Week 3 — Motion in 2D and Relative Motion in 2D

Overview

Lecture 3 extends 1D kinematics into two dimensions by decomposing motion into orthogonal x- and y-planes that share a common time. It covers projectile motion under gravity (no air resistance), conversion between polar and Cartesian vector forms, and relative velocity in 2D using the relation $\vec{v}={\vec{v}}^{\prime }+\vec{V}$ applied independently in each plane. Portfolio 2 is the assessed outcome of the workshop class that draws on this material.

Key concepts

• Marginal gains — small daily improvements compounding into large semester-long gains: total improvement $=(1+0.01{)}^n$ where $n$ is the number of compounding periods (slide 3).
• PDCA cycle — Plan, Do, Check, Act; iterate by identifying one study bottleneck, trying a change for a week, reviewing results, and standardising what works (slide 4).
• Problem-solving approach — Model, Visualise, Solve, Assess. Most time should go into visualising (drawing the diagram and labelling symbols) before any algebra (slide 7).
• Displacement — vector change in position; symbol $\vec{s}$ or $\vec{x}$; SI unit metres (m).
• Velocity — time-derivative of displacement, $\vec{v}=\frac{d\vec{x}}{dt}$; SI unit $\text{m/s}$ (slide 9).
• Acceleration — time-derivative of velocity, $\vec{a}=\frac{d\vec{v}}{dt}$; SI unit ${\text{m/s}}^2$ (slide 9).
• Average velocity — ${\vec{v}}_{\text{avg}}=\frac{s_f-s_i}{\Delta t}$, in m/s.
• Average acceleration — ${\vec{a}}_{\text{avg}}=\frac{v_f-v_i}{\Delta t}$, in ${\text{m/s}}^2$.
• Constant-acceleration kinematics — set of four equations linking the five variables $u,v,s,a,t$; valid only when $a$ is constant (slide 10). Knowing any three lets you solve for the other two.
• Gravitational acceleration near Earth’s surface — $g=9.81{\text{m/s}}^2$ (slide 10); often taken downward (negative on a y-up axis).
• Vector — polar form — magnitude and direction, e.g. “20 m/s at 30° CCW from $+x$” (slide 11).
• Vector — Cartesian form — component pair $(A_x,A_y)$, e.g. ”$10\text{m/s}$ in $x$ and $-6\text{m/s}$ in $y$” (slide 11).
• Quadrant correction — when computing direction $\theta ={\tan }^{-1}(A_y/A_x)$, add $180°$ if the vector points into quadrant 2 or 3 (slide 11).
• Orthogonal-plane decomposition — in 2D kinematics, the $x$ and $y$ motions are independent except they share the same $t$ (slide 12). This is the central problem-solving idea for the whole lecture.
• Projectile — an object under gravity only, so $a_x=0$ and $a_y=-g$ (sign depends on chosen positive axis).
• Relative velocity — velocity of object O measured from frame is the sum of its velocity in another frame plus that frame’s velocity: ${\vec{v}}_{OG}={\vec{v}}_{OM}+{\vec{v}}_{MG}$ (slide 18). Subscripts: O = object, M = medium/intermediate frame, G = ground.
• Vector notation $\hat{i},\hat{j}$ — unit vectors along $+x$ and $+y$ respectively, used to write components compactly.

Core formulas

Differential and integral relationships (slide 9):

$\vec{v}(t)=\int \vec{a}(t)dt,\vec{x}(t)=\int \vec{v}(t)dt$

$\vec{a}=\frac{d\vec{v}}{dt},\vec{v}=\frac{d\vec{x}}{dt}$

${\vec{a}}_{\text{avg}}=\frac{v_f-v_i}{\Delta t},{\vec{v}}_{\text{avg}}=\frac{s_f-s_i}{\Delta t}$

Constant-acceleration kinematics — 1D (slide 10):

$v=u+at$ $v^2=u^2+2as$ $s=\frac{1}{2}a{t}^2+ut$ $s=\left(\frac{u+v}{2}\right)t$

with $u$ = initial velocity [m/s], $v$ = final velocity [m/s], $a$ = acceleration [m/s²], $s$ = displacement [m], $t$ = time [s]. All of $u,v,a,s$ are vector quantities — be careful with sign conventions.

Polar ↔ Cartesian conversions (slide 11):

$A_x=A\cos \theta ,A_y=A\sin \theta$

$A=\sqrt{A_x^2+A_y^2}$

$\theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)\text{(quadrants 1, 4)}$

$\theta =180°+{\tan }^{-1}\left(\frac{A_y}{A_x}\right)\text{(quadrants 2, 3)}$

Angle measured from each axis (slide 12):

$\text{From }x\text{-axis:}{k}_x=k\cos \theta ,k_y=k\sin \theta$ $\text{From }y\text{-axis:}{k}_x=k\sin \varphi ,k_y=k\cos \varphi$

Constant-acceleration kinematics — 2D (slide 13). Time is shared between planes (no subscript on $t$):

$v_x=u_x+a_{x}t,v_y=u_y+a_{y}t$ $v_x^2=u_x^2+2a_x{s}_x,v_y^2=u_y^2+2a_y{s}_y$ $s_x=\frac{1}{2}{a}_x{t}^2+u_{x}t,s_y=\frac{1}{2}{a}_y{t}^2+u_{y}t$ $s_x=\left(\frac{u_x+v_x}{2}\right)t,s_y=\left(\frac{u_y+v_y}{2}\right)t$

Relative velocity in 2D (slide 18). Per plane, plus a vector form using $\hat{i},\hat{j}$:

${\vec{v}}_x={\vec{v}}_x^{\prime }+{\vec{V}}_x⟺{\vec{v}}_{OG,x}={\vec{v}}_{OM,x}+{\vec{v}}_{MG,x}$

${\vec{v}}_y={\vec{v}}_y^{\prime }+{\vec{V}}_y⟺{\vec{v}}_{OG,y}={\vec{v}}_{OM,y}+{\vec{v}}_{MG,y}$

In compact vector form: $\vec{v}\hat{i}={\vec{v}}^{\prime }\hat{i}+\vec{V}\hat{i}$ and $\vec{v}\hat{j}={\vec{v}}^{\prime }\hat{j}+\vec{V}\hat{j}$.

General workflow (slide 14): (1) define coordinate axes with positive directions, (2) draw a vector motion diagram, (3) split vectors into $x$ and $y$ components and list known $a,u,v,s$ in each plane, (4) solve in each plane with 1D kinematics linked by $t$.

Worked examples

Example 1 — Horizontal rifle shot (slide 15, handwritten notes pp. 1-2)

A rifle is aimed horizontally at a target $50\text{m}$ away. The bullet hits $2\text{cm}$ below the target.

Setup with $+x$ to the right (toward target), $+y$ downward, so $a_y=+9.8{\text{m/s}}^2$.

	x-plane	y-plane
$a$	$0$ (ignore air resistance)	$9.8{\text{m/s}}^2$
$s$	$50\text{m}$	$0.02\text{m}$
$u$	$u_x=v_x=?$	$u_y=0$

(a) Flight time. Use $s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ with $u_y=0$:

$t=\sqrt{\frac{2s_y}{a_y}}=\sqrt{\frac{2\times 0.02}{9.8}}=0.064\text{s}$

(b) Muzzle speed. Since $a_x=0$, $s_x=u_{x}t$, so

$u_x=\frac{s_x}{t}=\frac{50}{0.064}=781.25\text{m/s}$

Speed is the magnitude: $|\vec{u}|=\sqrt{u_x^2+u_y^2}=\sqrt{781.25^2+0^2}=781.25\text{m/s}$.

(c) Impact velocity. Horizontal component is unchanged ($v_x=781.25\text{m/s}$). Vertical:

$v_y=u_y+a_{y}t=0+9.8\times 0.064=0.6272\text{m/s}$

$|\vec{v}|=\sqrt{781.25^2+0.6272^2}\approx 781.25\text{m/s}$

$\theta ={\tan }^{-1}\left(\frac{0.6272}{781.25}\right)=0.046°\text{below horizontal}$

Example 2 — Projectile off a cliff (slide 16, handwritten notes pp. 3-5)

Initial speed $|u|=185\text{m/s}$ at $71°$ above horizontal; flight time $40.0\text{s}$; no air resistance. Axes: $+x$ horizontal, $+y$ up; gravity $a_y=-9.8{\text{m/s}}^2$.

Component initial velocities:

$u_x=185\cos 71°=60\text{m/s},u_y=185\sin 71°=175\text{m/s}$

(a) Speed at $t=21.0\text{s}$. $v_x$ is constant at $60\text{m/s}$ (since $a_x=0$).

$v_{y,21}=u_y+a_y{t}_{21}=175+(-9.8)(21)=-30.8\text{m/s}$

$|{\vec{v}}_{21}|=\sqrt{60^2+(-30.8{)}^2}=67.44\text{m/s}$

(b) Velocity just before impact at $t=40.0\text{s}$.

$v_{x,40}=60\text{m/s}$ $v_{y,40}=175+(-9.8)(40)=-217\text{m/s}$ $|{\vec{v}}_{40}|=\sqrt{60^2+(-217{)}^2}=225\text{m/s}$ $\theta ={\tan }^{-1}\left(\frac{217}{60}\right)=74.5°\text{below horizontal}$

(c) Cliff height. $s_y$ at $t=40\text{s}$:

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2=175(40)+\frac{1}{2}(-9.8)(40{)}^2=-840\text{m}$

The projectile finishes $840\text{m}$ below its launch point, so the cliff height is $840\text{m}$.

Example 3 — Plane in a crosswind (slide 19, handwritten notes p. 6)

An observer on the ground sees a plane moving due north at $220\text{km/h}$ and a cross-breeze blowing west at $40\text{km/h}$. Find the velocity of the plane relative to the wind.

Set $+y$ = north, $+x$ = east. Define O = object (plane), M = medium (air/wind), G = ground.

Plane relative to ground: ${\vec{v}}_{OG}=0\hat{i}+220\hat{j}\text{km/h}$.

Wind relative to ground (toward west): ${\vec{v}}_{MG}=-40\hat{i}+0\hat{j}\text{km/h}$.

Using ${\vec{v}}_{OG}={\vec{v}}_{OM}+{\vec{v}}_{MG}$, rearrange to ${\vec{v}}_{OM}={\vec{v}}_{OG}-{\vec{v}}_{MG}$, then split per plane:

$v_{OM,x}=v_{OG,x}-v_{MG,x}=0-(-40)=+40\text{km/h}$ $v_{OM,y}=v_{OG,y}-v_{MG,y}=220-0=+220\text{km/h}$

${\vec{v}}_{OM}=40\hat{i}+220\hat{j}\text{km/h}$

The plane is heading slightly east of north relative to the air at $\sqrt{40^2+220^2}\approx 224\text{km/h}$, bearing ${\tan }^{-1}(40/220)\approx 10.3°$ east of north.

Things to practise

Tutorial 3 (Wolfson Ch. 2 problems). Outline solutions for each from Tutorial 3_Solutions.pdf:

• Exercise 1 — Displacement from velocity graphs. Given $v_x$ and $v_y$ vs $t$ graphs, find position at $t=3.0\text{s}$. Use the fact that area under a $v$-$t$ graph = displacement. From the graphs: $s_x=\frac{1}{2}(3)(30)=45\text{m}$ (triangle), $s_y=\frac{1}{2}(3)(30-10)+3(10)=60\text{m}$ (triangle + rectangle). Then $|\vec{s}|=\sqrt{45^2+60^2}=75\text{m}$ at $\theta ={\tan }^{-1}(60/45)=53°$ above the $+x$-axis. Vector form $\vec{s}=45\hat{i}+60\hat{j}\text{m}$.
• Exercise 2 — Furious 7 car jump. Car launches horizontally at $72\text{km/h}=20\text{m/s}$ and drops $15\text{m}$. Find horizontal distance. Use $y$-plane to get time: $t=\sqrt{2|s_y|/g}=\sqrt{30/9.8}=1.75\text{s}$. Then $s_x=u_{x}t=20\times 1.75=35\mathbf{m}$.
• Exercise 3 — Slingshot to climbers. Launch angle $70°$; target is $390\text{m}$ horizontal, $270\text{m}$ above. Find launch speed $|u|$. Two equations, two unknowns. Use $\tan \theta =\sin \theta /\cos \theta$ trick: $390=|u|\cos 70°t$ and $270=|u|\sin 70°t-\frac{1}{2}(9.81)t^2$. Substituting gives $270=390\tan 70°-\frac{1}{2}(9.81)t^2$, so $t=\sqrt{2(390\tan 70°-270)/9.81}=12.79\text{s}$. Then $|u|=390/(\cos 70°\times 12.79)=89.16m/s$.
• Exercise 4 — Plane in wind. $1600\text{km}$ flight in $110\text{min}$ (= $1.83\text{h}$), so ground speed $=873\text{km/h}$ due south, i.e. $\vec{v}=0\hat{i}-873\hat{j}$. Plane heading is $15°$ west of south (azimuth $255°$ CCW from $+x$), so ${\vec{v}}^{\prime }=1200\cos 255°\hat{i}+1200\sin 255°\hat{j}$. Wind $\vec{V}=\vec{v}-{\vec{v}}^{\prime }$ gives $\vec{V}\approx 310.6\hat{i}+286.4\hat{j}\text{km/h}$, magnitude $423km/h$ at $43°$ CCW from $+x$ (i.e. roughly north-east).

Also: Mastering Physics, Motion in 2D set; Wolfson Chapter 2 additional reading; Portfolio 2 is assessed in the Workshop class — study Week 2 eContent + Lecture + this Tutorial to prepare (slide 21).

Common pitfalls

• Sign of $g$. Whether gravity contributes $+9.81$ or $-9.81$ depends on whether you chose $+y$ up or down. In Example 1 ($+y$ down) it was $+9.8$; in Example 2 ($+y$ up) it was $-9.8$. Choose once, then be consistent.
• Quadrant of ${\tan }^{-1}$. Calculator returns angles in $(-90°,90°)$. Add $180°$ for quadrants 2 or 3 (slide 11) — otherwise direction is flipped.
• Mixing planes too early. The $x$ and $y$ equations are independent; only $t$ links them. Don’t substitute $u_y$ into an $x$-equation.
• Speed vs velocity. “Speed” is the magnitude $|\vec{v}|$; “velocity” requires magnitude and direction (handwritten notes p. 2).
• Units. Convert km/h to m/s (divide by 3.6) before using $g=9.81{\text{m/s}}^2$; convert minutes to hours when needed. The lecture flags “Unit Consistency Checkpoints” — check units at every major algebraic step (slide 6), not just at the end.
• Constant-acceleration assumption. The four kinematics equations only apply when $a$ is constant. For non-constant $a$, integrate $a(t)$ and $v(t)$.
• Air resistance. Every example here ignores it. Real problems may not.
• Sanity check. “Assess” step (slide 7): does the result have plausible magnitude, correct sign, sensible units? E.g. a 78 m/s muzzle velocity passes; a cliff height of $-840$ m means displacement below launch (consistent with falling into the sea).

Source citations

• Slides 1-22 of EGD102-Physics/Lecture3_CTP1.pdf (lecture deck, CTP1 2026). Key slides: 9-10 (kinematics review), 11 (vectors), 12-14 (2D kinematics framework), 15-16, 19 (examples), 17-18 (relative velocity), 21 (Week 3 learning activities incl. Portfolio 2).
• EGD102-Physics/EGD102 - Lecture3 - Notes.pdf (handwritten worked solutions for Examples 1-3, pp. 1-6).
• EGD102-Physics/Tutorial 3.pdf (Exercises 1-4, slides 4, 9, 10, 13).
• EGD102-Physics/Tutorial 3_Solutions.pdf (full solutions for Exercises 1-4, pp. 1-5).
• Textbook reference: Wolfson, R. 2020. Essential University Physics, Vol. 1, 4th ed. SI Units, Chapter 2 (slide 22 of lecture deck).