Ultimate Guide to Rectilinear Resisted Motion in HSC Mathematics Extension 2

Introduction

In previous topics, we modelled motion in idealized, frictionless vacuums. However, in the real world, objects moving through a fluid (like a car driving through air or a submarine moving through water) experience drag. Rectilinear Resisted Motion introduces mathematical models for this resistance. In HSC Mathematics Extension 2, you must apply advanced calculus to derive the exact equations for a particle's velocity and displacement when it is slowed down by a resistive force.

Executive Summary

This comprehensive guide covers the core concepts of rectilinear (straight-line) resisted motion:

Newton's Laws and Resistance: Establishing the fundamental equation of motion $m\ddot{x} = -R(v)$ , where resistance is proportional to a power of speed ( $R = kv$ or $R = kv^2$ ).
Velocity as a Function of Time: Using $a = \frac{dv}{dt}$ and integration (often involving logarithms) to find $v(t)$ .
Velocity as a Function of Displacement: Using $a = v\frac{dv}{dx}$ and integration to find $v(x)$ .
Displacement as a Function of Time: Integrating the velocity function to find exactly where the particle is at any given moment $x(t)$ .

What is this about?

When you slide a block across a rough table, it eventually stops due to friction. When you coast on a bicycle, air resistance slowly brings you to a halt. In this topic, we model a particle moving horizontally with an initial velocity, but with no driving force pushing it forward. The only force acting on it is a resistive force that directly opposes its motion. Because this resistance changes depending on how fast the object is moving, the acceleration is not constant. You will use the integration techniques learned earlier in Extension 2 (like partial fractions or inverse trig functions) to solve the resulting differential equations.

Main Content

1. Setting up the Equation of Motion

Consider a particle of mass $m$ projected horizontally along the $x$ -axis with an initial velocity $u$ . The only external force acting horizontally is a resistive force $R$ that opposes the motion. According to the syllabus, the magnitude of this resistance is proportional to a power of the speed, typically $R = kv$ or $R = kv^2$ (where $k > 0$ ).

Using Newton's Second Law ( $F = ma$ ), and taking the direction of motion as positive: $F_{\text{net}} = -R$ $m\ddot{x} = -kv^n$

From this starting point, we can substitute our different expressions for acceleration to find velocity and displacement.

2. Velocity as a Function of Time, $v(t)$

To find how velocity changes over time, we use $a = \frac{dv}{dt}$ . $m\frac{dv}{dt} = -kv^n$

We solve this using the separation of variables: $\frac{dv}{v^n} = -\frac{k}{m} dt$

Example for $R = kv$ ( $n=1$ ): $\int \frac{1}{v} dv = \int -\frac{k}{m} dt$ $\ln(v) = -\frac{k}{m}t + C$

Using the initial condition (at $t=0$ , $v=u$ ): $\ln(u) = C$ $\ln(v) - \ln(u) = -\frac{k}{m}t$ $\ln\left(\frac{v}{u}\right) = -\frac{k}{m}t$ $v = u e^{-\frac{k}{m}t}$

This tells us that the velocity decays exponentially over time!

3. Velocity as a Function of Displacement, $v(x)$

To find how velocity relates to distance travelled, we use $a = v\frac{dv}{dx}$ . $m \left(v\frac{dv}{dx}\right) = -kv^n$

Again, we separate the variables: $\frac{v}{v^n} dv = -\frac{k}{m} dx$ $\frac{1}{v^{n-1}} dv = -\frac{k}{m} dx$

By integrating both sides and using the initial condition (at $x=0$ , $v=u$ ), you can establish the relationship between the particle's speed and how far it has travelled.

4. Displacement as a Function of Time, $x(t)$

Once you have established $v(t)$ in Step 2, you can find $x(t)$ simply by realizing that $v = \frac{dx}{dt}$ and integrating your expression with respect to time.

Taking our previous result $v = u e^{-\frac{k}{m}t}$ : $\frac{dx}{dt} = u e^{-\frac{k}{m}t}$ $\int dx = \int u e^{-\frac{k}{m}t} dt$ $x = -\frac{mu}{k} e^{-\frac{k}{m}t} + C$

Using the initial condition (at $t=0$ , $x=0$ ): $0 = -\frac{mu}{k}(1) + C \implies C = \frac{mu}{k}$ $x(t) = \frac{mu}{k} \left(1 - e^{-\frac{k}{m}t}\right)$

Notice something fascinating here: as $t \to \infty$ , the term $e^{-\frac{k}{m}t} \to 0$ . Therefore, $x \to \frac{mu}{k}$ . This means the particle will never travel further than $\frac{mu}{k}$ metres, even given an infinite amount of time!

Simple Worked Example

Question: A particle of unit mass ( $m=1$ ) is projected horizontally with an initial speed of $10 \text{ m/s}$ . It experiences a resistance of magnitude $2v^2$ , where $v$ is its speed. Find an expression for the particle's velocity $v$ in terms of its displacement $x$ .

Solution: Step 1: Set up the equation of motion. $F = ma$ $-2v^2 = 1 \cdot a$

Step 2: Choose the correct acceleration formula. Since we want velocity in terms of displacement $x$ , we use $a = v\frac{dv}{dx}$ . $v\frac{dv}{dx} = -2v^2$

Step 3: Separate variables and integrate. Divide both sides by $v$ (since $v \ne 0$ while it's moving): $\frac{dv}{dx} = -2v$ $\frac{1}{v} dv = -2 dx$ $\int \frac{1}{v} dv = \int -2 \, dx$ $\ln(v) = -2x + C$

Step 4: Find the constant of integration. At $x = 0$ , the initial velocity is $v = 10$ . $\ln(10) = -2(0) + C \implies C = \ln(10)$

Step 5: Write the final expression. $\ln(v) = -2x + \ln(10)$ $\ln(v) - \ln(10) = -2x$ $\ln\left(\frac{v}{10}\right) = -2x$ $\frac{v}{10} = e^{-2x}$ $v = 10e^{-2x}$

mini-FAQ page

Q: Do I need to memorize the final equations for $v(t)$ and $x(t)$ ? A: Absolutely not. The HSC syllabus explicitly states you must derive these expressions from Newton's laws. Questions will frequently change the initial conditions, the mass, or the resistance function ( $kv$ vs $kv^2$ vs $kv^3$ ), which entirely changes the final integrated formula.

Q: Why is the resistance always negative in the initial equation? A: We define the direction of motion as the positive direction. Because resistance acts in the opposite direction to motion, the force vector must be negative.

Common mistakes to avoid

Messing up the integration of $\frac{1}{v}$ : Remember that $\int \frac{1}{v} dv = \ln|v| + C$ . However, $\int \frac{1}{v^2} dv \ne \ln(v^2)$ . Instead, use power rules: $\int v^{-2} dv = -v^{-1} + C = -\frac{1}{v} + C$ .
Forgetting the absolute value in logarithms: While $v$ is usually positive in rectilinear motion (since it just slows down and doesn't turn around), it's a good habit to recognize $\ln|v|$ just in case the coordinate system is flipped.
Skipping the separation of variables: You cannot just integrate $v\frac{dv}{dx} = -kv^2$ with respect to $x$ directly on both sides without moving the $v$ terms to the $dv$ side first.

Practice on Vu's Maths Hub

Mastering resisted motion requires absolute confidence in your calculus and integration skills. Prepare for the HSC with our detailed resources:

Practice deriving kinematic equations from scratch with the HSC Mechanics Booklet.
Strengthen your integration (including partial fractions and log laws) with the HSC Integrals Booklet.
Take on the hardest Extension 2 calculus challenges in the HSC Last Resorts Booklet.

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Want to master Mechanics and lock in top marks for HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, detailed video explanations, and expert advice to help you ace your exams!