Quantities of Motion

Syllabus mapping

PY-11-01 — Fundamentals of Mechanics. Covers: scalar and vector quantities used in describing motion (distance, displacement, speed, velocity, acceleration); representing vectors graphically; adding and subtracting vectors, including resolving a vector into perpendicular components; distinguishing instantaneous and average speed/velocity.

Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.

You need to know: the difference between a scalar and a vector, how to add/subtract vectors (graphically and by components), and the difference between average and instantaneous speed/velocity — everything else in Mechanics builds on this.

Assumed knowledge check

Before this page makes sense, you should be comfortable with:

• Vectors — adding, subtracting, resolving into components
• Trigonometry — SOH-CAH-TOA, needed to resolve vectors

Core

Scalars vs vectors

A scalar is fully described by a magnitude (a number and a unit) alone. A vector needs a magnitude and a direction.

Scalar	Vector
Distance	Displacement
Speed	Velocity
—	Acceleration
Time, mass, energy	Force

• Distance — total path length travelled. Always positive, never decreases.
• Displacement, \(\vec{s}\) — straight-line distance and direction from start to finish. Can be zero even after a long trip, if you end up back where you started.

Worked example

A student walks 8 m east, then 6 m north.

• Distance travelled: \(8 + 6 = 14\ \text{m}\)
• Displacement: the straight-line vector from start to end. Magnitude by Pythagoras:

\[ |\vec{s}| = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{m} \]

Direction: \(\theta = \tan^{-1}\left(\dfrac{6}{8}\right) \approx 37^\circ\) north of east.

So the displacement is 10 m, N37°E — distance and displacement give different answers from the same trip.

Speed vs velocity

• Speed, \(v\) — rate of change of distance. Scalar.
• Velocity, \(\vec{v}\) — rate of change of displacement. Vector (has direction).

\[ \text{average speed} = \frac{\text{distance travelled}}{\text{time taken}} \qquad \vec{v}_{av} = \frac{\Delta \vec{s}}{\Delta t} \]

Matches your data sheet

This is written exactly as it appears on the HSC Physics data sheet — subscript av, not "avg". Small thing, but worth matching so the notation looks familiar in the exam.

Average vs instantaneous:

• Average velocity is measured over an interval — total displacement divided by total time.
• Instantaneous velocity is the velocity at one specific moment — what a speedometer shows right now, not the trip average.

For an object with changing speed, these are usually different. A car's speedo doesn't read its average speed for the whole journey — it reads instantaneous speed at that instant.

Vector diagrams and addition

🖼️ Diagram needed: vector addition shown both tip-to-tail and by components, plus a labelled example of a displacement vector with its angle from a reference direction. TODO — placeholder until sourced/drawn.

Vectors are drawn as arrows: length represents magnitude (to scale), the arrow direction represents direction. Two common ways to add vectors:

Graphical (tip-to-tail)Component method

Draw the first vector, then start the second vector's tail at the first vector's tip. The resultant is the single vector from the start of the first to the tip of the last. Good for building intuition and rough checks, but precision is limited by how carefully you draw and measure.

Resolve each vector into perpendicular (usually x and y) components, add the components separately, then recombine.

For a vector \(\vec{A}\) of magnitude \(A\) at angle \(\theta\) from the x-axis:

\[ A_x = A\cos\theta \qquad A_y = A\sin\theta \]

To add \(\vec{A}\) and \(\vec{B}\):

\[ R_x = A_x + B_x \qquad R_y = A_y + B_y \]

\[ |\vec{R}| = \sqrt{R_x^2 + R_y^2} \qquad \theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right) \]

This is the method that scales to exam-style problems — always show the component working, not just a final number.

Worked example — component method

A hiker walks 5 km at 30° north of east, then 3 km due north. Find the total displacement.

Vector 1 (\(30^\circ\) from east): \(A_x = 5\cos30^\circ = 4.33\ \text{km}\), \(A_y = 5\sin30^\circ = 2.50\ \text{km}\)

Vector 2 (due north): \(B_x = 0\), \(B_y = 3\ \text{km}\)

Resultant components: \(R_x = 4.33\ \text{km}\), \(R_y = 5.50\ \text{km}\)

\[ |\vec{R}| = \sqrt{4.33^2 + 5.50^2} \approx 7.00\ \text{km} \]

\[ \theta_R = \tan^{-1}\left(\frac{5.50}{4.33}\right) \approx 52^\circ \text{ north of east} \]

Advanced

Combining three or more vectors, and non-right-angle problems. The component method extends directly to any number of vectors — sum all the x-components, sum all the y-components, then recombine. This is also the standard approach when vectors aren't conveniently perpendicular to begin with: resolve everything onto a common x–y axis first, even if none of the original vectors line up with those axes.

Vector subtraction — \(\vec{A} - \vec{B}\) is \(\vec{A} + (-\vec{B})\): reverse \(\vec{B}\)'s direction (flip the sign of its components) and add as normal. This matters for change-in-velocity problems (\(\Delta \vec{v} = \vec{v}_f - \vec{v}_i\)) later in Momentum and Energy, and comes back explicitly in relative velocity in Motion Relationships — being solid on subtraction now avoids a stall later.

Extension

Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.

📎 Depth study idea

Instantaneous velocity is formally the derivative of displacement with respect to time, \(\vec{v} = \dfrac{d\vec{s}}{dt}\) — the average-velocity formula is really just the gradient of a secant line on a displacement–time graph, and instantaneous velocity is what that gradient approaches as the time interval shrinks to zero (the gradient of the tangent). Students doing Extension 1/2 Maths will recognise this immediately as a limit definition of a derivative. Worth a genuine depth study if you want to connect kinematics to calculus explicitly — e.g. deriving the SUVAT equations from \(a = \frac{dv}{dt}\) by integration, rather than taking them as given — see the worked derivation in the Extension box on Motion Relationships.

Video/visual resources

🖥️ PhET Simulation — Vector Addition

If the simulation doesn't load (some school networks block embedded content), open it directly: PhET — Vector Addition

🖥️ PhET Simulation — Moving Man

Good fit for the speed-vs-velocity and instantaneous-vs-average content above — drive the man's position/velocity/acceleration directly and watch the live graphs respond.

If the simulation doesn't load (some school networks block embedded content), open it directly: PhET — Moving Man

• 🎥 Khan Academy — Intro to vectors and scalars
• 🎥 Khan Academy — Vector components from magnitude and direction
• 🎥 Khan Academy — Adding and subtracting vectors
• 🎥 Physics High — Understanding vectors

Check yourself

1. A cyclist rides 400 m east, then 300 m south. Calculate (a) the distance travelled and (b) the magnitude and direction of the displacement.

   Answer

   (a) Distance \(= 400 + 300 = 700\ \text{m}\)

   (b) \(|\vec{s}| = \sqrt{400^2 + 300^2} = 500\ \text{m}\), direction \(\theta = \tan^{-1}(300/400) \approx 37^\circ\) south of east.

2. Explain why a car's average velocity over a full lap of a circular track is always zero, even though its average speed is not.

   Answer

   Displacement is start-to-finish in a straight line — a full lap ends where it started, so the displacement vector is zero, making average velocity zero. Distance travelled (the full lap length) is not zero, so average speed is not zero. This shows why speed and velocity aren't interchangeable.

3. A boat's velocity has components \(v_x = 6\ \text{m s}^{-1}\) and \(v_y = 8\ \text{m s}^{-1}\). Find the magnitude and direction of the resultant velocity.

   Answer

   \(|\vec{v}| = \sqrt{6^2 + 8^2} = 10\ \text{m s}^{-1}\)

   \(\theta = \tan^{-1}(8/6) \approx 53^\circ\) from the x-axis (whichever direction that axis represents)