Modelling Motion with Differential Equations

Problem Statement

The fundamental bridge between physics and advanced mathematics in the HSC Mechanics syllabus is the translation of Newton's Second Law ($F = ma$) into a differential equation. Because acceleration $a$ is the derivative of velocity ($\frac{dv}{dt}$) and the second derivative of displacement ($\frac{d^2x}{dt^2}$), any statement about forces acting on a body is inherently a differential equation.

Solving these differential equations allows us to predict the future state of a system—its velocity and position at any given time.

Consider a skydiver of mass $m$ falling vertically. The force of gravity $mg$ acts downwards, and the air resistance acts upwards. The air resistance is modelled as being proportional to the square of the skydiver's velocity, so $R = mkv^2$, where $k$ is a positive constant and $v$ is the velocity at time $t$. The skydiver falls from rest.

(a) Show that the equation of motion for the skydiver can be written as the differential equation: $\frac{dv}{dt} = g - kv^2$

(b) Find an expression for velocity $v$ as a function of time $t$. (You may use the standard integral $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right|$).

(c) Show that as $t \to \infty$, the skydiver approaches a terminal velocity of $\sqrt{\frac{g}{k}}$.

Hints

• Part (a): Define your coordinate system (e.g., let downwards be positive). Sum the forces to find $F_{net}$ and equate this to $m \frac{dv}{dt}$ using Newton's Second Law.
• Part (b): This is a separable first-order differential equation. Separate the variables $v$ and $t$ so that all $v$ terms are on one side with $dv$, and $dt$ is on the other. Integrate both sides. You will need to rewrite $g - kv^2$ in the form $a^2 - x^2$ to use the provided standard integral.
• Part (c): Take the limit of your velocity expression as $t \to \infty$. Consider what happens to exponential terms like $e^{-ct}$ as $t$ becomes very large.

Solutions

Part (a): Forming the Differential Equation

1. Let downwards be the positive direction.
2. Forces acting on the skydiver:
   • Weight (downwards): $W = mg$
   • Air resistance (upwards): $R = -mkv^2$
3. The net force is $F = mg - mkv^2$.
4. By Newton's Second Law, $F = ma = m \frac{dv}{dt}$. $m \frac{dv}{dt} = mg - mkv^2$
5. Divide by the mass $m$: $\frac{dv}{dt} = g - kv^2$ This is our differential equation of motion.

Part (b): Solving for Velocity

1. Separate the variables: $\frac{dt}{dv} = \frac{1}{g - kv^2}$ $t = \int \frac{1}{g - kv^2} \, dv$
2. To use the standard integral, factor out $k$ from the denominator: $t = \frac{1}{k} \int \frac{1}{\frac{g}{k} - v^2} \, dv$
3. Now this matches the form $\frac{1}{a^2 - x^2}$ with $a = \sqrt{\frac{g}{k}}$ and $x = v$. Apply the standard integral: $t = \frac{1}{k} \left( \frac{1}{2\sqrt{\frac{g}{k}}} \right) \ln \left| \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} \right| + C$ $t = \frac{1}{2\sqrt{gk}} \ln \left| \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} \right| + C$
4. Apply initial conditions to find $C$: The skydiver falls from rest, so at $t=0$, $v=0$. $0 = \frac{1}{2\sqrt{gk}} \ln \left| \frac{\sqrt{\frac{g}{k}} + 0}{\sqrt{\frac{g}{k}} - 0} \right| + C$ $0 = \frac{1}{2\sqrt{gk}} \ln |1| + C$ Since $\ln(1) = 0$, $C = 0$.
5. Now we have: $t = \frac{1}{2\sqrt{gk}} \ln \left( \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} \right)$ (Absolute values are removed because $v$ starts at 0 and will not exceed terminal velocity $\sqrt{g/k}$).
6. Rearrange to make $v$ the subject. Multiply by $2\sqrt{gk}$: $2t\sqrt{gk} = \ln \left( \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} \right)$
7. Exponentiate both sides: $e^{2t\sqrt{gk}} = \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v}$
8. Let $A = \sqrt{\frac{g}{k}}$ and $E = e^{2t\sqrt{gk}}$ for algebraic simplicity: $E = \frac{A + v}{A - v}$ $E(A - v) = A + v$ $EA - Ev = A + v$ $EA - A = v + Ev$ $A(E - 1) = v(1 + E)$ $v = A \frac{E - 1}{E + 1}$
9. Substitute $A$ and $E$ back in: $v(t) = \sqrt{\frac{g}{k}} \left( \frac{e^{2t\sqrt{gk}} - 1}{e^{2t\sqrt{gk}} + 1} \right)$

Part (c): Terminal Velocity Limit

1. We want to find $\lim_{t \to \infty} v(t)$.
2. Look at the expression for velocity: $v(t) = \sqrt{\frac{g}{k}} \left( \frac{e^{2t\sqrt{gk}} - 1}{e^{2t\sqrt{gk}} + 1} \right)$
3. As $t \to \infty$, the terms $e^{2t\sqrt{gk}}$ become overwhelmingly large.
4. Divide numerator and denominator by $e^{2t\sqrt{gk}}$ to evaluate the limit: $v(t) = \sqrt{\frac{g}{k}} \left( \frac{1 - e^{-2t\sqrt{gk}}}{1 + e^{-2t\sqrt{gk}}} \right)$
5. As $t \to \infty$, $e^{-2t\sqrt{gk}} \to 0$.
6. Therefore, the limit is: $\lim_{t \to \infty} v(t) = \sqrt{\frac{g}{k}} \left( \frac{1 - 0}{1 + 0} \right) = \sqrt{\frac{g}{k}}$ This matches our expected terminal velocity where $\frac{dv}{dt} = 0 \implies g = kv^2$.

Takeaways

• Translating Physics to Maths: Newton's Second Law is the ultimate tool for generating differential equations in mechanics. The forces define the right hand side, and mass times acceleration defines the left.
• Separation of Variables: This is the most common technique for solving first-order ODEs in mechanics. Practice separating $v$ and $x$, or $v$ and $t$, accurately.
• Limiting Behaviours: Understanding the physical meaning of mathematical limits (like $t \to \infty$) is crucial. A terminal velocity is simply the horizontal asymptote of the velocity-time graph.

Further Readings

• HSC Differential Equations: https://vumaths.com/booklets/hsc-differential-equations/
• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/

Connect with me

Tackling differential equations can feel daunting, but it's the most rewarding part of Extension 2 Mechanics. If you need more structured practice, check out the booklets on Vu's Maths Hub. You can also watch my video explanations of similar problems on my YouTube channel. Follow my Instagram for bite-sized tips, and read my extended thoughts on the syllabus on Substack. Let's connect!