Comprehensive Study Notes – One-Dimensional Kinematics & Related Topics

Branches of Classical Mechanics

Classical mechanics is traditionally divided into three interacting branches.

1. Kinematics – describes motion without reference to its causes.

2. Statics – studies bodies in equilibrium (resultant force $\vec F_{\text{res}} = 0$).

3. Dynamics – analyses motion while explicitly taking the forces that generate it into account.

Throughout these notes we focus on kinematics in one dimension and its extensions, while frequently referring to dynamic causes for clarity and continuity.

The Material Point (Particle) Idealisation

• A material point (punto materiale) is a body with non-zero mass whose geometrical dimensions are negligible compared with the displacements under study.
• All kinematic variables are consequently functions of a single coordinate along a chosen axis.

One–Dimensional Motion — Fundamental Quantities

Displacement

For an initial coordinate $x<em>i$ at time $t</em>i$ and a final coordinate $x<em>f$ at time $t</em>f$,
$\Delta x = x<em>f - x</em>i$
$\Delta t = t<em>f - t</em>i$

Average Velocity

$v<em>m = \frac{\Delta x}{\Delta t}\qquad \left[\frac{L}{T}\right] = \text{m\,s}^{-1}$ Graphically it equals the slope $\tan \theta$ of the chord joining two points $P(x</em>i,t<em>i)$ and $Q(x</em>f,t_f)$ on the $x(t)$ plot.

Instantaneous Velocity

Taking the limit as the time interval shrinks to zero,
$v(t) = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$
The instantaneous value equals the slope of the tangent drawn to the $x(t)$ curve at the considered point.

Average Acceleration

$a<em>m = \frac{\Delta v}{\Delta t} = \frac{v</em>f - v<em>i}{t</em>f - t_i}\qquad \left[\frac{L}{T^2}\right] = \text{m\,s}^{-2}$

Instantaneous Acceleration

$a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
Physically it is the rate of change of velocity per unit time.

Interpreting Graphs

1. $x(t)$ – position–time: slope $(dx/dt)$ gives $v(t)$.

2. $v(t)$ – velocity–time: slope gives $a(t)$, area under the curve returns displacement.

3. $a(t)$ – acceleration–time: area equals the change in velocity.

On an $x(t)$ plot:
• Positive slope → motion in positive direction.
• Zero slope → instantaneous rest.
• Negative slope → motion backwards.

Examples (Slide-6): points $P, Q, R, S$ respectively have vP>0,\; vQ>0,\; vR=0,\; vS<0.

Canonical Motions in One Dimension

1. Uniform Rectilinear Motion (MRU)

Condition: $a=0,\; v=\text{constant}=v<em>0$ Equation of motion: $x(t)=x</em>0+v<em>0 t$ Graphs: $x(t)$ is a straight line of slope $v</em>0$; $v(t)$ is a horizontal line; $a(t)=0$.

2. Uniformly Accelerated Rectilinear Motion (MUA)

Condition: constant acceleration $a$.

Fundamental relations derived from $a=\text{const}$:
$v(t)=v<em>0+at$ $x(t)=x</em>0+v<em>0 t+\tfrac12 a t^2$ Velocity–position relation (independent of $t$): $v^2=v</em>0^2+2a\,(x-x_0)$
(“Poisson formula”).

Graphical representation: $v(t)$ is a straight line; area under $v(t)$ from 0 to $t$ yields displacement.

Worked Example – Automobile with Three Phases (Slide-10)

An initially stationary car executes four sequential stages:

1. Acceleration with constant $a$ until reaching a cruising velocity.

2. Motion at that constant velocity.

3. Braking with negative constant acceleration (deceleration).

4. Comes to rest and remains stationary.
   The complete $v(t)$ plot is trapezoidal; corresponding $x(t)$ consists of a cubic-like continuous curve whose slope follows the velocity profile, while $a(t)$ presents two rectangular pulses of opposite sign separated by an interval at zero.

Tests (Slides-11, 21)

Typical short-answer questions asked during tutorials:

1. Define average vs instantaneous velocity.

2. Define average vs instantaneous acceleration; relate an infinitesimal displacement $dx$ to acceleration $a$.

3. Sketch a generic law $x(t)$ and draw tangents to illustrate three instantaneous velocities.

4. Explain the link between instantaneous velocity and the trigonometric tangent in the $x(t)$ diagram.

5. Criteria for MRU and MUA, and graphical representations of $x(t),v(t),a(t)$ for each.

Vertical Motion Under Uniform Gravity

Take downward as positive; gravitational acceleration $g\approx 9.81\,\text{m\,s}^{-2}$.

1. Free fall from rest:
   $x(t)=h-\tfrac12 g t^2$
   Total fall time $T=\sqrt{2h/g}$.
   Velocity as a function of time: $v(t)=g t$ (directed downward).
   Velocity as a function of height: $v(x)=\sqrt{2g(h-x)}$.

2. Vertical launch downward with initial speed $v1$:
   $x(t)=h - v</em>1 t - \tfrac12 g t^2$
   $v(t)=v<em>1+g t$ $v(x)=\sqrt{v</em>1^2+2g(h-x)}$

3. Vertical launch upward with speed $v2$:
   Upward is chosen positive, hence acceleration $a=-g$.
   Motion equations:
   $v(t)=v</em>2-g t$
   $x(t)=v<em>2 t - \tfrac12 g t^2$ Time to reach the peak: $t</em>M=v<em>2/g$; maximum height $x</em>M=\tfrac12 v<em>2^2/g$. Return time equals ascent time; total flight $2t</em>M$.

General energy–like integral:
$a(x)=\frac{dv}{dt} = \frac{dv}{dx}\,\frac{dx}{dt}=v\frac{dv}{dx}$
Therefore
$\int<em>{x</em>1}^{x<em>2} a(x)\,dx = \tfrac12 (v</em>2^2-v_1^2)$.

Problem-Solving Strategy (Slide-16)

1. Translate words into symbols, list knowns and unknowns.

2. Choose the simplest reference frame and coordinate axis.

3. First solve symbolically; perform dimensional consistency checks.

4. Substitute numerical data only at the end and quote the answer with proper units.

Selected Complex Examples

1. Marathon Runner & Pigeon (Slide-17, code M-3-106)

Runner speed: $v<em>r=15\,\text{km/h}$. Distance to finish when bird departs: $d=7.5\,\text{km}$. Bird speed: $v</em>b=30\,\text{km/h}$.
Because $v<em>b=2v</em>r$, the bird shuttles between runner and finish until the runner arrives. The total time to finish is $t=d/v<em>r=0.5\,\text{h}$. In that duration the pigeon always flies at $30\,\text{km/h}$ ⇒ total distance flown $s</em>b=v_b t=15\,\text{km}$.

2. Keys Dropped in an Ascending Lift (Slide-17, code M-3-224)

Lift ascending at constant $V<em>1$. Keys released with initial upward velocity $V</em>1$ relative to Earth. Equations in Earth frame:
$x<em>{\text{elev}}=V</em>1 t$
$x<em>{\text{keys}}=-\tfrac12 g t^2+V</em>1 t + d$
( $d$ is height of pocket above elevator floor). Setting equality for contact $x<em>{\text{keys}}=x</em>{\text{elev}}$ delivers $t=\sqrt{2d/g}$.
The keys’ trajectory relative to the elevator is a simple parabola opening downward whose apex is at $t=t/2$. Both floor and keys share the same absolute altitude at contact time, matching the qualitative task request.

3. Two Particles, Braking Condition (Slide-20, “no rear-end collision”)

Given v1>v2, the leading particle decelerates at constant $-a$. Time to match speeds:
$t<em>F=\frac{v</em>1-v<em>2}{a}$ Necessary condition to avoid collision obtained imposing equal positions at general $t$, then substituting $t</em>F$:
$d=\frac{(v<em>1-v</em>2)^2}{2a}$
If initial distance $d$ satisfies the above, collision is averted; otherwise, impact occurs before full braking.

Variable Acceleration & General Relations

If acceleration depends on time only: $a(t)$ ⇒ integrate twice.
If acceleration depends on position: $a(x)$ ⇒ use differential identity
$a(x)=v\frac{dv}{dx}$ leading to
$\int a(x)\,dx=\tfrac12 (v^2-v_0^2)$.

Worked illustration (Slide-27):
For $a= -A x - B$ with A,B>0, initial v0>0, the particle stops when $\tfrac12 A x^2 + Bx = \tfrac12 v</em>0^2$
Giving
$x<em>{\text{stop}} = \frac{-B + \sqrt{B^2 + A v</em>0^2}}{A}$.

Motion with Resistive or Driving Forces – Exponential Damping

Assume proportionality $a=-k v$ (linear drag):
$v(t)=v<em>0 e^{-kt}$ $x(t)=\frac{v</em>0}{k}\bigl(1-e^{-kt}\bigr)$
If a constant driving term $h$ is added: $a= h - k v$ ⇒
$v(t)=\frac{h}{k}\bigl(1-e^{-kt}\bigr)+v<em>0 e^{-kt}$ $x(t)=\frac{h}{k} t + \frac{h}{k^2}\bigl(e^{-kt}-1\bigr)+\frac{v</em>0}{k}\bigl(1-e^{-kt}\bigr)$
Such models approximate viscous damping or terminal-velocity scenarios.

Simple Harmonic Motion (SHM)

Definition

SHM arises when the acceleration is proportional to the negative of displacement:
$a(t) = -\omega^2 x(t)$
Solution:
$x(t)=X<em>0 \sin(\omega t + \varphi)$ Where • $X</em>0$ – amplitude (maximum excursion).
• $\omega$ – angular frequency $[\text{rad\,s}^{-1}]$.
• $T = 2\pi/\omega$ – period.
• $f = 1/T$ – ordinary frequency.

Derived Quantities

Velocity:
$v(t)=\frac{dx}{dt}=\omega X<em>0 \cos(\omega t + \varphi)$ Acceleration: $a(t)=-\omega^2 X</em>0 \sin(\omega t + \varphi) = -\omega^2 x(t)$.

Typical physical realisation: block attached to an ideal spring (no friction), governed by Hooke’s law $F=-k(x-x_0)$ ⇒ $\omega=\sqrt{k/m}$.
Graphical depiction reveals sinusoidal $x(t)$, cosine-shifted $v(t)$, and oppositely phased $a(t)$.

Zenon’s Paradox & Infinite Geometric Series (Slides-37/38)

Scenario: a fly shuttles between two trains approaching one another. By transforming into a frame where one train is stationary, total distance equals
$s = v<em>{\text{fly}} t</em>{\text{meet}}$
while the recursive path lengths form a geometric series whose sum remains finite despite the infinite number of trips. Slides show manipulation:
$\sum<em>{n=1}^{\infty} \Bigl(\frac{V-v}{V+v}\Bigr)^n = \frac{V-v}{V+v-v}=\text{finite}$ (exact algebra depends on naming conventions). The moral: an infinite sequence of ever-decreasing intervals may converge to a finite result, illustrating the geometric-series formula \sum{n=0}^{\infty} q^n = \frac{1}{1-q}\quad (|q|<1).

Rapid-Fire Concept Checks (Slide-28)

1. Throwing an object downward does not alter gravitational acceleration: $a=g$ always (air resistance neglected).

2. Two stones released successively in free fall:
   • Velocity difference remains constant (both accelerate equally).
   • Spatial separation increases because the earlier stone has been accelerating longer.
   • The time interval between their ground impacts equals the release interval (symmetry of uniform acceleration).

Formula Compendium – Quick Reference (Test-7 request)

• Uniform acceleration summary:
\boxed{\begin{aligned}
a &= \text{const}\
v(t)&=v0+at\ x(t)&=x0+v0 t+\tfrac12 a t^2\ v^2 &= v0^2 + 2a (x-x_0)
\end{aligned}}

• Free fall (downward positive):
$x(t)=\tfrac12 g t^2,\quad v(t)=g t,\quad T=\sqrt{2h/g}$.

Practical & Philosophical Implications

• Choosing a suitable reference frame simplifies mathematics and conceptual understanding (elevator example).
• Representation by graphs fosters intuition: slopes and areas translate directly into physically measurable quantities.
• Finite results emerging from infinite processes (Zeno) foreshadow concepts in calculus and series convergence.
• Understanding kinematic foundations is essential before introducing Newton’s Second Law, energy methods, or non-inertial frames in dynamics.

Synthesis

Kinematics provides the language and tools to quantify motion using displacement, velocity, and acceleration, independent of the forces that cause it. From simple constant-velocity cases through uniformly accelerated motion, vertical free fall, damped motion, and harmonic oscillations, the same calculus-based definitions apply. Graphical interpretation, dimensional checks, and systematic problem-solving are recurring themes, preparing the ground for deeper dynamic analysis.