Ultimate Guide to Forces and Further Motion in a Straight Line

Introduction

At the heart of classical mechanics lies the relationship between force, mass, and motion. In HSC Mathematics Extension 2, Forces and Further Motion in a Straight Line builds upon the foundational kinematics you learned in Extension 1 and physics. Instead of simply dealing with constant acceleration, you are now required to manipulate differential equations where acceleration is a function of displacement or velocity. This topic bridges the mathematical purity of calculus with the physical reality described by Sir Isaac Newton.

Executive Summary

This comprehensive guide covers the core concepts of forces and straight-line motion:

Expressions for Acceleration: Moving beyond $a = \frac{dv}{dt}$ to derive and apply $a = v\frac{dv}{dx}$ and $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ .
Newton's Laws of Motion: Understanding and applying $F=ma$ (where force can be constant or non-constant).
Vector Resolutions: Recognizing forces as vector quantities and resolving concurrent forces into perpendicular components in 2D and 3D contexts.
Solving Kinematic Problems: Integrating acceleration equations to find velocity in terms of displacement, and vice-versa.

What is this about?

In basic kinematics, acceleration is typically a constant (like gravity, $g \approx 9.8 \text{ m/s}^2$ ) or a simple function of time. However, in many real-world scenarios—such as a rocket pushing through varying atmospheric density, or a block attached to a spring—the force (and therefore the acceleration) depends on where the object is (displacement) or how fast it is moving (velocity).

To solve these problems, we need alternative calculus identities for acceleration. By mastering these identities and resolving forces using vector geometry, you can mathematically model complex physical systems.

Main Content

1. Expressions for Acceleration

By definition, velocity $v$ is the rate of change of displacement $x$ , and acceleration $a$ is the rate of change of velocity $v$ .

$v = \frac{dx}{dt} = \dot{x}$
$a = \frac{dv}{dt} = \ddot{x}$

Deriving $a = v\frac{dv}{dx}$ : Using the chain rule of differentiation: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$ Since $\frac{dx}{dt} = v$ , we get: $a = v\frac{dv}{dx}$

Deriving $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ : If we differentiate $\frac{1}{2}v^2$ with respect to $x$ using the chain rule: $\frac{d}{dx}\left(\frac{1}{2}v^2\right) = \frac{d}{dv}\left(\frac{1}{2}v^2\right) \cdot \frac{dv}{dx}$ $= v \cdot \frac{dv}{dx}$ Since we just established that $v\frac{dv}{dx} = a$ , we arrive at: $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$

When to use which formula?

Use $a = \frac{dv}{dt}$ when acceleration is given as a function of time ( $t$ ).
Use $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ or $a = v\frac{dv}{dx}$ when acceleration is given as a function of displacement ( $x$ ).
If acceleration is a function of velocity ( $v$ ), you can use $a = \frac{dv}{dt}$ to link $v$ and $t$ , or $a = v\frac{dv}{dx}$ to link $v$ and $x$ .

2. Newton's Laws of Motion

Newton's three laws form the foundation of classical mechanics:

First Law (Inertia): An object remains at rest or in uniform motion unless acted upon by a net external force.
Second Law ( $F=ma$ ): The net force acting on an object is equal to its mass multiplied by its acceleration.
Third Law (Action-Reaction): For every action, there is an equal and opposite reaction.

In Extension 2, the Second Law is our primary tool. Because $F = ma$ , we can substitute our calculus expressions for acceleration: $F = m\frac{dv}{dt} = m\left(v\frac{dv}{dx}\right) = m\frac{d}{dx}\left(\frac{1}{2}v^2\right)$

This allows us to find the velocity of a particle if we know the net force acting on it.

3. Vector Resolutions and Concurrent Forces

Force is a vector quantity, meaning it has both magnitude and direction. When multiple forces act on a single body (concurrent forces), we must find the net force (resultant force) by adding the vectors.

To do this algebraically, we resolve forces into perpendicular components (usually horizontal $x$ and vertical $y$ , and occasionally $z$ for 3D contexts). If a force $F$ acts at an angle $\theta$ to the horizontal:

Horizontal component: $F_x = F \cos \theta$
Vertical component: $F_y = F \sin \theta$

Vector Projections: To determine how much of a force vector $\underset{\sim}{F}$ acts in the direction of another vector $\underset{\sim}{u}$ , we use the scalar projection: $\text{Projection of } \underset{\sim}{F} \text{ onto } \underset{\sim}{u} = \frac{\underset{\sim}{F} \cdot \underset{\sim}{u}}{|\underset{\sim}{u}|}$

Simple Worked Example

Question: A particle of mass 2 kg moves in a straight line along the $x$ -axis. The only force acting on the particle is given by $F = -4x$ Newtons. If the particle starts from rest at $x = 3$ metres, find an expression for its velocity $v$ in terms of its displacement $x$ .

Solution: Step 1: Set up the equation using Newton's Second Law. $F = ma$ $-4x = 2a$ $a = -2x$

Step 2: Choose the correct expression for acceleration. Since acceleration is a function of displacement ( $x$ ), we use $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ . $\frac{d}{dx}\left(\frac{1}{2}v^2\right) = -2x$

Step 3: Integrate both sides with respect to $x$ . $\int \frac{d}{dx}\left(\frac{1}{2}v^2\right) dx = \int -2x \, dx$ $\frac{1}{2}v^2 = -x^2 + C$

Step 4: Use initial conditions to find the constant $C$ . The particle starts from rest ( $v = 0$ ) at $x = 3$ . $\frac{1}{2}(0)^2 = -(3)^2 + C$ $0 = -9 + C \implies C = 9$

Step 5: Write the final expression. $\frac{1}{2}v^2 = 9 - x^2$ $v^2 = 18 - 2x^2$ $v = \pm \sqrt{18 - 2x^2}$

Note: This specific motion, where acceleration is proportional and opposite to displacement ( $a = -kx$ ), is actually Simple Harmonic Motion, which leads directly into the next topic of the syllabus!

mini-FAQ page

Q: Do I need to write out the derivation for $a = v\frac{dv}{dx}$ in an exam? A: Usually, no. Unless the question specifically asks you to "Derive the expression," you can quote $a = v\frac{dv}{dx}$ or $a = \frac{d}{dx}(\frac{1}{2}v^2)$ directly.

Q: What happens if mass isn't given in a question? A: Often, if the force is given as a generic expression (like "force is proportional to velocity"), you can let the mass be $m$ and the constant of proportionality be $mk$ . The mass $m$ will subsequently cancel out from both sides of $F=ma$ .

Common mistakes to avoid

Integrating with the wrong variable: A very common error is seeing $a = -2x$ and integrating to get $v = -x^2 + C$ . This implies you integrated with respect to $t$ ( $v = \int a dt$ ), which is mathematically invalid since $-2x$ is a function of $x$ , not $t$ . Always substitute the correct differential form of $a$ first.
Forgetting the $\pm$ when taking the square root: When you solve $v^2 = 18 - 2x^2$ , the velocity can be positive (moving right) or negative (moving left). Don't blindly assume velocity is always positive.

Practice on Vu's Maths Hub

Mechanics is a highly practical topic that requires you to build physical intuition through repetition and problem-solving.

Master all variations of kinematic equations in our HSC Mechanics Booklet.
Strengthen the integration techniques required for mechanics with the HSC Integrals Booklet.
Tackle challenging exam-style physics applications in the HSC Last Resorts Booklet.

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Want to conquer mechanics and guarantee top marks in HSC Mathematics Extension 2? Head over to Vu's Maths Hub for in-depth booklets, rigorous worked solutions, and expert advice to help you ace your exams!