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Module 16 · L02 of 16 ~45 min ⚡ +90 XP available

Acceleration as a Function of x: $v\dfrac{dv}{dx}$

When the acceleration depends on position rather than time — gravity from a planet, a spring restoring force, a particle dragged by a position-dependent field — the chain rule gives a powerful re-write: $a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$. This single identity turns hard kinematics into separable integration in $x$.

Today's hook — A particle has acceleration $a = -4x$. At $x = 0$ its speed is $6$ m s$^{-1}$. Before reading on, write down (a) the identity that lets you replace $a$, (b) the resulting DE in $v$ and $x$, and (c) the maximum displacement reached. Compare your answers after card 05.

0/5QUESTS

Recall — your gut answer first

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A particle has acceleration $a = -2x$. Before checking — try to integrate $a = dv/dt$ directly. What goes wrong? Now apply the chain rule to write $a$ in terms of $v$ and $dv/dx$. Sketch your reasoning below.

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The two moves when $a = a(x)$

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When acceleration is given as a function of position, two habits unlock the problem: replace $a$ using the chain rule $a = v\dfrac{dv}{dx}$ (or equivalently $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$), then separate variables in $x$ and $v$ so each side has only one variable.

The chain-rule trick: starting from $a = \dfrac{dv}{dt}$ and using $\dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$, we eliminate $t$ entirely. Equivalently, $v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, so integrating with respect to $x$ gives $\tfrac{1}{2}v^2 = \int a(x)\,dx + C$.

$a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$

$\tfrac{1}{2}v^2 = \displaystyle\int a(x)\,dx + C$

Why $v\dfrac{dv}{dx}$ works

It comes straight from the chain rule: $\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. No new physics — just calculus that eliminates $t$.

$v^2$ is the natural variable

Because $v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, integrating gives $v^2$ as a function of $x$. To find a max-displacement, set $v^2 = 0$ — no need to solve for $v(t)$.

Sign of $v$ matters

$v^2$ is non-negative, so $v = \pm\sqrt{\cdot}$. Choose the sign based on the direction of motion at the instant of interest. Don't just take the positive root by default.

What you'll master

Know

Key facts

$a = v\dfrac{dv}{dx}$ — the chain-rule identity
$v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$
$\tfrac{1}{2}v^2 = \displaystyle\int a(x)\,dx + C$
For $a = -kx$ ($k > 0$): SHM with $v^2 = k(A^2 - x^2)$

Understand

Concepts

Why $v\dfrac{dv}{dx}$ eliminates $t$ when $a$ depends on $x$
Why $v^2$ (not $v$) is the convenient intermediate variable
How initial conditions in the form ($x_0, v_0$) fix the constant of integration

Can do

Skills

Rewrite $\ddot{x} = a(x)$ as $v\dfrac{dv}{dx} = a(x)$ and integrate
Find $v^2$ in terms of $x$ for $a = -kx$, $a = -k/x^2$, and other forms
Solve for maximum displacement (where $v = 0$) and maximum speed

Key terms

Chain-rule identity$\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. Replaces time derivative with a position derivative.

$v\dfrac{dv}{dx}$ formExpresses acceleration in terms of $v$ and $x$. Useful whenever $a$ is given as $a(x)$.

$\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$ formEquivalent to $v\dfrac{dv}{dx}$ by the chain rule on $\tfrac{1}{2}v^2$. Lets you integrate directly with respect to $x$.

Separable DEA differential equation that can be rewritten so each side depends on only one variable, then integrated side-by-side.

Restoring acceleration$a = -kx$ ($k > 0$): acceleration directed toward the origin, proportional to displacement. Produces simple harmonic motion.

Inverse-square acceleration$a = -k/x^2$: acceleration directed toward the origin, proportional to $1/x^2$. Models gravity from a point mass.

MEX-M1NESA outcome (Applications of Calculus to Mechanics): uses calculus to model and solve problems involving the motion of a particle in a straight line, including when acceleration is expressed as a function of position.

Deriving $a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$

core concept

Start from $a = \dfrac{dv}{dt}$. By the chain rule, since $v$ depends on $x$ which depends on $t$,

$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\,\frac{dv}{dx}.$$

Now observe that $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = v\dfrac{dv}{dx}$ by the chain rule again. So $a = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$. Integrating both sides with respect to $x$:

$$\frac{1}{2}v^2 = \int a(x)\,dx + C.$$

Worked through the hook: $a = -4x$ with $v = 6$ when $x = 0$. Apply $v\dfrac{dv}{dx} = -4x$, so $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -4x$. Integrate: $\tfrac{1}{2}v^2 = -2x^2 + C$. Use $v = 6$ at $x = 0$: $\tfrac{1}{2}(36) = 0 + C \Rightarrow C = 18$. So $v^2 = 36 - 4x^2$. Maximum displacement is where $v = 0$: $x^2 = 9$, $|x| = 3$. The particle oscillates between $x = -3$ and $x = 3$ — simple harmonic motion.

Connecting to energy. The identity $\tfrac{1}{2}v^2 = \int a(x)\,dx + C$ is exactly the work–energy theorem in disguise: $a = F/m$, so $\tfrac{1}{2}mv^2 - \int F\,dx = $ constant. NESA doesn't require energy language, but recognising the link helps with physics-style questions.

$a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$ · Integrate w.r.t. $x$: $\tfrac{1}{2}v^2 = \int a(x)\,dx + C$ · Fix $C$ using the value of $v$ at a known $x$ · For $a = -kx$: $v^2 = k(A^2 - x^2)$ where $A$ is the amplitude

Pause — copy $a = v\,dv/dx = d(\tfrac{1}{2}v^2)/dx$, the integration-w.r.t.-$x$ result $\tfrac{1}{2}v^2 = \int a(x)\,dx+C$, and the SHM application $v^2 = k(A^2-x^2)$ into your book.

Quick check: A particle has acceleration $a = -9x$, and $v = 12$ when $x = 0$. Which equation correctly gives $v^2$ in terms of $x$?

Choosing which form to use

core concept

We just saw the identity $a = v\,dv/dx = d(\tfrac{1}{2}v^2)/dx$, which converts $a = a(x)$ into a separable ODE in $v^2$ and $x$. That raises a question: given that $a$ can depend on $t$, $v$, or $x$, which DE form should you use for each case? This card answers it → $a = a(t)$: integrate w.r.t. $t$; $a = a(v)$: separate variables; $a = a(x)$: use $v\,dv/dx$.

The three faces of acceleration each suit a different problem:

$a = \dfrac{d^2x}{dt^2}$ — when $a$ is given as $a(t)$, integrate directly with respect to $t$.
$a = \dfrac{dv}{dt}$ — when $a$ depends on $v$ only, separate variables: $\dfrac{dv}{a(v)} = dt$.
$a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$ — when $a$ depends on $x$, integrate with respect to $x$ to get $v^2$ as a function of $x$.

Choosing the wrong form is a common time-sink. The decisive rule: look at which variable $a$ depends on first, then pick the form whose integration variable is that variable.

Common mistake. Trying to integrate $a = -kx$ "directly with respect to $t$" goes nowhere because $x$ itself depends on $t$ in an unknown way. The $v\dfrac{dv}{dx}$ trick is precisely the technique that lets you make progress when $t$ is hidden.

$a = a(t)$: integrate w.r.t. $t$ · $a = a(v)$: separate variables, $\int dv/a(v) = \int dt$ · $a = a(x)$: use $v\,dv/dx = a(x)$, integrate w.r.t. $x$ · The form you pick is dictated by what $a$ depends on

Pause — copy the three-case classification ($a = a(t)$: integrate; $a = a(v)$: separate; $a = a(x)$: use $v\,dv/dx$) and the rule that the form of $a$ dictates the DE choice into your book.

Did you get this? True or false: the identity $v\dfrac{dv}{dx}$ is most useful when the acceleration is given as a function of position $x$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · LINEAR RESTORING (a = -kx)

A particle moves with $a = -16x$ m s$^{-2}$. At $x = 0$ it has speed $8$ m s$^{-1}$. Find $v^2$ as a function of $x$, and find the amplitude (max displacement).

Replace $a$ using the identity: $v\dfrac{dv}{dx} = -16x$, i.e. $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -16x$.

The form $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$ lets you integrate immediately with respect to $x$.

PROBLEM 2 · INVERSE-SQUARE (a = -k/x²)

A particle of mass 1 kg is attracted toward the origin with $a = -\dfrac{8}{x^2}$ m s$^{-2}$ for $x > 0$. It starts from rest at $x = 4$ m. Find $v^2$ as a function of $x$, and the speed when it reaches $x = 1$.

Use $v\dfrac{dv}{dx} = -\dfrac{8}{x^2}$, i.e. $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -8x^{-2}$.

Inverse-square forms integrate cleanly to $1/x$. Recognise this before doing the integral.

PROBLEM 3 · HSC-STYLE APPLICATION

A particle moves in a straight line with acceleration $a = -\dfrac{1}{(x+1)^2}$ m s$^{-2}$ for $x \geq 0$. Initially the particle is at $x = 0$ with velocity $v = 1$ m s$^{-1}$ (away from the origin). Show that $v^2 = \dfrac{2}{x+1} - 1$, and find the maximum value of $x$ reached.

Apply the identity: $v\dfrac{dv}{dx} = -\dfrac{1}{(x+1)^2}$, i.e. $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -(x+1)^{-2}$.

Translate the given acceleration into the integrable form. Write $(x+1)^{-2}$ to make the integration step obvious.

Fill the gap: The chain rule gives $\dfrac{dv}{dt} = v\dfrac{d\,\rule{0.6em}{0.5pt}\,}{d\,\rule{0.6em}{0.5pt}\,}$, and this equals $\dfrac{d}{dx}\!\left(\tfrac{1}{2}\,\rule{0.6em}{0.5pt}\,^2\right)$. Fill: $v\dfrac{d}{dx}$ of with respect to , which equals $\dfrac{d}{dx}$ of $\tfrac{1}{2}$ times squared.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Integrating $a(x)$ with respect to $t$

If $a = -kx$ and you try $\int -kx\,dt$, you're stuck because $x(t)$ is unknown. The whole point of $v\dfrac{dv}{dx}$ is to switch the integration variable to $x$ — where progress is possible.

Trap 02

Forgetting the $\tfrac{1}{2}$

The identity is $a = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, not $\dfrac{d}{dx}(v^2)$. After integration, $\tfrac{1}{2}v^2 = \int a\,dx + C$. Forgetting the factor of $\tfrac{1}{2}$ doubles your answer for $v^2$.

Trap 03

Defaulting to $v > 0$ when taking the root

Once you have $v^2 = f(x)$, the sign of $v$ depends on the direction of motion. A particle moving from $x = 5$ back toward $x = 0$ has $v < 0$ — take the negative root. Always read the physical setup.

Did you get this? True or false: for a particle with $a = -kx$ ($k > 0$) starting from rest at $x = A$, the equation $v^2 = k(A^2 - x^2)$ holds for all subsequent positions.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A particle has $a = -25x$. At $x = 0$, $v = 10$. Find $v^2$ as a function of $x$ and the amplitude of motion.

A particle has acceleration $a = 2x + 3$. It starts from rest at $x = 0$. Find $v^2$ in terms of $x$ and the speed when $x = 3$.

A particle moves with $a = -\dfrac{2}{x^3}$ for $x > 0$. It starts from rest at $x = 1$. Find $v^2$ in terms of $x$ and decide whether the particle ever reaches the origin.

A particle has $a = -4x + 8$. At $x = 0$, $v = 0$. Find $v^2$ in terms of $x$ and the value(s) of $x$ where $v = 0$ again.

(HSC-style) A spacecraft moves along a line with $a = -\dfrac{32}{(x+4)^2}$ (taking the centre of the planet as origin). Initially $x = 0$ and $v = 4$ (moving in the positive direction). Find $v^2$ as a function of $x$ and the farthest position reached.

Odd one out: Three of these are equivalent expressions for the acceleration when $v$ and $x$ are related by motion along a straight line. Which one is NOT?

Revisit your thinking

Earlier you tackled $a = -4x$ with $v = 6$ at $x = 0$ — finding the right identity, the resulting DE, and the maximum displacement.

The right move is $a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, giving $\tfrac{1}{2}v^2 = -2x^2 + C$. Initial condition fixes $C = 18$, so $v^2 = 36 - 4x^2$. Maximum displacement is where $v = 0$: $|x| = 3$. The crucial insight is that when acceleration depends on $x$, time disappears from the problem — you work entirely in terms of $x$ and $v$, and recover the amplitude immediately from $v^2 = 0$. This $v$-on-$v$ identity is one of the most useful tools in Module 16.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A particle moves with acceleration $a = -9x$. At $x = 0$, $v = 6$. Find $v^2$ as a function of $x$. (2 marks)

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ApplyBand 43 marks

Q2. A particle has acceleration $a = -\dfrac{18}{x^2}$ ($x > 0$) and starts from rest at $x = 6$. Find $v^2$ as a function of $x$ and the speed when the particle reaches $x = 2$. (3 marks)

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AnalyseBand 53 marks

Q3. A particle moves with $a = 2x - 6$. At $x = 1$, $v = 0$. (a) Find $v^2$ in terms of $x$. (b) Find the other position at which the particle is momentarily at rest. (c) Find the maximum speed and the position at which it occurs. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\tfrac{1}{2}v^2 = -\tfrac{25}{2}x^2 + C$; $v(0) = 10 \Rightarrow C = 50$. So $v^2 = 100 - 25x^2$. Amplitude: $v = 0 \Rightarrow x^2 = 4$, $A = 2$ m.

2. $\tfrac{1}{2}v^2 = x^2 + 3x + C$; $v = 0$ at $x = 0 \Rightarrow C = 0$. So $v^2 = 2x^2 + 6x$. At $x = 3$: $v^2 = 18 + 18 = 36$, $|v| = 6$ m s$^{-1}$.

3. $\tfrac{1}{2}v^2 = \tfrac{1}{x^2} + C$ (since $\int -2x^{-3}\,dx = x^{-2}$); $v = 0$ at $x = 1 \Rightarrow 0 = 1 + C \Rightarrow C = -1$. So $v^2 = \tfrac{2}{x^2} - 2$. As $x \to 0^+$, $v^2 \to \infty$ — the particle can reach the origin (mathematically; physically it would accelerate without bound).

4. $\tfrac{1}{2}v^2 = -2x^2 + 8x + C$; $v = 0$ at $x = 0 \Rightarrow C = 0$. So $v^2 = -4x^2 + 16x = 4x(4 - x)$. $v = 0$ again at $x = 4$.

5. $\tfrac{1}{2}v^2 = \dfrac{32}{x+4} + C$; $v = 4$ at $x = 0 \Rightarrow 8 = 8 + C \Rightarrow C = 0$. So $v^2 = \dfrac{64}{x+4}$. This is always $> 0$ for $x \geq 0$, so $v$ never reaches zero — the spacecraft has escape velocity and travels to infinity.

Q1 (2 marks): $v\dfrac{dv}{dx} = -9x \Rightarrow \tfrac{1}{2}v^2 = -\tfrac{9}{2}x^2 + C$ [1]. $v(0) = 6 \Rightarrow C = 18$, so $v^2 = 36 - 9x^2$ [1].

Q2 (3 marks): $\tfrac{1}{2}v^2 = \dfrac{18}{x} + C$ [1]. $v = 0$ at $x = 6 \Rightarrow C = -3$, so $v^2 = \dfrac{36}{x} - 6$ [1]. At $x = 2$: $v^2 = 18 - 6 = 12$, $|v| = 2\sqrt{3}$ m s$^{-1}$ [1].

Q3 (3 marks): (a) $\tfrac{1}{2}v^2 = x^2 - 6x + C$; $v = 0$ at $x = 1 \Rightarrow 0 = 1 - 6 + C \Rightarrow C = 5$. So $v^2 = 2x^2 - 12x + 10 = 2(x-1)(x-5)$ [1]. (b) $v = 0$ also at $x = 5$ [1]. (c) Max speed occurs at $x$ where $\dfrac{d(v^2)}{dx} = 4x - 12 = 0$, i.e. $x = 3$. But $v^2(3) = 2(2)(-2) = -8 < 0$ — the particle does not reach $x = 3$ between the rests. The motion is confined to where $v^2 \geq 0$: from the factor form $v^2 = 2(x-1)(x-5)$, this requires $x \leq 1$ or $x \geq 5$. The given motion stays in one regime, and max speed within that regime is at the endpoint — for $x \leq 1$ the speed grows as $x$ decreases; no finite max (open boundary) [1].

Boss battle · The Chain-Rule Champion

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Five timed questions on $v\dfrac{dv}{dx}$, restoring forces, inverse-square gravity, and finding max displacement. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering quick $v\dfrac{dv}{dx}$ questions. Lighter alternative to the boss.

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Module overview · Extension 2 · Checkpoint 1