What Maths Should You Revise Before Year 11 Physics?

You don't need to master all of Year 10 mathematics before starting Year 11 Physics. But you do need to be comfortable rearranging equations, reading graphs, working with angles, and handling numbers written in scientific notation. These skills appear repeatedly in physics, often when you're also trying to understand a new physical idea.

The good news is that the maths used in Year 11 Physics is quite manageable if you have the foundations in place. You don't need to complete Mathematics Advanced or Extension 1 before studying Physics. You should, however, be prepared to use basic algebra, trigonometry, and vectors with confidence.

This guide covers the six areas worth revising before Year 11: algebra, graphs, trigonometry, vectors, units, and scientific notation. It also includes a diagnostic checklist and practice questions so you can identify what needs attention before your first physics lessons.

## Do you need to be good at maths to study Year 11 Physics?

You need to be willing to use mathematics, but you don't need to be an outstanding mathematician.

Physics uses mathematics to describe relationships between measurable quantities. For example, if a car accelerates from rest, you might calculate how far it travels in a certain time. If a ball is thrown at an angle, you might separate its horizontal and vertical motion. If a circuit contains several resistors, you might use algebra to calculate the current.

The mathematics usually helps you describe what is happening physically. It isn't a separate collection of complicated tricks.

A useful way to think about your preparation is this:

- **Essential foundations:** algebra, substitution, rearranging equations, units, scientific notation, and reading graphs.
- **Very useful foundations:** right-angled trigonometry, ratios, gradients, and basic vector addition.
- **Helpful but not essential before starting:** more advanced algebraic techniques, logarithms, calculus, and complex trigonometric identities.

You will develop some mathematical skills as you study physics. You don't have to arrive knowing everything.

01Algebra: rearranging equations

Algebra is probably the most important mathematical skill to get comfortable with before Physics.

Physics uses equations to describe relationships between quantities. You’ll often be given an equation and asked to calculate a quantity that isn’t already isolated.

For example, consider:

\[ v=u+at \]

Here:

• \(v\) = final velocity, in metres per second (m/s)
• \(u\) = initial velocity, in metres per second (m/s)
• \(a\) = acceleration, in metres per second squared (m/s²)
• \(t\) = time, in seconds (s)

Suppose you know the initial velocity, final velocity, and time, but need to calculate acceleration.

You need to rearrange the equation:

\[ v=u+at \]

Subtract \(u\) from both sides:

\[ v-u=at \]

Divide both sides by \(t\):

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

That’s all equation rearrangement involves. You’re performing the same operation on both sides until the unknown quantity is by itself.

Before starting Physics, make sure you can:

• Substitute numbers into equations without losing brackets.
• Rearrange simple equations to isolate an unknown.
• Work confidently with fractions and negative numbers.
• Calculate squares and square roots.
• Solve basic linear equations.

You don’t need complicated algebraic techniques. Being accurate with straightforward algebra is much more useful than knowing an advanced technique but making mistakes with signs.

02Graphs: understanding gradients and relationships

Physics uses graphs constantly.

You’ll use them to describe how an object’s position changes, how quickly it moves, and how its motion changes over time.

Imagine watching a car travel along a straight road. You record its position every two seconds.

You could put time on the horizontal axis and position on the vertical axis. Each measurement becomes a point on the graph.

The important skill is understanding what the graph tells you, rather than simply plotting the points.

Understanding the gradient

The gradient measures how quickly the vertical quantity changes relative to the horizontal quantity.

The formula is:

\[ \text{Gradient}=\frac{\text{change in }y}{\text{change in }x} \]

For a position-time graph:

\[ \text{Gradient}= \frac{\text{change in position}}{\text{change in time}} \]

This gives velocity.

For a velocity-time graph:

\[ \text{Gradient}= \frac{\text{change in velocity}}{\text{change in time}} \]

This gives acceleration.

The units also tell you what you’re calculating.

There is another useful relationship: the area under a velocity-time graph represents displacement.

You don’t need to master every type of physics graph before Year 11. Start by understanding gradients, intercepts, and how to read the axes.

A common graph misconception

Suppose a position-time graph is a straight line that slopes upwards.

Does this mean the object is accelerating?

No. A straight line has a constant gradient, which means the object has constant velocity.

A steeper straight line means a greater velocity. Acceleration occurs when the velocity changes, which appears as a changing gradient on a position-time graph.

Make sure you can:

• Identify the independent and dependent variables.
• Read values from both axes.
• Calculate the gradient between two points.
• Interpret what a positive, negative, or zero gradient means.
• Recognise the difference between a constant gradient and a changing gradient.

03Trigonometry: working with angles

Trigonometry becomes particularly useful when Physics involves forces, motion, or other quantities acting at an angle.

You don’t need to revise every trigonometric identity from Year 10. Focus on right-angled triangles.

Imagine pulling a suitcase with a strap angled upwards. The pull has two effects:

• It pulls the suitcase forwards.
• It also pulls upwards.

Trigonometry allows us to calculate how much of the pull acts in each direction.

The three basic trigonometric ratios

For a right-angled triangle, remember SOH CAH TOA:

\[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \]

\[ \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \]

\[ \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \]

The hypotenuse is always the longest side, opposite the right angle.

The opposite and adjacent sides depend on which angle you’re considering.

You should also be comfortable with Pythagoras’ theorem:

\[ c^2=a^2+b^2 \]

This allows you to calculate the third side of a right-angled triangle when you know the other two.

Understanding your calculator

One small detail can cause some very frustrating mistakes.

Make sure your calculator is set to degrees when working with angles measured in degrees.

For example:

\[ \sin(30^\circ)=0.5 \]

If your calculator is in radians mode, you’ll get a different result.

Before starting Physics, practise using:

• Sine, cosine, and tangent.
• Pythagoras’ theorem.
• Inverse trigonometric functions, such as \(\tan^{-1}\).
• Angles measured from the horizontal or vertical.

You don’t need to memorise complicated trigonometric identities.

04Vectors: understanding magnitude and direction

This is one of the most useful mathematical ideas to revise before Year 11 Physics.

Consider two people walking 5 metres.

One walks 5 metres east. The other walks 5 metres north.

Have they travelled the same distance? Yes.

Have they moved in the same direction? No.

That difference is important in physics.

A scalar is a quantity with magnitude only. A vector has both magnitude and direction.

For example, saying a car travels at 20 m/s gives its speed.

Saying it travels at 20 m/s east gives its velocity.

Adding vectors

You cannot always add vectors like ordinary numbers.

If two people push a box in the same direction, their forces add.

If they push in opposite directions, the forces subtract.

For perpendicular vectors, you can use Pythagoras’ theorem to calculate the resultant magnitude.

For example, if one vector has a horizontal component of 3 units and a vertical component of 4 units, its magnitude is:

\[ R=\sqrt{3^2+4^2} \]

\[ R=\sqrt{25}=5 \]

You can also calculate the direction using:

\[ \tan\theta=\frac{\text{vertical component}} {\text{horizontal component}} \]

Therefore:

\[ \theta=\tan^{-1}\left(\frac{4}{3}\right) \]

\[ \theta=53.1^\circ \]

The vector has a magnitude of 5 units and points 53.1° north of east.

Breaking a vector into components

The reverse process is called resolving a vector into components.

Suppose a force \(F\) acts at an angle \(\theta\) above the horizontal.

Its horizontal and vertical components are:

\[ F_x=F\cos\theta \]

\[ F_y=F\sin\theta \]

Here:

• \(F_x\) is the horizontal component.
• \(F_y\) is the vertical component.
• \(F\) is the original force.
• \(\theta\) is the angle measured from the horizontal.

The reason cosine is used horizontally is that the horizontal component is adjacent to the angle. The vertical component is opposite the angle, so we use sine.

Be careful: if the angle is measured from the vertical instead, the sine and cosine relationships swap.

05Units: converting measurements correctly

A physics calculation can be mathematically perfect and still produce the wrong answer if you use inconsistent units.

Imagine calculating the speed of a car using a distance of 2 kilometres and a time of 10 seconds.

You cannot simply substitute 2 and 10 into the equation and call the answer metres per second.

You need to convert the distance into metres first.

Physics generally uses the International System of Units (SI).

Some important units to recognise are:

Conversions worth memorising

For example:

\[ 72\text{ km/h}=\frac{72}{3.6} \]

\[ =20\text{ m/s} \]

A useful habit is to convert everything into the appropriate SI units before substituting values into an equation.

There is one extra detail worth remembering: area and volume conversions involve squared and cubed units.

For example:

\[ 1\text{ cm}=0.01\text{ m} \]

But:

\[ 1\text{ cm}^2=0.0001\text{ m}^2 \]

The conversion factor must also be squared when converting area.

Why units help you check your answers

Units aren’t just something you add at the end of a calculation.

They help you identify mistakes.

Suppose you calculate acceleration using a change in velocity divided by time:

\[ a=\frac{\text{m/s}}{\text{s}} \]

This gives:

\[ \text{m/s}^2 \]

That makes sense because acceleration is measured in metres per second squared.

If your answer has units of m/s instead, you’ve probably calculated velocity rather than acceleration.

06Scientific notation: working with very large and small numbers

Physics deals with quantities that can be extremely large or extremely small.

The distance between planets is enormous. The dimensions of atoms are tiny.

Writing all these numbers using ordinary decimal notation becomes inconvenient.

Scientific notation solves this problem.

It expresses a number in the form:

\[ a\times10^n \]

where \(a\) is a number greater than or equal to 1 and less than 10, and \(n\) is an integer.

For example:

\[ 450\\,000=4.5\times10^5 \]

\[ 0.00032=3.2\times10^{-4} \]

A positive power of ten represents a large number. A negative power represents a small number.

Multiplying and dividing powers of ten

You should be comfortable with two basic rules.

When multiplying powers of ten, add the exponents:

\[ 10^3\times10^4=10^7 \]

When dividing powers of ten, subtract the exponents:

\[ \frac{10^6}{10^2}=10^4 \]

These rules are particularly useful when substituting scientific notation into physics equations.

Common prefixes

You’ll encounter these prefixes regularly:

For example:

\[ 5.0\text{ mA}=5.0\times10^{-3}\text{ A} \]

The lowercase m in mA means milli, not metres.

Also, learn how to enter scientific notation into your calculator. Many calculators use an EXP or EE button instead of requiring you to type out the entire power of ten.

07Your pre-Physics maths diagnostic checklist

You don’t need to be perfect at every skill before starting Year 11. However, you should be able to attempt the following tasks independently.

Use this checklist to identify what you should revise.

Algebra

• I can substitute numbers into an equation using brackets correctly.
• I can rearrange an equation to make a different variable the subject.
• I can work with fractions, negative numbers, and powers.
• I can calculate squares and square roots.

Graphs

• I can identify the independent and dependent variables.
• I can read values from a graph.
• I can calculate the gradient between two points.
• I understand what a positive, negative, or zero gradient represents.
• I can interpret the units of a gradient.

Trigonometry

• I can use sine, cosine, and tangent in right-angled triangles.
• I can apply Pythagoras’ theorem.
• I can use inverse trigonometric functions to find angles.
• I know how to switch my calculator into degree mode.

Vectors

• I understand the difference between scalar and vector quantities.
• I can add vectors pointing in the same or opposite directions.
• I can calculate the magnitude of perpendicular vectors.
• I can resolve an angled vector into horizontal and vertical components.
• I can describe a vector’s direction clearly.

Units

• I can convert kilometres, centimetres, and millimetres into metres.
• I can convert grams into kilograms.
• I can convert between km/h and m/s.
• I understand why units must be consistent within an equation.
• I can identify the correct units for common physical quantities.

Scientific notation

• I can express large and small numbers using powers of ten.
• I can multiply and divide numbers in scientific notation.
• I understand common metric prefixes.
• I can enter scientific notation into my calculator correctly.

If you struggle with algebra, units, or scientific notation, start there. These skills appear across almost every area of physics.

If vectors and trigonometry are unfamiliar, give those extra attention before studying mechanics.

08Do you need Mathematics Advanced or Extension 1 for Year 11 Physics?

No HSC Mathematics Advanced or Mathematics Extension 1 course is listed as a statewide NESA prerequisite for entering Year 11 Physics.

There are three different things worth separating.

1. NESA course eligibility

NESA’s published ACE Rules set out the general requirements for entering Preliminary courses. These include satisfactory completion of Year 10 and the required courses of study for the RoSA.

The rules do not specify Mathematics Advanced or Mathematics Extension 1 as a statewide entry requirement for Physics.

You can check the current NESA course eligibility rules.

2. Your school’s subject-selection policy

Individual schools may provide their own recommendations or apply local subject-selection processes.

A school might recommend Mathematics Advanced because it provides useful preparation for Physics. Another school may have different advice based on its students and course arrangements.

Check your school’s subject-selection handbook and speak with the Physics head teacher before making your decision.

A school’s recommendation should not be confused with a statewide NESA prerequisite.

3. University admission requirements

University requirements are a separate issue.

Some university degrees have formal subject prerequisites. Others list assumed knowledge, which means the university expects you to have studied a particular subject but does not necessarily require it for admission.

For example, the UNSW 2027 Domestic Undergraduate Student Guide lists Mathematics Advanced or Mathematics Extension 1 as assumed knowledge for its Bachelor of Science, depending on the chosen major. UNSW also explains that assumed knowledge is not a formal prerequisite for admission, although it strongly recommends filling any knowledge gaps.

The University of Sydney similarly distinguishes between formal prerequisites and assumed knowledge. Some of its courses require a Band 4 result in Mathematics Advanced, while assumed knowledge is not required to receive an offer.

These are university-specific requirements, not requirements for entering HSC Physics.

You can check the UNSW assumed knowledge guidance and the University of Sydney prerequisite and assumed knowledge explanation.

Admission information checked 8 November 2026. The UNSW example relates to 2027 undergraduate entry. Always check the requirements for your intended university course and admission year, as these can change.

For a more detailed discussion of the relationship between HSC Mathematics and Physics, read Do You Need Maths Advanced for HSC Physics?.

What about Physics scaling?

You might have heard that Physics scales well and therefore automatically improves your ATAR.

That’s not how scaling works.

UAC calculates the ATAR using scaled HSC marks. Scaling considers the performance of students studying each course across their other HSC subjects. It is not a fixed bonus awarded to everyone who studies Physics.

Physics has traditionally scaled strongly, but that doesn’t guarantee a particular scaled mark or ATAR outcome.

UAC specifically advises students to choose subjects based on their interests, demonstrated abilities, and future study plans rather than predicted scaling advantages.

You can read the official explanation at UAC: How your ATAR is calculated.

Choose Physics because you’re interested in understanding the physical world and are prepared to develop the necessary skills, not because you expect an automatic ATAR advantage.

If you’re still deciding whether Physics belongs in your HSC subject combination, read Should I Choose HSC Physics? A Practical Year 10 Guide.

09Which Physics syllabus will you study?

This is particularly important if you’re entering Year 11 in 2027.

As at 8 November 2026, NESA’s replacement Physics 11-12 Syllabus (2025) is scheduled to begin implementation in 2027.

For students starting Year 11 in 2027:

• Year 11 teaching begins in Term 1, 2027.
• The new Year 12 course begins in Term 4, 2027.
• The first HSC examination for the replacement syllabus is in 2028.

The new Year 11 course covers three focus areas:

1. Fundamentals of mechanics
2. Waves
3. Electricity and magnetism

These areas make regular use of the mathematical skills covered in this guide, particularly algebra, graphs, trigonometry, vectors, and unit conversions.

You can read the official NESA Physics 11-12 Syllabus (2025).

Students who started Year 11 in 2026 remain on the Physics Stage 6 Syllabus (2017) for Year 12 in 2027.

This distinction matters when looking for study resources. The old Module 1-8 structure belongs to the 2017 syllabus. It should not be treated as the structure of the replacement 2025 syllabus.

You can find the 2017 Physics syllabus here.

10How should you prepare before starting Physics?

You don’t need to spend your entire summer learning Year 11 Physics.

A more useful approach is to identify your mathematical weaknesses and fix them before they become obstacles.

I’d suggest the following process.

Step 1: Complete the diagnostic checklist.

Work through each skill without looking at notes or examples.

If you cannot rearrange a basic equation, don’t skip straight to vector calculations. Fix the algebra first.

Step 2: Practise using physics-style questions.

Maths exercises often give you a formula and ask you to calculate an answer.

Physics questions require an extra step: you must work out which formula or mathematical relationship describes the situation.

Practise identifying the known quantities, the unknown quantity, and the relationship connecting them.

Step 3: Get comfortable showing your working.

Develop a consistent approach:

1. Write down the relevant equation.
2. Define the quantities you are using.
3. Convert units if necessary.
4. Rearrange the equation.
5. Substitute values.
6. Calculate the answer with units.
7. Check whether the answer makes physical sense.

Step 4: Practise explaining your answer.

If you calculate that an object accelerates at 2.5 m/s², you should be able to explain what that means.

Physics isn’t just about obtaining a number. You need to understand what the number tells you about the situation.

11Questions and solutions

Try these questions without looking at the solutions first. They progress from straightforward calculations to problems that combine several mathematical skills.

Question 1

A car travels 240 metres in 12 seconds.

Calculate its average speed in:

a) metres per second

b) kilometres per hour.

Solution 1

The car’s average speed is 20 m/s, or 72 km/h.

Step 1: Recall the average speed equation.

\[ v=\frac{d}{t} \]

Here, \(d\) is distance in metres and \(t\) is time in seconds.

Step 2: Substitute the values.

\[ v=\frac{240}{12} \]

\[ \boxed{v=20\text{ m/s}} \]

Step 3: Convert the answer into kilometres per hour.

To convert m/s into km/h, multiply by 3.6.

\[ v=20\times3.6 \]

\[ \boxed{v=72\text{ km/h}} \]

The car’s average speed is 20 metres per second, equivalent to 72 kilometres per hour.

The important skill here is knowing which units the question requires and converting between them correctly.

Question 2

A student records the position of an object moving along a straight track. The following measurements are obtained:

Assume the object moves at constant velocity between successive measurements.

a) Calculate the velocity during each of the four time intervals.

b) Calculate the average velocity over the entire eight seconds.

c) Calculate the average speed over the entire eight seconds.

d) Explain why average speed and average velocity are different in this situation.

Solution 2

The four interval velocities are 3 m/s, 3 m/s, 0 m/s, and -4 m/s. The average velocity is 0.5 m/s in the positive direction, while the average speed is 2.5 m/s.

Step 1: Calculate the velocity during each interval.

The gradient of a position-time graph gives velocity.

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

For the first interval, from 0 to 2 seconds:

\[ v_1=\frac{6-0}{2-0} =3\text{ m/s} \]

For the second interval, from 2 to 4 seconds:

\[ v_2=\frac{12-6}{4-2} =3\text{ m/s} \]

For the third interval, from 4 to 6 seconds:

\[ v_3=\frac{12-12}{6-4} =0\text{ m/s} \]

For the final interval, from 6 to 8 seconds:

\[ v_4=\frac{4-12}{8-6} =-4\text{ m/s} \]

The negative sign indicates that the object is moving in the opposite direction to the chosen positive direction.

Step 2: Calculate average velocity.

Average velocity depends on total displacement, not total distance travelled.

The object starts at 0 metres and finishes at 4 metres.

Therefore:

\[ \Delta x=4-0=4\text{ m} \]

The total time is 8 seconds.

\[ v_{\text{avg}}=\frac{\text{displacement}}{\text{time}} \]

\[ v_{\text{avg}}=\frac{4}{8} \]

\[ \boxed{v_{\text{avg}}=0.5\text{ m/s}} \]

Step 3: Calculate average speed.

Average speed uses total distance travelled.

During the first interval, the object travels 6 metres.

During the second interval, it travels another 6 metres.

It then remains stationary before travelling 8 metres backwards.

Therefore:

\[ d_{\text{total}}=6+6+0+8 \]

\[ d_{\text{total}}=20\text{ m} \]

Average speed is:

\[ v_{\text{speed}}=\frac{20}{8} \]

\[ \boxed{v_{\text{speed}}=2.5\text{ m/s}} \]

Step 4: Explain the difference.

Average speed measures the total distance travelled per second.

Average velocity measures the net displacement per second, including direction.

The object travels 20 metres but only finishes 4 metres from its starting position.

This is why the two answers are different.

The important lesson is that distance and displacement are not interchangeable, even though they are both measured in metres.

Question 3

Three ropes are attached to a ring.

• The first rope pulls with a force of 60.0 N east.
• The second rope pulls with a force of 80.0 N north.
• The third rope pulls with a force of 40.0 N west.

All three forces act in the same horizontal plane.

Calculate:

a) The horizontal and vertical components of the resultant force.

b) The magnitude of the resultant force.

c) The direction of the resultant force, measured anticlockwise from east.

Solution 3

The resultant force is 82.5 N at 76.0° anticlockwise from east.

Step 1: Calculate the horizontal component.

The two horizontal forces act in opposite directions.

Taking east as positive:

\[ F_x=60.0-40.0 \]

\[ \boxed{F_x=20.0\text{ N east}} \]

Step 2: Calculate the vertical component.

There is only one vertical force.

\[ \boxed{F_y=80.0\text{ N north}} \]

Step 3: Calculate the magnitude of the resultant.

The horizontal and vertical components are perpendicular, so use Pythagoras’ theorem.

\[ F_R=\sqrt{F_x^2+F_y^2} \]

Substitute:

\[ F_R=\sqrt{20.0^2+80.0^2} \]

\[ F_R=\sqrt{6800} \]

\[ \boxed{F_R=82.5\text{ N}} \]

Step 4: Calculate the direction.

Use tangent because we know the opposite and adjacent sides of the right-angled triangle.

\[ \tan\theta=\frac{80.0}{20.0} \]

\[ \theta=\tan^{-1}(4) \]

\[ \boxed{\theta=76.0^\circ} \]

The resultant force points 76.0° anticlockwise from east.

A common mistake would be to add all three force magnitudes:

\[ 60+80+40=180\text{ N} \]

That is incorrect because the forces do not all act in the same direction.

You must account for direction before calculating the resultant.

Question 4

An electrical sensor measures a charge of \(3.6\times10^{-3}\) C passing through a component in \(1.2\times10^{-2}\) seconds.

The electric current is calculated using:

\[ I=\frac{Q}{t} \]

where \(I\) is current in amperes, \(Q\) is charge in coulombs, and \(t\) is time in seconds.

a) Calculate the current in amperes, expressing your answer in scientific notation.

b) The sensor is then adjusted so that three times as much charge passes through the component in twice the original time. Calculate the new current.

Solution 4

The original current is 0.30 A, and the new current is 0.45 A.

Step 1: Substitute the original values.

\[ I=\frac{3.6\times10^{-3}} {1.2\times10^{-2}} \]

Step 2: Separate the numbers and powers of ten.

\[ I=\frac{3.6}{1.2} \times10^{-3-(-2)} \]

\[ I=3.0\times10^{-1}\text{ A} \]

Therefore:

\[ \boxed{I=0.30\text{ A}} \]

Step 3: Calculate the new charge and time.

The charge is tripled:

\[ Q_{\text{new}}=3Q \]

The time is doubled:

\[ t_{\text{new}}=2t \]

Step 4: Substitute these relationships into the current equation.

\[ I_{\text{new}}=\frac{3Q}{2t} \]

Since \(Q/t=I\):

\[ I_{\text{new}}=\frac{3}{2}I \]

\[ I_{\text{new}}=1.5\times0.30 \]

\[ \boxed{I_{\text{new}}=0.45\text{ A}} \]

The current increases by a factor of 1.5, rather than tripling.

This is a useful example of why understanding ratios matters. Increasing the charge increases the current, but increasing the time reduces the rate at which charge flows.

Both changes must be considered.

Question 5

A small cart moves along a horizontal surface. Its position is measured using two perpendicular coordinates:

• \(x\) measures displacement east or west.
• \(y\) measures displacement north or south.

The cart moves in three straight-line segments at constant velocity within each segment.

Its measured positions are:

Answer the following.

a) Calculate the total distance travelled by the cart.

b) Calculate its displacement vector and the magnitude of that displacement.

c) Calculate its average speed.

d) Calculate its average velocity, including its magnitude and direction.

e) Now suppose a second student receives only the east-west position measurements, \(x\), and has no information about the north-south motion. They calculate:

\[ \frac{6-0}{10}=0.60\text{ m/s} \]

They claim this is the cart’s average speed.

Is their conclusion correct? Explain why or why not. Your explanation must include a different possible motion that produces exactly the same east-west position readings but a different average speed.

Solution 5

The cart travels 23 m, has a displacement magnitude of 7.81 m, an average speed of 2.30 m/s, and an average velocity of 0.781 m/s at 39.8° north of east. The second student’s conclusion is incorrect because the east-west position data alone cannot determine the total distance travelled.

Step 1: Calculate the distance travelled in each segment.

From the first position to the second:

\[ d_1=12\text{ m} \]

From the second position to the third:

\[ d_2=5\text{ m} \]

From the third position to the fourth:

\[ d_3=6\text{ m} \]

The total distance is:

\[ d_{\text{total}}=12+5+6 \]

\[ \boxed{d_{\text{total}}=23\text{ m}} \]

Step 2: Calculate the displacement vector.

Displacement depends only on the starting and finishing positions.

The cart starts at \((0,0)\) and finishes at \((6,5)\).

Therefore, its displacement is:

• 6 metres east.
• 5 metres north.

In component form:

\[ \Delta\vec r=(6,5)\text{ m} \]

Step 3: Calculate the magnitude of the displacement.

The horizontal and vertical displacements are perpendicular.

Use Pythagoras’ theorem:

\[ |\Delta\vec r|=\sqrt{6^2+5^2} \]

\[ =\sqrt{61} \]

\[ \boxed{|\Delta\vec r|=7.81\text{ m}} \]

Step 4: Calculate average speed.

Average speed is total distance divided by total time.

The total time is 10 seconds.

\[ v_{\text{speed}}=\frac{23}{10} \]

\[ \boxed{v_{\text{speed}}=2.30\text{ m/s}} \]

Step 5: Calculate average velocity.

Average velocity is displacement divided by total time.

Its horizontal component is:

\[ v_x=\frac{6}{10}=0.60\text{ m/s east} \]

Its vertical component is:

\[ v_y=\frac{5}{10}=0.50\text{ m/s north} \]

Therefore, the average velocity vector has components of 0.60 m/s east and 0.50 m/s north.

Its magnitude is:

\[ v_{\text{avg}}= \sqrt{0.60^2+0.50^2} \]

\[ =\sqrt{0.61} \]

\[ \boxed{v_{\text{avg}}=0.781\text{ m/s}} \]

The direction is:

\[ \theta=\tan^{-1}\left(\frac{0.50}{0.60}\right) \]

\[ \boxed{\theta=39.8^\circ} \]

The average velocity is therefore 0.781 m/s at 39.8° north of east.

Step 6: Evaluate the second student’s conclusion.

The second student’s answer of 0.60 m/s is not the average speed.

It is the east-west component of the average velocity.

This distinction is important because the cart also moves north during part of its journey.

Here is the interesting part.

Suppose the cart had remained on the east-west axis for the entire journey instead. Its north-south coordinate would always be zero.

It could still produce exactly the same east-west readings:

• At 0 seconds, \(x=0\).
• At 4 seconds, \(x=12\).
• At 8 seconds, \(x=12\).
• At 10 seconds, \(x=6\).

However, its total distance would now be:

\[ d=12+0+6=18\text{ m} \]

Its average speed would be:

\[ v_{\text{speed}}=\frac{18}{10} \]

\[ \boxed{v_{\text{speed}}=1.80\text{ m/s}} \]

Compare the two possible motions:

Both motions produce exactly the same east-west position readings, but their average speeds are different.

Therefore, knowing only the east-west position data is not enough to calculate average speed.

This is the deeper idea: a single coordinate graph cannot necessarily describe an object’s entire motion in two dimensions.

The tempting mistake is to assume that the horizontal change in position tells you everything about how far the object travelled. It doesn’t. You also need to know what happened in the perpendicular direction.

This is why vectors, graphs, and the distinction between speed and velocity need to work together.

12What should you learn next?

Once you’re comfortable with these six mathematical skills, you’re in a much better position to start Year 11 Physics.

Your next step is to apply them to mechanics.

Start with displacement, velocity, and acceleration. Learn how to interpret motion graphs, calculate changes in velocity, and resolve forces into components.

From there, the same mathematical foundations will help you understand waves and electricity.

You don’t need to arrive at your first Physics lesson knowing every equation. You need to be able to understand a relationship, work with the mathematics, and explain what your answer means.

That’s the foundation worth building before Year 11.