Week 2 In-Depth — Why Kinematics Works

Read the cheatsheet first. This note explains why the SUVAT equations look the way they do, walks through one full worked example per topic, and finishes with an exam-style end-to-end. Every step is on the page — no hand-waving.

1 — Where the SUVAT equations come from

The four SUVAT equations are not handed down from the sky. They are direct consequences of two facts:

• $v=dx/dt$ — velocity is the derivative of position.
• $a=dv/dt$ — acceleration is the derivative of velocity.

If $a$ is constant, both equations integrate cleanly, and you get all four SUVATs in two lines. The lecturer does this geometrically (from areas under graphs) instead of analytically — the result is the same.

Geometric derivation (from the handwritten notes, pp. 1–2)

Step 1 — plot $a$ vs $t$. Constant $a$ is a horizontal line at height $a$. From $t_0=0$ to $t_1=t$, the area under this line is a rectangle of width $t$ and height $a$:

$\Delta v=\text{area}=at.$

But $\Delta v=v-u$, so

$\boxed{v=u+at}.$

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

Step 2 — plot $v$ vs $t$. Because $v=u+at$ is linear in $t$, the $v$-$t$ graph is a straight line from $(0,u)$ to $(t,v)$. Displacement is the area under that line. Split it into a triangle on top of a rectangle:

$A_1=\frac{1}{2}\cdot t\cdot (v-u)\text{(triangle)}$

$A_2=t\cdot u\text{(rectangle)}$

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

Substitute $v=u+at$ to eliminate $v$:

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

Step 3 — eliminate $t$ from steps 1 & 2. From step 1, $t=(v-u)/a$. Substitute into $s=\frac{1}{2}t(u+v)$ (an equivalent form of $A_1+A_2$):

$s=\frac{1}{2}\cdot \frac{v-u}{a}\cdot (u+v)=\frac{v^2-u^2}{2a}.$

Rearrange:

$\boxed{v^2=u^2+2as}.$

Step 4 — the average form. From step 2’s $A_1+A_2$ before substitution,

$s=\frac{1}{2}t(u+v)=\boxed{\left(\frac{u+v}{2}\right)t}.$

So the four SUVATs are not four separate facts — they are four algebraic faces of the same two-line graphical derivation. If you remember the picture, you can re-derive any one of them.

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2 — Why “Model -> Visualise -> Solve -> Assess” actually saves time

It looks like bureaucratic ritual. It isn’t. The pattern of mistakes the lecturer warns about all stem from skipping Visualise:

• “I used $g=+9.81$ but the rock was going up” — no sign convention on the sketch.
• “I wrote $u$ for two different things” — no symbol legend on the sketch.
• “I picked the wrong SUVAT” — didn’t list known vs unknown before reaching for a formula.

The Visualise step exists so the Solve step is mechanical. If your sketch labels every quantity with a symbol and a sign, the algebra is mostly bookkeeping. Slide 9 says spend most of your time here. Take that literally.

Worked example with all the steps — Rock to a Frisbee (slide 17)

A Frisbee is lodged $6.1\text{m}$ above the ground. Hitting it at $\ge 3.0\text{m/s}$ will dislodge it. You release a rock from $1.1\text{m}$ above ground. What is the minimum launch speed (upward)?

Model. Object moves vertically in 1D under gravity only.

Visualise. Pick “up” positive. Sketch start point (rock in hand, $y_0=1.1$), end point (Frisbee, $y_1=6.1$). Label $u$ (unknown, upward), $v=+3.0$ at Frisbee, $a=-9.81$, $s=y_1-y_0=5.0$.

Solve. Three of five known: $v$, $a$, $s$. Want $u$. No $t$ available, no $t$ wanted — use

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

Rearrange:

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

Physically the rock launches upward, so $u>0$:

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

Assess. A $10\text{m/s}$ overarm throw is plausible (Olympic baseball pitchers reach ~$40\text{m/s}$). Units are m/s. Sign matches the sketch.

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3 — Free fall is just SUVAT with $|a|=g$

There is no separate “free-fall toolkit.” Free fall is the constant-acceleration case with $|a|=9.81{\text{m/s}}^2$ pointing toward Earth. The only thing that changes is your sign:

Axis choice	$a$ in equations
”Up” positive	$a=-9.81$
”Down” positive	$a=+9.81$

Both work. Be consistent.

Worked example — Braking motorist (slide 18)

A motorist brakes at $6.3{\text{m/s}}^2$, hits a stalled car at $18\text{km/h}$, skid marks $34\text{m}$. (a) Speed at start of braking? (b) Time elapsed?

Visualise. Take direction of travel as positive. $u$ unknown (positive), $v=18\text{km/h}=5\text{m/s}$, $a=-6.3$ (decelerating), $s=34$.

(a) Three known are $v,a,s$. No $t$ — use $v^2=u^2+2as$:

$u=\sqrt{v^2-2as}=\sqrt{25-2(-6.3)(34)}=\sqrt{25+428.4}=\sqrt{453.4}\approx 21.29\text{m/s}.$

In familiar units, $21.29\text{m/s}\times 3.6\approx 76.6\text{km/h}$.

(b) Now $u,v,a$ known; want $t$. Use $v=u+at$:

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

Assess. $76\text{km/h}$ down to $18\text{km/h}$ in $2.6\text{s}$ on a $34\text{m}$ skid — plausible numbers for a real collision.

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4 — Why relative motion has its own equation

Velocity is observer-dependent. The passenger walking forward in a train is moving at $1.5\text{m/s}$ from another passenger’s perspective, and $21.5\text{m/s}$ from a person standing on the platform. Both numbers are correct; they’re measured in different frames.

The relative-velocity equation just adds the frames:

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

Read the subscripts like fractions: $OG=OM+MG$, the inner letters cancel. O is the object, M is the moving frame, G is the ground.

Why bother switching frames?

Because the physics doesn’t change between inertial frames (this is Galilean relativity), but the arithmetic does. Pick the frame where the problem looks simplest.

Worked example — Following too close (slide 20)

Your car B at $85\text{km/h}$ is $10\text{m}$ behind car A at $60\text{km/h}$. You brake at $4.2{\text{m/s}}^2$. (a) Do you collide? (b) Closest approach?

Frame switch. Move into A’s frame. A is now stationary. B’s initial velocity becomes its velocity relative to A:

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

A’s acceleration is zero in any inertial frame, so B’s deceleration relative to A is still $-4.2{\text{m/s}}^2$.

Reframe the question. Does B cover the $10\text{m}$ gap before its relative velocity reaches zero?

Use $v^2=u^2+2as$ with final $v=0$:

$s=\frac{0-u_{BA}^2}{2a}=\frac{-(6.94{)}^2}{2(-4.2)}=\frac{-48.16}{-8.4}\approx 5.73\text{m}.$

$5.73<10$, so B stops short of A. No collision.

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

Compare with the brute-force method

Solving in the ground frame requires:

1. Find B’s stopping distance from $u_B=23.61$ to $v_B=u_A=16.67$ (i.e. when they’re moving the same speed) — about $33.27\text{m}$.
2. Find the time elapsed — about $1.65\text{s}$.
3. Find how far A travels in that time — $27.54\text{m}$.
4. Compare: B advances $33.27-27.54=5.73\text{m}$ on A. Gap was $10\text{m}$. Closest approach $=4.27\text{m}$.

Same answer, eight lines longer. The frame switch saved you that work. That’s the point of relative motion.

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5 — How these topics connect

Three threads, one underlying calculus:

Thread	What’s really going on
Kinematics + SUVAT	The constant-$a$ special case of integrating $a(t)\to v(t)\to x(t)$.
Free fall	The constant-$a$ special case with $
Relative motion	The same kinematic equations applied in a different inertial frame. Galilean relativity guarantees the physics survives the frame change.

So once you have $v=dx/dt$, $a=dv/dt$, and “the laws are the same in every inertial frame,” everything in this week falls out by integration.

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6 — Exam-style sample, end-to-end

Tutorial 2, Exercise 3 — A springbok’s legs accelerate it at $39{\text{m/s}}^2$ over $0.50\text{m}$ before take-off. (a) Maximum height of the leap. (b) Total time in the air.

Model. Two stages. Stage 1: leg push, constant acceleration $a_1=+39{\text{m/s}}^2$ over $s_1=0.50\text{m}$. Stage 2: airborne, constant acceleration $a_2=-g=-9.81{\text{m/s}}^2$. Use “up” positive throughout.

Visualise. Three labelled points: start of push (at rest), end of push / start of flight (velocity $v_1$), apex (velocity $0$). Label every quantity on the sketch.

Stage 1 — find take-off speed $v_1$.

Known: $u_1=0$, $a_1=39$, $s_1=0.50$. Want $v_1$. No $t$ needed — use

$v_1^2=u_1^2+2a_1{s}_1=0+2(39)(0.50)=39.$

$v_1=\sqrt{39}\approx 6.24\text{m/s}.$

Stage 2a — find max height $h$ above take-off.

Treat take-off as the new starting instant: $u_2=v_1=6.24$, $a_2=-9.81$, at apex $v=0$, want $s=h$.

$0=u_2^2+2a_{2}h⟹h=\frac{-u_2^2}{2a_2}=\frac{-39}{-19.62}\approx 1.99\text{m}.$

Above the ground, $h_{\text{total}}=0.50+1.99\approx \boxed{2.49\text{m}}$ (if “max height” is measured from where the feet started). If “max height” is from take-off, $h\approx 1.99\text{m}$. Read the question carefully.

Stage 2b — time in the air.

Time up: $v=u+at⟹0=6.24+(-9.81)t⟹t_{\text{up}}\approx 0.636\text{s}$.

By symmetry, time down (from apex back to take-off level) equals time up: $t_{\text{down}}\approx 0.636\text{s}$.

$\boxed{t_{\text{air}}\approx 1.27\text{s}}.$

Assess. A springbok leaping $\sim 2\text{m}$ in $\sim 1.3\text{s}$ matches nature documentaries. Units are correct. Signs match the sketch.

Pattern to memorise. Variables -> sketch -> sign convention -> three-of-five -> SUVAT pick -> substitute -> classify -> answer with units. That layout is the template for every constant-acceleration question on the paper.

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Where to look next

• The cheatsheet for fast revision and the quiz.
• The lecture reconstruction for the slide-by-slide narrative.
• Tutorial 2 PDF for four more practice problems (mini-golf, lake depth, springbok, jet exhaust).