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L16 of 16 · Final · End of Course ~50 min ⚡ +100 XP available

Module 16 Synthesis & Exam Technique

Every Module 16 mechanics problem follows the same three steps: recognise the form of acceleration (a function of $t$, of $x$, or of $v$); set up the differential equation using $a = dv/dt$, $a = v\,dv/dx$, or $a = d^2x/dt^2$; and solve with the calculus toolkit from Module 15. This final lesson distils that workflow, gives you an exam-day decision tree, and runs three closing examples that span the breadth of M16, SHM, resisted motion, and circular motion.

Today's hook, A particle moves so that $a = -k v^2$ ($k$ constant) and is released from rest at $x = 0$ with initial speed $v_0$. Before reading on, decide which form of $a$ this is (function of $t$, $x$, or $v$?) and which form of the DE you would use to find $v$ as a function of $x$.

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Recall, your gut answer first

+5 XP warm-up The three identities at the heart of Module 16: $a = \dfrac{dv}{dt}$, $a = v\dfrac{dv}{dx}$, $a = \dfrac{d^2x}{dt^2}$.

Recall, your gut answer first

+5 XP warm-up

The three identities at the heart of Module 16: $a = \dfrac{dv}{dt}$, $a = v\dfrac{dv}{dx}$, $a = \dfrac{d^2x}{dt^2}$. Before checking for each form of $a$ below, decide which identity gives a separable DE that you would actually try first: (i) $a = -g$; (ii) $a = -\omega^2 x$; (iii) $a = -kv$; (iv) $a = -kv^2$ and we want $v$ as a function of $x$.

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The three moves for any M16 problem

+5 XP to read Every Module 16 question can be unlocked by the same disciplined sequence: recognise the form of $a$ (function of $t$, $x$, or $v$?), choose the matching identity for the differential equation, then solve carefully and apply initial conditions.

The three moves for any M16 problem

+5 XP to read

Every Module 16 question can be unlocked by the same disciplined sequence: recognise the form of $a$ (function of $t$, $x$, or $v$?), choose the matching identity for the differential equation, then solve carefully and apply initial conditions. Half the marks in a long mechanics question come from explicitly stating which form of $a$ is being used and writing the DE before integrating.

The recognise-set up-solve workflow: (1) read the problem and state $a$ as a function of $t$, $x$, or $v$; (2) write the DE, $\frac{dv}{dt}$ for "what is $v(t)$?", $v\frac{dv}{dx}$ for "what is $v(x)$?", $\frac{d^2x}{dt^2}$ for SHM/equilibrium analyses; (3) separate, integrate, apply ICs, check units and limits.

$a(t)$: integrate w.r.t. $t$ · $a(x)$: use $v\frac{dv}{dx}$ · $a(v)$: choose $\frac{dv}{dt}$ for $v(t)$ or $v\frac{dv}{dx}$ for $v(x)$

M16 acceleration decision guide migrated to comparison-table

$a = \dfrac{dv}{dt} = v\dfrac{dv}{dx} = \dfrac{d^2x}{dt^2}$

State the form of $a$ explicitly

Markers reward the line "Since $a$ is a function of $x$, use $a = v\,dv/dx$." Writing this once signals you understand the choice of DE, and protects you if a later algebra step goes wrong.

Initial conditions are mark gold

Always write "When $t = 0$, $x = \ldots$, $v = \ldots$" before integrating. Substitute the IC into the constant of integration in the same step, sloppy ICs lose more marks than any algebra slip.

Sanity-check end points

After solving, ask: as $t \to \infty$ does $v$ tend to a sensible limit? At an equilibrium $x$, is $a = 0$? Does dimensional analysis work? A 10-second check catches sign errors and stray factors.

What you'll master

Know

Key facts

Three identities: $a = dv/dt$, $a = v\,dv/dx$, $a = d^2x/dt^2$
SHM: $\ddot x = -\omega^2 (x - c)$ ⇒ $x = c + A\cos(\omega t + \phi)$, period $2\pi/\omega$
Resisted motion: $m\ddot x = -mg - mkv^n$ (upward) or $m\ddot x = mg - mkv^n$ (downward)
Circular motion (uniform): centripetal $a = v^2/r$ towards centre

Understand

Concepts

Why the form of $a$ dictates which DE to write
Why SHM emerges whenever the restoring force is linear in displacement
Why terminal velocity is found by setting $a = 0$ in the resisted-motion DE

Can do

Skills

Diagnose $a$ as a function of $t$, $x$, or $v$ and pick the correct DE
Solve SHM, resisted motion, and circular problems within a 15-minute exam slot
Apply initial conditions cleanly and sanity-check answers at limits

Key terms

Equation of motionA differential equation expressing Newton's second law. In M16: $m\ddot x = F$, where $F$ may depend on $t$, $x$, or $v$.

Three forms of acceleration$a = dv/dt$ (natural when $a$ depends on $t$ or you want $v(t)$); $a = v\,dv/dx$ (natural when $a$ depends on $x$ or you want $v(x)$); $a = d^2x/dt^2$ (natural for SHM and equilibrium analysis).

Simple harmonic motion (SHM)Motion satisfying $\ddot x = -\omega^2 (x - c)$. Solution: $x = c + A\cos(\omega t + \phi)$. Period $T = 2\pi/\omega$; amplitude $A$ and phase $\phi$ from ICs.

Resisted motionOne-dimensional motion under gravity plus a velocity-dependent resistive force, typically $kv$ or $kv^2$. The DE is solved by separating variables.

Terminal velocityThe constant velocity reached when net force is zero. Found by setting $\ddot x = 0$ in the equation of motion.

Initial condition (IC)The known value of $x$, $v$ or $t$ used to determine the constant of integration. Always state ICs in words before substituting.

MEX-M1NESA outcome (Applications of Calculus to Mechanics): applies calculus to mechanics in one and two dimensions, including SHM, resisted motion and circular motion.

The M16 workflow: recognise · set up · solve

Almost every Module 16 problem can be tackled in three numbered steps.

The M16 workflow: recognise · set up · solve

core concept

Almost every Module 16 problem can be tackled in three numbered steps.

Recognise. Read the problem and write $a$ explicitly. Is it a function of $t$ (e.g., $a = -g$, $a = \sin t$), of $x$ (e.g., $a = -\omega^2 x$), or of $v$ (e.g., $a = -kv$, $a = -g - kv^2$)?
Set up the DE. Pick the identity that gives the simplest separable equation:
- $a$ depends on $t$: integrate $a = \dfrac{dv}{dt}$ directly.
- $a$ depends on $x$: use $a = v\dfrac{dv}{dx}$, separate $v\,dv = a\,dx$.
- $a$ depends on $v$ and you want $v(t)$: use $\dfrac{dv}{dt} = a(v)$, separate $\dfrac{dv}{a(v)} = dt$.
- $a$ depends on $v$ and you want $v(x)$: use $v\dfrac{dv}{dx} = a(v)$, separate $\dfrac{v\,dv}{a(v)} = dx$.
Solve and apply ICs. Integrate. State the IC in words. Substitute to find the constant. Simplify. Check: does the answer behave sensibly as $t \to \infty$ or at endpoints?

Why this workflow always wins. Markers want to see you (a) state the form of $a$, (b) write the DE explicitly, (c) state the IC, (d) integrate cleanly. These are the four most reliable mark-gates in any Module 16 question, and they cost nothing in extra time.

Step 1: $a$ as function of $t$, $x$, or $v$? · Step 2: pick the DE identity matching the form of $a$ and the target variable · Step 3: separate, integrate, IC, simplify, sanity-check · The four mark-gates: form of $a$ stated · DE written · IC stated · clean integration

Pause, copy the M16 three-step workflow (recognise form of $a$, write and separate DE, IC + simplify), the four mark-gates, and the sanity-check instruction into your book.

Quick check: A particle has $a = -kv$ (with $k > 0$) and starts from $x = 0$ with $v = v_0$. You want to find $v$ as a function of $x$. Which DE form gives the cleanest separation?

HSC exam approach for Module 16

We just saw the M16 workflow: identify what depends on, select the matching DE identity, separate and integrate, apply ICs,...

HSC exam approach for Module 16

core concept

We just saw the M16 workflow: identify what $a$ depends on, select the matching DE identity, separate and integrate, apply ICs, and sanity-check, with four mark-gates (form of $a$, DE written, IC stated, clean integration). That raises a question: what does the HSC marker look for in an exam response specifically? This card answers it → always state $a$ first, write the chosen DE form ($dv/dt$ or $v\,dv/dx$), state the IC in words before substituting, find terminal velocity by setting $a = 0$, and close with a one-line sanity check.

Long mechanics questions (typically 5–8 marks) in HSC Extension 2 reward structure as much as algebra. A reliable layout:

Line 1, state $a$. "$a = $ \ldots, a function of $t$/$x$/$v$".
Line 2, choose the DE. "Since $a$ depends on $x$, use $a = v\,dv/dx$."
Lines 3-5, separate & integrate. Show the separated form and the integrated form before substituting ICs.
Line 6, apply ICs. "When $t = 0$, $x = x_0$ and $v = v_0$ ⇒ $C = \ldots$".
Line 7, answer. Substitute back; state the final relation clearly.
Line 8, sanity check. One sentence: "As $t \to \infty$, $v \to v_T$, as expected."

For circular motion, replace step 2 by "Newton II radially: $F_{\text{net}} = mv^2/r$" and step 3 by an FBD or component equation.

$$\text{State } a \;\Rightarrow\; \text{Write DE} \;\Rightarrow\; \text{Separate} \;\Rightarrow\; \text{Integrate} \;\Rightarrow\; \text{Apply IC} \;\Rightarrow\; \text{Check}$$

Common-time error. Students often confuse $v(t)$ with $v(x)$. Read the question, does it ask "how fast after $T$ seconds?" ($v(t)$) or "how fast after travelling $D$ metres?" ($v(x)$)? The choice of DE depends on which one you want.

Always state $a$ before choosing the DE · $v(t)$ ⇒ $a = dv/dt$; $v(x)$ ⇒ $a = v\,dv/dx$ · Write IC in words before substituting · Terminal velocity: set $a = 0$ and solve for $v$ · One-line sanity check at the end of every long question

Pause, copy the five HSC habits for Module 16: state $a$ before choosing DE; $v(t) \Rightarrow dv/dt$, $v(x) \Rightarrow v\,dv/dx$; state IC in words first; terminal velocity via $a=0$; one-line sanity check at the end into your book.

Did you get this? True or false: to find the terminal velocity of a body falling under gravity with resistance $kv$, you set $\ddot x = 0$ in the equation of motion and solve for $v$.

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

PROBLEM 1 · SHM (a as a function of x)

A particle moves so that $\ddot x = -4(x - 3)$. At $t = 0$ it is at $x = 7$ with $\dot x = 0$. Find $x$ as a function of $t$, state the amplitude and period, and find the speed when $x = 3$.

Recognise: $a$ depends on $x$ ⇒ SHM about the centre $x = 3$ with $\omega^2 = 4$, so $\omega = 2$. General solution: $x = 3 + A\cos(2t + \phi)$.

Identify $\omega$ from the coefficient, $c$ from the equilibrium shift. Period and amplitude follow.

PROBLEM 2 · RESISTED MOTION (a as a function of v)

A body of unit mass falls from rest under gravity and a resistive force $kv$ per unit mass ($k > 0$). So $\ddot x = g - kv$ (taking downward positive). Find $v$ as a function of $t$, and find the terminal velocity.

Recognise: $a$ depends on $v$. Want $v(t)$ ⇒ use $dv/dt = g - kv$. Separate: $\dfrac{dv}{g - kv} = dt$.

When the target is $v(t)$ and $a$ depends on $v$, $dv/dt = a(v)$ is the right identity. Separate immediately.

PROBLEM 3 · CIRCULAR MOTION (radial Newton II)

A car of mass $m$ rounds a banked frictionless curve of radius $r$ at the designed speed $v$. The bank angle is $\theta$. Derive $\tan\theta = v^2/(rg)$, then find the normal reaction $N$ from the road in terms of $m$, $g$, and $\theta$.

Recognise: this is uniform circular motion (constant speed). Two equations only: vertical equilibrium $N\cos\theta = mg$; horizontal Newton II $N\sin\theta = mv^2/r$.

No friction, so $N$ supplies both vertical support and horizontal centripetal component. State this before writing the equations.

Fill the gap: A body of unit mass has $\ddot x = g - kv$. Its terminal velocity is $v_T =$ , found by setting in the equation of motion.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using $dv/dt$ when $a$ depends on $x$

If $a = -\omega^2 x$, you cannot integrate $\int -\omega^2 x\,dt$ directly without knowing $x(t)$. The right move is $a = v\,dv/dx$, giving $v\,dv = -\omega^2 x\,dx$, integrable straight away.

Trap 02

Forgetting to state the form of $a$

Long mechanics questions allocate marks for explicitly stating "$a$ is a function of $v$, so use $dv/dt$". Skipping this line costs a structural mark even if the algebra is right.

Trap 03

Sign errors in resisted motion

Resistance always opposes motion. For a body rising under gravity with resistance $kv^2$: $\ddot x = -g - kv^2$ (both decelerating). For falling: $\ddot x = g - kv^2$. Many students copy one form for both phases and lose all the marks.

Did you get this? True or false: for SHM of the form $\ddot x = -\omega^2 x$, the speed at the centre is $A\omega$, where $A$ is the amplitude.

Revisit your thinking

Activities · practice with the ideas

A particle satisfies $\ddot x = -9(x - 2)$ with $x(0) = 5$ and $\dot x(0) = 0$. Write $x(t)$, state the period and amplitude, and the maximum speed.

A particle has $a = -kv$ (with $k > 0$) and at $t = 0$ has $v = v_0$. Find $v$ as a function of $t$ and as a function of $x$ (taking $x = 0$ at $t = 0$).

A body of unit mass falls from rest with $\ddot x = g - kv^2$. Find $v$ as a function of $x$ (target: $v(x)$, not $v(t)$).

A particle moves in a horizontal circle of radius $r = 0.4$ m on a string of length $\ell$, the string making $40^\circ$ with the vertical. Find $\ell$ and the speed.

Diagnose, for each form of $a$, which DE identity you would use and write the separated form: (a) $a = -kx$; (b) $a = -kv$ for $v(t)$; (c) $a = -kv^2$ for $v(x)$; (d) $a = g - kv$ for $v(t)$.

Odd one out: Three of these are sensible first moves for a Module 16 mechanics question. Which one is NOT?

Revisit your thinking

Earlier you diagnosed several forms of $a$ and chose the matching DE identity.

The whole of Module 16 is built on the three identities $a = dv/dt = v\,dv/dx = d^2x/dt^2$, applied with discipline: state the form of $a$, choose the right DE, separate, integrate, apply ICs, check. Whether the problem is SHM, resisted motion, or circular motion, the workflow is the same, only the specific algebra changes. That habit is what carries marks in the HSC.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

Multiple choice

+5 XP per correct · +25 XP all-correct

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 so that $\ddot x = -16(x - 1)$. State the centre of motion, the angular frequency $\omega$, and the period. (2 marks)

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ApplyBand 43 marks

Q2. A body of unit mass moves under $\ddot x = -g - kv^2$ (rising; $k > 0$) and is projected upward from $x = 0$ with speed $u$. Find $v$ as a function of $x$ (i.e., on the way up). (3 marks)

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AnalyseBand 53 marks

Q3. A particle of mass $m$ on a light string of length $\ell$ moves as a conical pendulum at angle $\theta$ to the vertical with period $T$. Show that $T = 2\pi\sqrt{\ell\cos\theta/g}$ and explain in one sentence why this is independent of $m$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\omega = 3$, centre $c = 2$. With $x(0) = 5$, $\dot x(0) = 0$: $A = 3$, $\phi = 0$. $x(t) = 2 + 3\cos(3t)$. Period $2\pi/3$. Amplitude $3$. Max speed $A\omega = 9$.

2. $v(t)$: $dv/dt = -kv \Rightarrow \ln(v/v_0) = -kt \Rightarrow v = v_0 e^{-kt}$. $v(x)$: $v\,dv/dx = -kv \Rightarrow dv = -k\,dx \Rightarrow v = v_0 - kx$ (valid until $v = 0$).

3. $v\,dv/dx = g - kv^2$. Let $u = g - kv^2$, $du = -2kv\,dv$. So $-du/(2k) = u\,dx/v$, easier: $v\,dv/(g - kv^2) = dx$, integrate: $-(1/(2k)) \ln|g - kv^2| = x + C$. IC $v = 0$ at $x = 0$: $C = -(1/(2k))\ln g$. Result: $g - kv^2 = g e^{-2kx}$, i.e., $v^2 = (g/k)(1 - e^{-2kx})$.

4. $\ell = r/\sin 40^\circ = 0.4/\sin 40^\circ \approx 0.622$ m. $v^2 = rg\tan 40^\circ = 0.4(9.8)(0.839) \approx 3.29$, so $v \approx 1.81$ m s$^{-1}$.

5. (a) $v\,dv = -kx\,dx$. (b) $dv/v = -k\,dt$. (c) $dv/v = -k\,dx$. (d) $dv/(g - kv) = dt$.

Q1 (2 marks): Centre $x = 1$; $\omega = 4$ [1]; period $T = 2\pi/4 = \pi/2$ [1].

Q2 (3 marks): $a$ depends on $v$, want $v(x)$, so use $v\,dv/dx = -(g + kv^2)$ [1]. Separate and integrate: $(1/(2k))\ln(g + kv^2) = -x + C$ [1]. IC $v = u$ at $x = 0$: $v^2 = \big((g + ku^2)e^{-2kx} - g\big)/k$ [1].

Q3 (3 marks): Vertical $T_s\cos\theta = mg$ [1]; radial $T_s\sin\theta = m\omega^2\ell\sin\theta$, hence $\omega^2 = g/(\ell\cos\theta)$ [1]; period $= 2\pi/\omega = 2\pi\sqrt{\ell\cos\theta/g}$; mass cancels because both force equations are proportional to $m$ [1].

Consolidate and move on

Sit the module quiz, then close the lesson off.

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

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End of course, mark complete

This lesson completes Module 16 and the entire Mathematics Extension 2 course. From induction and complex numbers through proof, integration, vectors, and mechanics, you now have the full Extension 2 toolkit. Tick the box when you've finished practice and review, and take a moment to look back at how far you've come.

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