Week 1 In-Depth — Why the Language of Motion Looks This Way

Read the cheatsheet first. This note explains why each concept exists and walks through one substantial worked example per topic, with every step on the page.

1 — Why the particle model exists

Real objects have size, shape, orientation, and internal structure. If you tried to track all of that for a flying ball or a sprinting runner, your equations would have hundreds of variables. The particle model throws all of it away and says: pretend the object is a single point with all its mass at that point.

You lose accuracy on questions about rotation, deformation, or air drag. You gain the ability to describe motion with three numbers per instant: a position $x$, a velocity $v$, and an acceleration $a$. The first half of EGD102 is built on that trade.

Why this works

A particle has no “front” or “back” — so its position is unambiguous. That means $\Delta x=x_f-x_i$ is a single number, not a collection of them, and $\bar{v}=\Delta x/\Delta t$ is well-defined. The moment you reintroduce shape (Week 2+), you’ll need angular quantities too. For now, one point is enough.

Worked example with all the steps — the slope of a piecewise $x$–$t$ graph

A displacement-vs-time plot has corners at $A(0,5)$, $B(4,2)$, $C(6,2)$, $D(10,7)$. Find the velocity on each segment and the average velocity over the whole interval.

Model. Object is a particle moving in 1D along the $x$-axis.

Visualise. Coordinate system: positive = right. Knowns are the four $(t,x)$ corners. Unknowns are the segment velocities $v_{AB}$, $v_{BC}$, $v_{CD}$ and the overall $\bar{v}$.

Solve. On any straight segment, velocity = slope of $x$ vs $t$:

$v_{AB}=\frac{s_B-s_A}{t_B-t_A}=\frac{2-5}{4-0}=-0.75m/s$

$v_{BC}=\frac{s_C-s_B}{t_C-t_B}=\frac{2-2}{6-4}=0m/s$

$v_{CD}=\frac{s_D-s_C}{t_D-t_C}=\frac{7-2}{10-6}=1.25m/s$

For the overall average, take the slope between first and last points:

$\bar{v}=\frac{s_D-s_A}{t_D-t_A}=\frac{7-5}{10-0}=0.2m/s$

Assess. Units are $m/s$ throughout. The signs make sense: the object moves backward (negative $v$) on $AB$, stops on $BC$, then moves forward more quickly on $CD$. The small positive $\bar{v}$ reflects that the net motion was tiny — mostly the object retraced its path.

The lesson is general: instantaneous velocity = local slope; average velocity = slope of the line between endpoints.

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2 — Why “scalar vs vector” is the central distinction

A scalar answers “how much?” A vector answers “how much, and which way?” If you treat a vector like a scalar (drop the direction), you can get the right magnitude with the wrong physics.

Why this works

Time, mass, and energy never have a “direction” — adding them is just adding numbers. But displacement, velocity, and acceleration are differences of positions, and positions live on an axis. Subtracting position vectors gives you both how far apart and which side. Lose the sign and you lose half the information.

The cleanest demonstration is the 400 m runner: total path length (a scalar — call it distance) is $400\mathrm{m}$, but net displacement (a vector) from start to finish is zero because the start and finish points are identical. Speed = $400/48.6\approx 8.23m/s$, velocity = $0/48.6=0m/s$. Same time, same path, different question.

Worked example with all the steps — the Olympic sprinter and the 400 m runner

(a) An Olympic sprinter completes $100\mathrm{m}$ in $10.49\mathrm{s}$. Find the average speed and the average velocity. (b) A runner completes one full $400\mathrm{m}$ lap of an oval in $48.6\mathrm{s}$. Find the average speed and the average velocity.

Model. Particle on a track.

Visualise. For (a): take the start line as origin $x_0=0$, positive = down the track. Knowns: $x_i=0$, $x_f=100\mathrm{m}$, $\Delta t=10.49\mathrm{s}$, $d=100\mathrm{m}$.

For (b): start = finish on the oval, so $x_i=0$ and $x_f=0$. Knowns: $d=400\mathrm{m}$ (path length), $\Delta t=48.6\mathrm{s}$.

Solve (a).

$\text{speed}=\frac{d}{t}=\frac{100}{10.49}=9.534\dots \approx 9.5m/s$

$\bar{v}=\frac{\Delta x}{\Delta t}=\frac{100-0}{10.49-0}=+9.5m/s(\text{right})$

Solve (b).

$\text{speed}=\frac{d}{t}=\frac{400}{48.6}=8.23m/s$

$\bar{v}=\frac{x_f-x_i}{\Delta t}=\frac{0-0}{48.6}=0m/s$

Assess. In (a), straight-line motion in one direction: speed and velocity have the same magnitude. In (b), the lap is closed: net displacement is zero, so average velocity is zero even though the runner is moving the whole time. This is not a trick — it is the entire reason scalar and vector get separate names.

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3 — Why a velocity-vs-time graph hides displacement under its curve

The cleanest formulation of kinematics says: $v(t)=dx/dt$. Differentiate position, get velocity. Run that backwards: integrate velocity, get position. Geometrically, an integral is an area — so the area under a $v$–$t$ curve is a displacement.

Why this works

Imagine slicing the $v$–$t$ graph into thin vertical strips of width $\Delta t$. Each strip has height $v(t)$ at some instant, so its area is $v(t)\Delta t$. But $v(t)\Delta t$ is approximately how far the particle moved during that strip. Sum all the strips — that’s the total displacement. Let the strips get arbitrarily thin and the sum becomes the integral.

The same argument reversed says: slope of $v$–$t$ is acceleration. A rate of change is a derivative, and a derivative is a slope. So:

$a=\frac{dv}{dt}v=\frac{dx}{dt}$ $x(t)=\int v(t)dtv(t)=\int a(t)dt$

You go up the chain (position → velocity → acceleration) by differentiating, and down it (acceleration → velocity → position) by integrating.

Worked example with all the steps — reading a piecewise $v$–$t$ graph

A particle has velocity $v=2m/s$ on $0\le t\le 2\mathrm{s}$, then $v$ increases linearly from $2m/s$ to $5m/s$ between $t=2\mathrm{s}$ and $t=4\mathrm{s}$. (a) Find the displacement over $0\le t\le 4\mathrm{s}$. (b) Draw the acceleration-vs-time graph.

Model. Particle in 1D.

Visualise. Sketch the $v$–$t$ graph: horizontal at $v=2$ from $t=0$ to $2$, then a straight line up to $v=5$ at $t=4$.

Solve (a) — displacement = area under the $v$–$t$ curve. Split the area into a rectangle on $[0,2]$ and a trapezium on $[2,4]$. Following the handwritten working (a $4\times 2$ rectangle from $0$ to $4$ plus a triangle on top from $2$ to $4$):

$$s_{0\to 4}=(4\times 2)+\frac{2\times (5-2)}{2}=8+3=11\mathrm{m}$$

(Equivalent split: rectangle on $[0,2]$ of area $2\times 2=4\mathrm{m}$, plus a trapezium on $[2,4]$ of area $\frac{1}{2}(2+5)(2)=7\mathrm{m}$. Same answer.)

Solve (b) — acceleration = gradient of $v$–$t$, in two segments:

• On $[0,2]\mathrm{s}$: $a=\frac{\Delta v}{\Delta t}=\frac{0}{2}=0m/{\mathrm{s}}^2$ (flat line, no change in velocity).
• On $[2,4]\mathrm{s}$: $a=\frac{v_4-v_2}{4-2}=\frac{5-2}{2}=1.5m/{\mathrm{s}}^2$.

The $a$–$t$ graph is therefore a step: zero from $0$ to $2\mathrm{s}$, then jumps to $1.5m/{\mathrm{s}}^2$ from $2\mathrm{s}$ to $4\mathrm{s}$.

Assess. Units: $m/{\mathrm{s}}^2$. Signs: positive acceleration matches a velocity that increases over time — consistent with the rising line. Magnitudes: $1.5m/{\mathrm{s}}^2$ is small (less than $g/6$), which fits the modest speed change shown.

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4 — How these connect

The four-step approach links every idea above:

Step	What it forces you to do
Model	Choose the particle model so you can use $x$, $v$, $a$ as single numbers.
Visualise	Pick axes (which way is positive?), so scalars and vectors get their signs straight. Define every symbol.
Solve	Apply $\bar{v}=\Delta s/\Delta t$ or $\bar{a}=\Delta v/\Delta t$, or read slope/area off a graph.
Assess	Check units, signs, and whether the magnitude is plausible.

The big single insight is that slope and area are the same idea read in two directions. Once you internalise that, motion graphs stop looking like a separate topic and start looking like a graphical version of $\bar{v}=\Delta s/\Delta t$.

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5 — Exam-style sample, end-to-end

An egg is dropped from a window. After $1.12\mathrm{s}$ of free fall, it is moving at $17.0m/s$ downward. On impact with the floor, it comes to rest in $0.121\mathrm{s}$. Find (a) the average acceleration during the fall, and (b) the average acceleration during the stop.

Model. Egg treated as a particle.

Visualise. Take downward = positive (override the default; declare this on the diagram). Knowns:

Symbol	Meaning	Value
$v_i$	initial velocity (rest)	$0m/s$
$v_f^{(\text{fall})}$	velocity just before impact	$+17.0m/s$
$\Delta t_{\text{fall}}$	fall duration	$1.12\mathrm{s}$
$v_f^{(\text{stop})}$	velocity at rest	$0m/s$
$\Delta t_{\text{stop}}$	stop duration	$0.121\mathrm{s}$

Unknowns: ${\bar{a}}_{\text{fall}}$, ${\bar{a}}_{\text{stop}}$.

Solve (a).

${\bar{a}}_{\text{fall}}=\frac{v_f^{(\text{fall})}-v_i}{\Delta t_{\text{fall}}}=\frac{17.0-0}{1.12}\approx +15.2m/{\mathrm{s}}^2$

Solve (b). Initial velocity for the stop is $+17.0m/s$, final is $0$:

${\bar{a}}_{\text{stop}}=\frac{0-17.0}{0.121}\approx -140.5m/{\mathrm{s}}^2$

Assess. Units: $m/{\mathrm{s}}^2$, correct.

• Sign in (a) is positive: velocity grows in the positive (downward) direction during the fall.
• Sign in (b) is negative: velocity decreases in the positive direction — the egg decelerates from $+17m/s$ down to $0$.
• Magnitudes: $15.2m/{\mathrm{s}}^2$ during free fall is plausible (close to $g\approx 9.8m/{\mathrm{s}}^2$). $140.5m/{\mathrm{s}}^2$ during impact is about $14g$ — extreme, but believable for a sudden stop on a hard floor.

Answer. ${\bar{a}}_{\text{fall}}\approx +15.2m/{\mathrm{s}}^2$, ${\bar{a}}_{\text{stop}}\approx -140.5m/{\mathrm{s}}^2$ (or $|{\bar{a}}_{\text{stop}}|\approx 140.5m/{\mathrm{s}}^2$ if asked for magnitude).

The template — Model, Visualise, Solve, Assess, with units and signs at every step — is the layout for every kinematics problem on the EGD102 paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The Tutorial 1 problems (train from rest; curved $x$–$t$ plot; basketball player on a $28.7\mathrm{m}$ court) for combining slope-reading, area-under-curve, and motion diagrams.
• Wolfson, Essential University Physics, Vol. 1, Ch. 2 — the textbook chapter cited on slide 35.
• Whatever you add to this folder — Lecture Atlas picks up any new Markdown note automatically.