of integration.
Correction: Whenever you integrate without limits, write \(+ \, c\) immediately and find its value using the stated initial conditions.
Mistake 3: Subtracting in the wrong order for relative velocity.
Correction: Remember that \(_{A}\mathbf{v}_B = \mathbf{v}_A - \mathbf{v}_B\). A quick check: the velocity of \(A\) relative to \(A\) must be \(\mathbf{v}_A - \mathbf{v}_A = \mathbf{0}\).
Mistake 4: Confusing speed with velocity.
Correction: If asked for velocity, your final answer should be a vector (\(p\mathbf{i} + q\mathbf{j}\)). If asked for speed, find the scalar magnitude \(\sqrt{p^2 + q^2}\).
7. Quick Review Summary
- Rate of change of displacement is velocity: \(v = \frac{\mathrm{d}x}{\mathrm{d}t}\)
- Rate of change of velocity is acceleration: \(a = \frac{\mathrm{d}v}{\mathrm{d}t} = v\frac{\mathrm{d}v}{\mathrm{d}x}\)
- 2D Speed: \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\)
- Velocity of \(A\) relative to \(B\): \(_{A}\mathbf{v}_{B} = \mathbf{v}_A - \mathbf{v}_B\)
- Closest approach: Minimize \(|\mathbf{r}_A - \mathbf{r}_B|^2\) by differentiating with respect to \(t\) and setting to zero.