Do You Need Calculus for HSC Physics?

A car's speed increases from 5 m/s to 17 m/s over four seconds. You can calculate its average acceleration using one subtraction and one division. No derivatives required.

**No, you do not need calculus to study NSW HSC Physics.** You do need to be comfortable with algebra, trigonometry, rearranging equations, interpreting graphs, and working with rates of change. These are important mathematical skills, but they are not the same as differentiation and integration.

The distinction matters because HSC Physics is mathematically demanding in places. Students sometimes assume that because university physics uses calculus, they need to learn it before starting the HSC course. That's unnecessary. Your time is better spent mastering the mathematics the syllabus actually uses.

## What mathematics does HSC Physics require?

Think about the difference between calculating a car's average speed over a journey and reading its speedometer at one instant.

Average speed uses the total distance travelled divided by the time taken. A speedometer gives an estimate of the car's speed at a particular moment. Both describe motion, but they answer slightly different questions.

HSC Physics regularly uses calculations involving changes over measured time intervals. Calculus provides a more advanced way of describing what happens at individual instants and how quantities change continuously.

The distinction becomes clearer when you look at the equations.

### The mathematics in the 2025 Physics syllabus

As at 8 November 2026, NESA's replacement [Physics 11-12 Syllabus (2025)](https://curriculum.nsw.edu.au/learning-areas/science/physics-11-12-2025/content) is scheduled for implementation from 2027. Its equations cover a substantial range of physical relationships, but they do not require formal differentiation or integration.

Here are representative examples of the mathematics involved.

| Mathematical skill | Examples from HSC Physics | What you need to do |
|---|---|---|
| Algebra | \(F=ma\), \(V=IR\) | Substitute values and rearrange equations |
| Powers and roots | \(E_k=\frac12 mv^2\), \(v=\sqrt{\frac{2GM}{r}}\) | Work with squares, square roots, and scientific notation |
| Trigonometry | \(F_x=F\cos\theta\), \(F_y=F\sin\theta\) | Resolve vectors into components |
| Ratios and proportions | \(v=f\lambda\), transformer relationships | Compare quantities and calculate unknowns |
| Graph interpretation | Displacement-time and velocity-time relationships | Interpret gradients, trends, and areas |
| Logarithms | \(\lambda=\frac{\ln 2}{T_{1/2}}\) | Evaluate logarithmic expressions using a calculator |

The last example deserves attention. The natural logarithm, \(\ln\), appears in radioactive decay calculations. This is logarithmic mathematics, not calculus. Seeing a logarithm in a physics equation does not mean you need to know how to differentiate or integrate it.

The syllabus also contains equations for circular motion, gravitational fields, electromagnetic induction, wave behaviour, and special relativity. These can look intimidating, but the mathematical operations are largely algebraic.

### What about rates of change?

This is where students sometimes get confused.

Consider the average acceleration equation:

\[
a=\frac{\Delta v}{\Delta t}
\]

Here:
- \(a\) is average acceleration, measured in metres per second squared (\(\text{m/s}^2\)).
- \(\Delta v\) is the change in velocity, measured in metres per second (\(\text{m/s}\)).
- \(\Delta t\) is the time interval, measured in seconds (s).

The symbol \(\Delta\) means a finite change. For example:

\[
\Delta v=v_{\text{final}}-v_{\text{initial}}
\]

You can calculate this using ordinary subtraction and division.

Calculus uses a different notation for instantaneous acceleration:

\[
a=\frac{dv}{dt}
\]

The symbol \(d\) represents an infinitesimally small change. Differentiation tells you how quickly velocity is changing at one particular instant.

You do not need to perform that differentiation to solve standard HSC Physics problems.

[IMAGE: A comparison of two velocity-time graphs. The first is a straight line representing constant acceleration, with its gradient labelled as acceleration and its trapezium area representing displacement. The second is a curved line representing changing acceleration, with a secant showing average acceleration and a tangent at one point illustrating instantaneous acceleration. Clearly distinguish the two mathematical ideas.]

### Why the constant-acceleration equations don't need calculus

Take the familiar equation:

\[
v=u+at
\]

where:
- \(v\) is final velocity in m/s
- \(u\) is initial velocity in m/s
- \(a\) is constant acceleration in m/s²
- \(t\) is elapsed time in seconds.

It comes directly from the average acceleration equation:

\[
a=\frac{v-u}{t}
\]

Rearrange it:

\[
at=v-u
\]

Therefore:

\[
v=u+at
\]

That's algebra. No differentiation.

The displacement equation can also be obtained using ordinary geometry. When acceleration is constant, the velocity-time graph is a straight line. Its area is a trapezium, and that area represents displacement.

Since average velocity is \(\frac{u+v}{2}\):

\[
s=\frac{u+v}{2}t
\]

Substituting \(v=u+at\) gives:

\[
s=ut+\frac12at^2
\]

Again, this is algebra and geometry rather than integration.

[IMAGE: A labelled velocity-time graph for an object undergoing constant acceleration. Show initial velocity u at time zero, final velocity v at time t, a straight rising line, and the trapezium beneath it representing displacement. Include the equivalent rectangle and triangle areas to illustrate the algebraic derivation.]

There is one important condition: these equations assume constant acceleration. If acceleration changes throughout the motion, you cannot automatically use them. Recognising when a formula applies is more important than knowing additional mathematical techniques.

### A closer look at the 2025 syllabus equations

It helps to examine the mathematics across all seven focus areas rather than judging the course by a few complicated-looking equations.

| Syllabus focus area | Representative mathematical relationships | Main mathematical skills |
|---|---|---|
| Fundamentals of mechanics | \(v=u+at\), \(F=ma\), \(E_k=\frac12mv^2\) | Algebra, vectors, graphs, and rearranging equations |
| Waves | \(v=f\lambda\), \(f=\frac1T\) | Ratios, substitution, and trigonometry |
| Electricity and magnetism | \(F=qE\), \(V=IR\), \(P=VI\) | Algebra, proportionality, and powers |
| Advanced mechanics | \(a_c=\frac{v^2}{r}\), \(F_g=\frac{GMm}{r^2}\) | Algebra, square roots, and circular relationships |
| Electromagnetism | \(\Phi=BA\cos\theta\), \(\varepsilon=-N\frac{\Delta\Phi}{\Delta t}\) | Trigonometry, ratios, and finite rates of change |
| Nature of light | \(d\sin\theta=m\lambda\) | Trigonometry and geometric relationships |
| Matter, energy and the cosmos | \(E=hf\), \(E=mc^2\), \(\lambda=\frac{\ln 2}{T_{1/2}}\) | Algebra, scientific notation, logarithms, and powers |

These are representative equations, not an exhaustive formula sheet. The important point is that the mathematical techniques remain within algebra, trigonometry, graph interpretation, and related numerical methods.

There is no requirement to differentiate functions, evaluate definite integrals, or solve differential equations.

### What calculus adds to physics

Calculus becomes useful when a quantity changes continuously and you want to describe its behaviour at an exact instant.

Consider a car that accelerates unevenly. Its velocity-time graph might be curved rather than a straight line.

You can still calculate its average acceleration between two measurements:

\[
a_{\text{average}}=\frac{\Delta v}{\Delta t}
\]

However, this doesn't tell you the acceleration at every moment.

Calculus provides the relationship:

\[
a(t)=\frac{dv}{dt}
\]

This means acceleration is the instantaneous rate of change of velocity.

Integration provides another useful relationship:

\[
\Delta s=\int_{t_1}^{t_2}v(t)\,dt
\]

It calculates displacement from the velocity function over a time interval.

[IMAGE: A curved velocity-time graph for a car with changing acceleration. Show two measured points connected by a secant line representing average acceleration, a tangent at a selected instant representing instantaneous acceleration, and the area beneath the curve representing displacement. Clearly distinguish the three quantities.]

These methods are powerful. They allow physicists to work with much more complicated motion than the constant-acceleration equations can describe.

But knowing they exist is different from being required to use them.

In HSC Physics, you should understand what gradients and areas represent. You may need to calculate a gradient using two points or estimate an instantaneous gradient using a tangent to a curve. You may also need to calculate areas using familiar geometric shapes.

None of this requires you to differentiate an equation or evaluate a formal integral.

## Which Physics syllabus applies to you?

This is particularly important in 2026 because NSW is changing its senior science syllabuses.

NESA is replacing the 2017 Physics Stage 6 Syllabus with the Physics 11-12 Syllabus (2025).

According to NESA's implementation schedule:

| Year | Physics syllabus arrangements |
|---|---|
| 2026 | Schools prepare to teach the replacement syllabus |
| Term 1, 2027 | New syllabus begins for Year 11 |
| 2027 | Year 12 continues with the 2017 syllabus |
| Term 4, 2027 | New syllabus begins for Year 12 |
| 2028 | First HSC examination under the 2025 syllabus |

Therefore:
- Students completing the HSC in 2026 or 2027 study the 2017 syllabus.
- Students starting Year 11 in 2027 study the 2025 syllabus and sit the first HSC examination for that course in 2028.

You can check the dates directly on [NESA's official Physics syllabus page](https://curriculum.nsw.edu.au/learning-areas/science/physics-11-12-2025/content).

This distinction also explains why some online study resources refer to Modules 1-8. Those module names belong to the 2017 syllabus. They should not be treated as the structure of the replacement course.

### Does NESA require Mathematics Advanced?

No. NESA does not set Mathematics Advanced, Mathematics Extension 1, or Mathematics Extension 2 as a prerequisite for HSC Physics.

The distinction is straightforward:

- Under the 2017 syllabus, Physics has no formal prerequisites.
- Under the 2025 syllabus, Year 11 Physics has no prerequisites.
- Under the 2025 syllabus, the prerequisite for Year 12 Physics is completion of Year 11 Physics.

Neither syllabus requires students to have studied calculus before enrolling in Physics.

You can confirm this in the official [2017 syllabus information](https://www.nsw.gov.au/education-and-training/nesa/curriculum/science/physics-stage-6-2017) and the [2025 syllabus course requirements](https://curriculum.nsw.edu.au/learning-areas/science/physics-11-12-2025/content).

### What if your school recommends Mathematics Advanced?

This is where you need to distinguish NESA requirements from individual school decisions.

A school may recommend Mathematics Advanced because it provides useful preparation for Physics. It may also have its own subject-selection advice or enrolment arrangements.

However, a recommendation from your school is not automatically a statewide NESA prerequisite.

If your school tells you that you need Mathematics Advanced, ask whether this is:
- A formal NESA requirement.
- A school-specific enrolment requirement.
- A recommendation based on the mathematical demands of the course.

Those are three different things.

## What mathematics should you actually practise?

If you want to prepare for HSC Physics, focus on the skills you will repeatedly use when solving problems.

You should be able to:

**1. Rearrange equations**

For example, given \(F=ma\), you should be comfortable finding any of the three variables:

\[
F=ma,\qquad a=\frac{F}{m},\qquad m=\frac{F}{a}
\]

This sounds basic, but complicated questions often become difficult because students struggle to rearrange an equation correctly.

**2. Work with scientific notation**

Physics regularly uses very large and very small quantities.

For example:

\[
3.0\times10^8\text{ m/s}
\]

is the speed of light in a vacuum.

You need to be comfortable with powers of ten, significant figures, and calculator notation.

**3. Use trigonometry**

You will encounter angles when resolving forces, analysing waves, and studying electromagnetic phenomena.

For a vector \(F\) at an angle \(\theta\) to the horizontal:

\[
F_x=F\cos\theta
\]

\[
F_y=F\sin\theta
\]

You should understand why the components use sine and cosine, rather than simply memorising the equations.

**4. Interpret graphs**

You need to recognise what the axes represent and what information the graph provides.

For example:
- A displacement-time gradient represents velocity.
- A velocity-time gradient represents acceleration.
- The area beneath a velocity-time graph represents displacement.

You should also recognise when a straight-line relationship indicates constant acceleration.

**5. Understand proportional relationships**

Suppose gravitational force follows:

\[
F=\frac{GMm}{r^2}
\]

If the distance \(r\) doubles, the force becomes one-quarter as large.

You should be able to recognise these relationships without calculating every possible value.

**6. Work with logarithms and exponentials**

These appear in radioactive decay and other relationships involving changing quantities.

You don't need to know how to differentiate exponential functions, but you should understand how to substitute values and use logarithms on your calculator.

[IMAGE: An illustrated physics study desk with an open exercise book showing a rearranged equation, a vector triangle with sine and cosine components, a velocity-time graph, and a scientific calculator displaying a power of ten. Each item should be recognisable as a practical HSC study skill.]

If some of these skills are unfamiliar, that doesn't mean Physics is unsuitable for you. It means you have specific mathematical foundations to strengthen before the course becomes more demanding.

For a closer look at the relationship between mathematics subject selection and Physics, see [Do You Need Maths Advanced for HSC Physics?](https://tutorgum.com/learn/resources/do-you-need-maths-advanced-for-hsc-physics/).

## Worked examples: solving HSC Physics problems without calculus

Let's apply the mathematics to two typical problems. Notice that neither requires differentiation or integration.

### Worked example: Finding acceleration and displacement

An electric scooter increases its velocity from 5.0 m/s to 17.0 m/s over 4.0 seconds. Assuming constant acceleration, calculate its acceleration and the displacement travelled during this interval.

[IMAGE: An electric scooter travelling along a straight, level path. Include initial velocity 5.0 m/s, final velocity 17.0 m/s, and a time interval of 4.0 s. Include a corresponding straight-line velocity-time graph with the initial and final values, without displaying the calculated acceleration or displacement.]

**Step 1**: Identify the known quantities.

\[
u=5.0\text{ m/s}
\]

\[
v=17.0\text{ m/s}
\]

\[
t=4.0\text{ s}
\]

We need to calculate acceleration first.

**Step 2**: Use the average acceleration equation.

\[
a=\frac{v-u}{t}
\]

Substitute the values:

\[
a=\frac{17.0-5.0}{4.0}
\]

\[
\boxed{a=3.0\text{ m/s}^2}
\]

**Step 3**: Calculate displacement using the constant-acceleration equation.

\[
s=ut+\frac12at^2
\]

Substitute:

\[
s=(5.0)(4.0)+\frac12(3.0)(4.0)^2
\]

\[
s=20+24
\]

\[
\boxed{s=44\text{ m}}
\]

The scooter travels 44 metres while accelerating at 3.0 m/s².

Notice the method: identify the variables, select the appropriate equation, rearrange if necessary, and substitute the values with their units.

No calculus is involved.

### Worked example: Calculating induced electromotive force

A coil contains 150 turns. The magnetic flux through each turn decreases from \(4.0\times10^{-4}\text{ Wb}\) to \(1.0\times10^{-4}\text{ Wb}\) over 0.030 seconds.

Calculate the magnitude of the average induced electromotive force (emf).

[IMAGE: A bar magnet being withdrawn from a 150-turn conducting coil connected to a voltmeter. Show the magnetic field passing through the coil and indicate that the magnetic flux decreases as the magnet moves away. Label the coil as 150 turns and show the two supplied flux measurements and time interval.]

**Step 1**: Identify the relevant equation.

Faraday's law gives the magnitude of the average induced emf:

\[
|\varepsilon|=N\frac{|\Delta\Phi|}{\Delta t}
\]

where:
- \(\varepsilon\) is induced emf in volts (V).
- \(N\) is the number of turns.
- \(\Delta\Phi\) is the change in magnetic flux per turn in webers (Wb).
- \(\Delta t\) is the time interval in seconds (s).

**Step 2**: Calculate the change in magnetic flux.

\[
\Delta\Phi=\Phi_{\text{final}}-\Phi_{\text{initial}}
\]

\[
=1.0\times10^{-4}-4.0\times10^{-4}
\]

\[
=-3.0\times10^{-4}\text{ Wb}
\]

The magnitude of the change is:

\[
|\Delta\Phi|=3.0\times10^{-4}\text{ Wb}
\]

**Step 3**: Substitute into Faraday's law.

\[
|\varepsilon|=
\frac{150(3.0\times10^{-4})}{0.030}
\]

\[
|\varepsilon|=\frac{0.045}{0.030}
\]

\[
\boxed{|\varepsilon|=1.5\text{ V}}
\]

The average induced emf has a magnitude of 1.5 V.

The negative sign in the full form of Faraday's law describes the direction of the induced emf, as determined by Lenz's law. It does not mean that the magnitude of the emf is negative.

Again, this calculation uses algebra and a finite rate of change. You do not need to differentiate the magnetic flux function.

## Does calculus matter for university physics?

Yes. This is where the distinction becomes important.

HSC Physics and university physics are different levels of study. Although the HSC course doesn't require calculus, many university physics and engineering subjects use it extensively.

For example, the University of Newcastle's 2026 Advanced Physics I course is explicitly described as calculus-based. Its stated assumed knowledge includes HSC Mathematics Advanced (Band 5 or 6) and HSC Physics.

This is a university subject's preparation requirement, not a prerequisite for studying HSC Physics.

### University prerequisites and assumed knowledge are different

When choosing HSC subjects, check the requirements of the degrees you might want to study.

UAC distinguishes three important categories:

| Category | Meaning |
|---|---|
| Course prerequisite | A subject or result you must have to qualify for admission |
| Assumed knowledge | Knowledge the university expects you to have, even if it isn't a formal admission requirement |
| Recommended studies | Subjects that provide useful preparation but aren't necessarily expected |

For students selecting subjects in 2026, UAC's [2026 Steps to Uni for Year 10 Students](https://uac.edu.au/assets/documents/year-10/year-10-booklet-2029.pdf) lists courses that participating institutions plan to offer in 2029.

Consider two examples from current university information.

**The University of Sydney**

For its 2029 Engineering course listings, the University of Sydney specifies a mathematics prerequisite of:
- Mathematics Advanced (Band 4), or
- Mathematics Extension 1 (Band 3), or
- Mathematics Extension 2 (Band E3).

Mathematics Extension 1 is also listed as assumed knowledge for several engineering streams.

These are university admission and preparation requirements. They are not NESA requirements for HSC Physics.

You can check the university's [mathematics prerequisite information](https://www.sydney.edu.au/study/applying/how-to-apply/undergraduate/mathematics-prerequisite.html).

**UNSW Sydney**

UNSW takes a different approach. It states that it does not have formal subject prerequisites for its degrees, but some degrees have assumed knowledge.

Students who haven't studied the assumed subjects may still be considered for admission, but could find the first-year content more difficult.

Read UNSW's [official explanation of assumed knowledge](https://www.unsw.edu.au/study/how-to-apply/undergraduate/entry-requirements/assumed-knowledge).

The lesson is simple: a university's mathematics requirements depend on the course and institution. Don't confuse them with the requirements of the NSW HSC Physics syllabus.

Also remember that university admission information can change. Check the requirements for your intended year of entry before finalising your HSC subjects.

## Should you choose Mathematics Advanced or Extension 1?

If you enjoy mathematics and are considering physics, engineering, or another mathematically demanding degree, Mathematics Advanced and Mathematics Extension 1 can be valuable choices.

They provide mathematical preparation that will become useful beyond the HSC.

However, you shouldn't choose an extension course simply because you think HSC Physics requires calculus.

A sensible approach is:

- Choose Mathematics Advanced or Extension courses if they suit your abilities, interests, and future study plans.
- Choose Physics because you are interested in understanding the physical world and are prepared to develop its mathematical skills.
- If you study a less advanced mathematics course, pay particular attention to algebra, trigonometry, vectors, and graph interpretation.
- If you have a specific university degree in mind, check its actual prerequisites and assumed knowledge before making your decision.

You should also avoid choosing Physics purely because of its perceived ATAR scaling.

UAC calculates the ATAR using scaled marks from the relevant 10 units of HSC courses. Scaling adjusts for differences between the academic achievement of students taking different courses. It isn't a fixed bonus awarded to everyone studying a particular subject.

A student still needs strong HSC results to achieve strong scaled marks. UAC explains this in its guide to [how the ATAR is calculated](https://uac.edu.au/future-applicants/atar/how-your-atar-is-calculated).

Choose your subjects based on your abilities, interests, and future plans rather than assuming a particular course will automatically improve your ATAR.

## Questions and solutions

The following questions test the mathematics you actually need for HSC Physics. They progress from straightforward calculations to more demanding reasoning about when a mathematical model is valid.

### Question 1

A cyclist travelling in a straight line increases their velocity from 6.0 m/s east to 18.0 m/s east over 3.0 seconds.

Calculate the cyclist's average acceleration.

### Solution 1

The average acceleration is **4.0 m/s² east**.

Use the average acceleration equation:

\[
a=\frac{v-u}{t}
\]

Substitute:

\[
a=\frac{18.0-6.0}{3.0}
\]

\[
a=\frac{12.0}{3.0}
\]

\[
\boxed{a=4.0\text{ m/s}^2\text{ east}}
\]

The positive direction is east, so the acceleration is also directed east.

This question only requires identifying the relevant variables and applying an algebraic relationship. It doesn't require calculus.

### Question 2

A runner moves along a straight track. Their velocity changes uniformly from 2.0 m/s at \(t=0\) to 8.0 m/s at \(t=4.0\) seconds. The runner then maintains a constant velocity of 8.0 m/s until \(t=7.0\) seconds.

Calculate:

**(a)** The total displacement over the seven seconds.

**(b)** The average velocity over the seven seconds.

[IMAGE: A velocity-time graph for a runner. The graph starts at (0 s, 2 m/s), increases linearly to (4 s, 8 m/s), and remains horizontal at 8 m/s until (7 s, 8 m/s). Label both axes and all three points. Do not include the calculated displacement or average velocity.]

### Solution 2

**(a) The total displacement is 44 m.**

**(b) The average velocity is 6.3 m/s in the direction of motion.**

The area beneath a velocity-time graph represents displacement.

Break the graph into two sections.

**Step 1**: Calculate the area of the first section.

The first section is a trapezium.

\[
A_1=\frac12(u+v)t
\]

\[
A_1=\frac12(2.0+8.0)(4.0)
\]

\[
A_1=20\text{ m}
\]

**Step 2**: Calculate the area of the second section.

The second section is a rectangle.

\[
A_2=vt
\]

\[
A_2=8.0(7.0-4.0)
\]

\[
A_2=24\text{ m}
\]

**Step 3**: Calculate the total displacement.

\[
s=A_1+A_2
\]

\[
s=20+24
\]

\[
\boxed{s=44\text{ m}}
\]

**Step 4**: Calculate average velocity.

Average velocity is displacement divided by elapsed time:

\[
v_{\text{average}}=\frac{\Delta s}{\Delta t}
\]

\[
v_{\text{average}}=\frac{44}{7.0}
\]

\[
\boxed{v_{\text{average}}=6.3\text{ m/s}}
\]

The important skill here is recognising that the area beneath a graph can represent a physical quantity.

The calculation uses rectangles and trapezia, not integration.

### Question 3

An ideal transformer has a primary voltage of 240 V and a secondary voltage of 1200 V. The current in the primary coil is 0.50 A.

Calculate:

**(a)** The current in the secondary coil.

**(b)** The ratio of secondary turns to primary turns.

[IMAGE: A clearly connected ideal transformer showing a primary coil and secondary coil wound around the same iron core. Label the primary voltage as 240 V, the secondary voltage as 1200 V, and the primary current as 0.50 A. Do not show the secondary current or turns ratio.]

### Solution 3

**(a) The secondary current is 0.10 A.**

**(b) The secondary-to-primary turns ratio is 5:1.**

Assume an ideal transformer with no energy losses.

**Step 1**: Use conservation of power.

For an ideal transformer:

\[
V_pI_p=V_sI_s
\]

where:
- \(V_p\) is primary voltage.
- \(I_p\) is primary current.
- \(V_s\) is secondary voltage.
- \(I_s\) is secondary current.

**Step 2**: Rearrange to make secondary current the subject.

\[
I_s=\frac{V_pI_p}{V_s}
\]

Substitute:

\[
I_s=\frac{240(0.50)}{1200}
\]

\[
\boxed{I_s=0.10\text{ A}}
\]

**Step 3**: Calculate the turns ratio.

The transformer voltage relationship is:

\[
\frac{V_s}{V_p}=\frac{N_s}{N_p}
\]

where \(N_s\) and \(N_p\) are the numbers of secondary and primary turns.

Substitute:

\[
\frac{N_s}{N_p}=\frac{1200}{240}
\]

\[
\boxed{N_s:N_p=5:1}
\]

The transformer increases voltage by a factor of five while reducing current by a factor of five.

This problem combines two equations and requires you to rearrange them correctly. Again, no calculus is involved.

### Question 4

A radioactive sample has an activity equal to 12.5% of its initial activity after 24 hours.

Assuming radioactive decay follows the usual exponential decay model, calculate:

**(a)** The half-life of the sample.

**(b)** Its decay constant in \(\text{h}^{-1}\).

### Solution 4

**(a) The half-life is 8.0 hours.**

**(b) The decay constant is \(0.0866\text{ h}^{-1}\).**

**Step 1**: Convert the remaining activity into a fraction.

\[
12.5\%=\frac{12.5}{100}=0.125
\]

Notice that:

\[
0.125=\frac18=\left(\frac12\right)^3
\]

The sample has therefore undergone three half-lives.

**Step 2**: Calculate the half-life.

\[
T_{1/2}=\frac{24}{3}
\]

\[
\boxed{T_{1/2}=8.0\text{ h}}
\]

**Step 3**: Calculate the decay constant.

The relationship between decay constant and half-life is:

\[
\lambda=\frac{\ln 2}{T_{1/2}}
\]

Substitute:

\[
\lambda=\frac{0.693}{8.0}
\]

\[
\boxed{\lambda=0.0866\text{ h}^{-1}}
\]

The decay constant describes the probability of decay per unit time for an individual unstable nucleus, within the exponential decay model.

Notice that this calculation uses a logarithm, \(\ln 2\). It does not require differentiation or integration.

This is a good example of why you shouldn't assume that every advanced-looking mathematical expression involves calculus.

### Question 5

A cart moves along a straight track in the positive direction.

Its measured velocities are:

| Time | Measured velocity |
|---|---:|
| 0 s | 0 m/s |
| 2 s | 8 m/s |
| 4 s | 16 m/s |

The measured displacement over the full four seconds is 36 m.

Assume all measurements are exact, the cart never reverses direction, and there are no measurement errors.

A student calculates the average acceleration over each two-second interval:

\[
a_1=\frac{8-0}{2}=4\text{ m/s}^2
\]

\[
a_2=\frac{16-8}{2}=4\text{ m/s}^2
\]

The student concludes that acceleration must have been constant throughout the four seconds.

They then use the constant-acceleration equation to calculate displacement:

\[
s=ut+\frac12at^2
\]

\[
s=0+\frac12(4)(4^2)=32\text{ m}
\]

However, the measured displacement is 36 m.

Is the student's conclusion justified?

Explain:
- Why the two average accelerations being equal does not necessarily establish constant acceleration.
- Why the 36 m measurement contradicts the constant-acceleration model.
- Whether a physically possible velocity-time graph could pass through all three measured points and still produce 36 m of displacement.

[IMAGE: A velocity-time graph with labelled axes and exactly three supplied data points: (0 s, 0 m/s), (2 s, 8 m/s), and (4 s, 16 m/s). Do not connect the points or draw a proposed curve. Include a separate annotation stating that total measured displacement is 36 m. Leave the velocity profile between measurements undetermined.]

### Solution 5

**No. The student's conclusion is incorrect. The measurements establish equal average accelerations over two intervals, but they do not prove that acceleration was constant throughout the motion.**

This is an important distinction between an average rate of change and an instantaneous rate of change.

**Step 1**: Understand what the measured velocities establish.

Between 0 and 2 seconds:

\[
a_{\text{average},1}
=\frac{8-0}{2}
=4\text{ m/s}^2
\]

Between 2 and 4 seconds:

\[
a_{\text{average},2}
=\frac{16-8}{2}
=4\text{ m/s}^2
\]

Both intervals have the same average acceleration.

However, this only tells us the overall change in velocity during each interval.

It doesn't tell us exactly how velocity changed between the measurements.

**Step 2**: Test the constant-acceleration assumption.

If acceleration really were constant at 4 m/s², the velocity-time graph would be a straight line.

Its area would be a triangle:

\[
s=\frac12\times4\times16
\]

\[
\boxed{s=32\text{ m}}
\]

This conflicts with the measured displacement of 36 m.

Because the measurements are assumed exact, constant acceleration cannot explain all the observations.

**Step 3**: Consider whether a different velocity profile is possible.

It is.

For example, consider a piecewise-linear velocity-time graph with these additional points:

| Time | Velocity |
|---|---:|
| 0 s | 0 m/s |
| 1 s | 8 m/s |
| 2 s | 8 m/s |
| 3 s | 12 m/s |
| 4 s | 16 m/s |

This profile passes through all three original measurements.

Its displacement can be calculated using trapezia and a rectangle.

From 0 to 1 second:

\[
A_1=\frac12(0+8)(1)=4\text{ m}
\]

From 1 to 2 seconds:

\[
A_2=8(1)=8\text{ m}
\]

From 2 to 3 seconds:

\[
A_3=\frac12(8+12)(1)=10\text{ m}
\]

From 3 to 4 seconds:

\[
A_4=\frac12(12+16)(1)=14\text{ m}
\]

Adding the areas:

\[
s=4+8+10+14
\]

\[
\boxed{s=36\text{ m}}
\]

[IMAGE: A velocity-time graph showing the original measured points at 0, 2, and 4 seconds, together with one possible piecewise-linear velocity profile passing through the additional points (1 s, 8 m/s) and (3 s, 12 m/s). Shade the four regions beneath the graph and label their areas 4 m, 8 m, 10 m, and 14 m. Clearly distinguish the proposed profile from the original measured data.]

This example demonstrates that the same measured velocities can be consistent with different acceleration histories.

The proposed profile has different gradients in different sections, so its acceleration is not constant. Yet its average acceleration over each original two-second interval is still 4 m/s².

**The crucial lesson:** equal average rates of change do not prove that a rate was constant at every instant.

The student has confused what the measurements establish with what the constant-acceleration model assumes.

To determine the actual instantaneous acceleration throughout the motion, we would need more information about the velocity function or additional measurements. Calculus could help if that continuous function were known, but recognising the limitation of the constant-acceleration model requires no calculus.

## What should you take away?

You do not need calculus for NSW HSC Physics, including the replacement 2025 syllabus.

You do need to develop confidence with algebra, trigonometry, vectors, graphs, logarithms, and the correct use of physical equations.

The most useful preparation is learning to recognise what a question is asking, choosing an equation whose assumptions match the situation, and checking that your answer makes physical sense.

As you progress through Physics, pay particular attention to the difference between average and instantaneous quantities, and between constant and changing acceleration. These ideas will help you understand when the familiar HSC equations apply and when a more advanced mathematical model is needed.

Calculus is a useful next step if you continue into university physics or engineering. It isn't a hurdle you need to clear before beginning HSC Physics.