Week 2 — Motion in 1D and Relative Motion in 1D

Overview

Week 2 launches the mechanics block of EGD102 with one-dimensional kinematics. The lecture sets up a disciplined problem-solving workflow (Model -> Visualise -> Solve -> Assess), defines the kinematic variables and their calculus relationships, derives the four constant-acceleration equations from an $a$-$t$ graph, and then extends to relative velocity in 1D so motion can be described from different reference frames. The Portfolio (3% / week) also begins in the Week 2 workshop, so this content is directly assessable.

Key concepts

• The Forgetting Curve (slide 5): Ebbinghaus (1885, replicated 2015) — you forget ~50% of new information within a day and ~90% within a week unless reviewed.
• Spaced repetition (slide 6): review new concepts within 24-48 h and at progressively longer intervals to reset the forgetting curve.
• Information relevance & social interaction (slide 7): tie new material to what you already know; work in groups to compare problem-solving approaches.
• Model (slide 9): simplify the real situation down to the essential physical features.
• Visualise (slide 9): draw a pictorial representation, label important quantities with symbols. The lecturer flags this as where you should spend most of your time.
• Solve (slide 9): only after modelling and visualising do you write equations; every symbol must already be defined on your sketch.
• Assess (slide 9): check that the answer is believable, has correct units and makes physical sense.
• Motion diagram (slide 10, optional): a sketch of the object at successive instants to build intuition for the motion.
• Coordinate system (slide 10): choose axes and origin to match the motion; for 1D pick either the $x$ or $y$ axis.
• Vector motion diagram (slide 11): show the object at the start, end and any point where motion behaviour changes; mark velocity and acceleration vectors at each key point.
• Standard kinematic symbols (slide 11): $u$, $v$, $s$, $t$, $a$ — every variable in the solution must be defined on the sketch first.
• Known information (slide 12): quantities you can read off the problem or derive quickly; convert to SI units.
• Desired unknown (slide 12): only list the unknowns needed to answer the question, not every unknown in sight.
• Displacement $s$ (or $x$): vector position change, units metres [m].
• Velocity $v$: time derivative of displacement, $v=dx/dt$, units [m/s]. A vector in 1D (sign indicates direction).
• Acceleration $a$: time derivative of velocity, $a=dv/dt$, units [m/s${}^2$]. A vector in 1D.
• Average velocity: $v_{\text{avg}}=(s_f-s_i)/\Delta t$ — total displacement over total time, units [m/s].
• Average acceleration: $a_{\text{avg}}=(v_f-v_i)/\Delta t$, units [m/s${}^2$].
• Integral relationship (slide 14): velocity is the area under an $a$-$t$ curve; displacement is the area under a $v$-$t$ curve.
• Derivative relationship (slide 14): acceleration is the slope of a $v$-$t$ graph; velocity is the slope of an $x$-$t$ graph.
• Constant acceleration: a special case where the SUVAT equations (the “4 key equations”) apply.
• Initial velocity $u$: velocity at $t=0$, units [m/s].
• Final velocity $v$: velocity at the chosen end time, units [m/s].
• Gravitational acceleration near Earth’s surface: $g=9.81{\text{m/s}}^2$, directed toward the centre of the Earth.
• Directionality rule (slide 16): $u,v,a,s$ are vectors — assign signs consistent with the chosen coordinate system; getting signs wrong is the most common error.
• Three-of-five rule (slide 16): there are 5 kinematic variables ($u,v,s,t,a$); if you know any 3, you can solve for the other 2.
• Reference frame: the coordinate system attached to an observer; motion looks different from frames in relative motion.
• Relative velocity (slide 19): the velocity of an object as measured by a chosen observer.
• Observer frame (G) / Moving frame (M) / Object (O): the three frames in the lecture’s relative-motion diagram — G is the ground observer, M is e.g. a train, O is the object (e.g. a person) inside M.

Core formulas

Calculus links between the kinematic quantities (slide 14):

$v(t)=\int a(t)dtx(t)=\int v(t)dt$

$a=\frac{dv}{dt}v=\frac{dx}{dt}$

Averages (slide 14):

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

The four constant-acceleration kinematic equations (slide 16):

$v=u+at$

$s=\frac{1}{2}a{t}^2+ut$

$v^2=u^2+2as$

$s=\left(\frac{u+v}{2}\right)t$

Where:

• $u$ = initial velocity [m/s]
• $v$ = final velocity [m/s]
• $a$ = (constant) acceleration [m/s${}^2$]
• $s$ = displacement [m]
• $t$ = elapsed time [s]

Free fall constant:

$g=9.81{\text{m/s}}^2$

(Take a sign convention; e.g. with “up” positive, $a=-g=-9.81{\text{m/s}}^2$.)

Relative velocity in 1D (slide 19):

$\vec{v}=\vec{v}{}^{\prime }+\vec{V}$

${\vec{v}}_{OG}={\vec{v}}_{OM}+{\vec{v}}_{MG}$

Where:

• $\vec{V}={\vec{v}}_{MG}$: velocity of the moving frame M with respect to the ground observer G.
• $\vec{v}{}^{\prime }={\vec{v}}_{OM}$: velocity of the object O with respect to the moving frame M.
• $\vec{v}={\vec{v}}_{OG}$: velocity of the object O with respect to the observer G.

Derivation: where the SUVAT equations come from (handwritten notes pp.1-2)

The lecturer derives the first two equations directly from graphs.

1. Plot constant $a$ vs $t$ — a horizontal line at height $a$. The change in velocity is the area under this rectangle:

   $\Delta v=(t_1-t_0)a=at(\text{taking }{t}_0=0)$

   $v-u=at\Rightarrow \boxed{v=u+at}$

   This is the equation of a straight line $y=mx+c$ with $y\leftrightarrow v$, slope $m\leftrightarrow a$, $x\leftrightarrow t$, intercept $c\leftrightarrow u$.

2. Plot $v$ vs $t$ — now a straight line from $u$ at $t_0=0$ up to $v$ at $t_1=t$. Displacement is the area under this line, split into a triangle ($A_1$) plus a rectangle ($A_2$):

   $A_1=\frac{1}{2}bh=\frac{1}{2}t(v-u)$

   $A_2=bh=tu$

   $s=A_1+A_2=\frac{1}{2}t(v-u)+ut(\text{Eq. 2})$

3. Substitute $v=u+at$ (Eq. 1) into Eq. 2 to eliminate $v$:

   $s=\frac{1}{2}t(u+at-u)+ut=\boxed{\frac{1}{2}a{t}^2+ut}$

The other two equations ($v^2=u^2+2as$ and $s=\frac{1}{2}(u+v)t$) are found by eliminating $t$ or $a$ between these two via substitution.

Worked examples

Example 1 — Rock to a Frisbee in a tree (slide 17, handwritten p.3)

A Frisbee is lodged in a tree branch 6.1 m above the ground. A rock thrown from below must be travelling at least 3.0 m/s to dislodge it. How fast must the rock be thrown upward if it leaves the hand 1.1 m above the ground?

Set-up — Take “up” as positive. The rock travels from release height to the Frisbee.

Knowns:

• $v=3.0\text{m/s}$ (minimum speed at the Frisbee)
• $s=6.1-1.1=5.0\text{m}$ (vertical displacement)
• $a=-g=-9.81{\text{m/s}}^2$ (gravity opposes motion)

Unknown: $u$ (and incidentally $t$).

Equation — choose the SUVAT that links $u,v,a,s$ (no $t$):

$v^2=u^2+2as$

$u=\sqrt{v^2-2as}$

$u=\sqrt{(3.0{)}^2-2(-9.81)(5.0)}=\sqrt{9+98.1}=\sqrt{107.1}$

$\boxed{u\approx 10.34\text{m/s (upward)}}$

Example 2 — Braking motorist (slide 18, handwritten p.4)

A motorist brakes at $6.3{\text{m/s}}^2$, hits a stalled car at $18\text{km/h}$, skid marks are 34 m long. (a) Initial speed when braking began? (b) Time from braking to collision?

Set-up — take the direction of travel as positive.

Knowns:

• $v=18\text{km/h}=\frac{18\times 1000}{60\times 60}=5\text{m/s}$
• $a=-6.3{\text{m/s}}^2$ (decelerating)
• $s=34\text{m}$

(a) Use $v^2=u^2+2as$ to solve for $u$:

$u=\sqrt{v^2-2as}=\sqrt{(5{)}^2-2(-6.3)(34)}=\sqrt{25+428.4}=\sqrt{453.4}$

$\boxed{u\approx 21.29\text{m/s}\approx 76.64\text{km/h}}$

(b) Use $v=u+at$:

$t=\frac{v-u}{a}=\frac{5-21.29}{-6.3}\approx \boxed{2.59\text{s}}$

Example 3 — Following too close, solved with relative motion (slide 20, handwritten pp.5-7)

Your car (B) is at 85 km/h, the car in front (A) is 10 m ahead at 60 km/h, you decelerate at $4.2{\text{m/s}}^2$. (a) Will you collide? (b) Closest approach?

Convert units:

• $u_A=v_A=60\text{km/h}=16.67\text{m/s}$, $a_A=0$
• $u_B=85\text{km/h}=23.61\text{m/s}$, $a_B=-4.2{\text{m/s}}^2$

Method 1 — direct kinematics (handwritten pp.5-6):

Distance B travels while braking (using $v_B^2=u_B^2+2a_B{s}_B$, with final $v_B$ once B has slowed to A’s speed at closest approach; here treating until B fully stops gives the lecturer’s number):

$s_B=\frac{v_B^2-u_B^2}{2a_B}=\frac{(16.67{)}^2-(23.61{)}^2}{2(-4.2)}\approx 33.27\text{m}$

Time to decelerate from 23.61 m/s to 16.67 m/s:

$t=\frac{v-u}{a}=\frac{16.67-23.61}{-4.2}\approx 1.65\text{s}$

Distance A travels in that time:

$\Delta s_A=v_A\Delta t=16.67\times 1.65\approx 27.54\text{m}$

For impact, B would need to cover $\Delta s_A+10=37.54\text{m}$. B only covers 33.27 m -> no impact.

Closest approach: $10-(s_B-\Delta s_A)=10-(33.27-27.54)=\boxed{4.27\text{m}}$.

Method 2 — relative-motion frame (handwritten p.7):

Switch to a frame moving with car A. In this frame A is stationary ($u_A=v_A=0$), and B’s initial velocity becomes the relative velocity:

$u_{BA}=u_{BG}-u_{AG}=85-60=25\text{km/h}=6.94\text{m/s}$

B’s acceleration is unchanged: $a_B=-4.2{\text{m/s}}^2$. In this frame B approaches a stationary A; the question becomes “does B cover the 10 m gap before its relative velocity reaches zero?”.

Using $v^2=u^2+2as$ with $v=0$:

$s_B=\frac{-u_B^2}{2a_B}=\frac{-(6.94{)}^2}{2(-4.2)}\approx 5.73\text{m}$

This is less than 10 m, so no impact. Closest distance:

$10-5.73=\boxed{4.27\text{m}}$

Same answer, much cleaner working — that’s the value of the relative-motion approach.

Things to practise

The Tutorial 2 slides give four practice problems (Wolfson Ch. 2). Work through all of them in the Model -> Visualise -> Solve -> Assess format:

• Exercise 1 — Mini-golf hole-in-one. Ball decelerates at $1.0{\text{m/s}}^2$ on horizontal sections and $6.0{\text{m/s}}^2$ on the slope. Find the slowest leaving speed that still drops it in the hole. Multi-segment SUVAT problem — solve each segment in reverse from the hole back to the club.
• Exercise 2 — Lead ball into a lake. Dropped from 5.0 m above water, then sinks at constant velocity equal to its entry speed; bottom is reached 3.0 s after release. Find lake depth. Two-stage problem: free fall to the surface, then constant velocity to the bottom.
• Exercise 3 — Springbok pronk. Legs accelerate at $39{\text{m/s}}^2$ over 50 cm before take-off. (a) Maximum height of the leap. (b) Time in the air. Sequence: find lift-off speed from the leg push, then treat the airborne phase as projectile motion under gravity.
• Exercise 4 — Boeing 747 exhaust. Plane cruises at 1050 km/h and ejects hot air rearward at 800 km/h relative to the plane. Find the air’s speed relative to the ground. Pure relative-velocity application.

These are also the kind of questions that appear in Portfolio 1 (workshop, 3% of unit grade) and on the end-of-semester exam.

Common pitfalls

• Signs first, numbers second. $u,v,a,s$ are vectors. Pick “up” or “right” as positive on your sketch before substituting, and use that sign convention consistently (slide 16 — flagged in red on the slide).
• Don’t skip the picture. The lecturer explicitly says “this is where you should be spending your time” (slide 9). Every variable in your equation must come off the sketch.
• Convert units before substituting. Example 2 needs km/h -> m/s; Exercise 3 needs cm -> m. Use SI units throughout (slide 12).
• Constant-acceleration only. The four key equations only work when $a$ is constant. For Exercise 1 and the lake-bottom phase of Exercise 2, you must split the motion into segments where $a$ is constant within each.
• “Three-of-five” check (slide 16). Before picking a formula, list which three of $\{u,v,s,t,a\}$ you know and which you want. Choose the equation that contains exactly those variables.
• $g=9.81{\text{m/s}}^2$ is a magnitude. Its sign in your equation depends on your axis choice ($+9.81$ if “down” is positive, $-9.81$ if “up” is positive).
• Square roots have two signs. In Example 1 the algebra gives $u=\pm 10.34\text{m/s}$; physically the rock is thrown upward, so take the positive root.
• Assess the answer. Are the units right? Is 76 km/h a believable initial speed for a road collision (Example 2)? If not, recheck.

Source citations

• Lecture slides: EGD102-Physics/Lecture2_CTP1.pdf — slides 5-7 (forgetting curve / study skills), slide 9 (problem-solving approach), slides 10-12 (pictorial depiction), slide 14 (kinematics calculus diagram), slides 15-16 (constant-acceleration derivation and 4 key equations), slide 17 (Example 1), slide 18 (Example 2), slide 19 (relative motion definition), slide 20 (Example 3), slide 22 (learning activities).
• Lecturer’s handwritten notes: EGD102-Physics/EGD102 - Lecture2 - Notes.pdf — pp.1-2 (graphical derivation of $v=u+at$ and $s=\frac{1}{2}a{t}^2+ut$), p.3 (Example 1 worked), p.4 (Example 2 worked), pp.5-6 (Example 3 by direct kinematics), p.7 (Example 3 by relative motion).
• Tutorial 2: EGD102-Physics/Tutorial 2.pdf — Exercises 1-4 (slides 5, 6, 7, 9).
• Textbook context (acknowledged on slide 23): Wolfson, Essential University Physics, Vol. 1, 4th Ed., Ch. 2; Murre & Dros (2015) PLoS ONE 10(7): e0120644 (Ebbinghaus replication).