KINEMATICS – MOTION ALONG A STRAIGHT LINE

Agenda

• Differentiate and relate the following quantities:
  • Distance
  • Displacement
  • Speed
  • Velocity
  • Acceleration
• Analyze motion graphs with respect to:
  • Displacement–time
  • Velocity–time
  • Acceleration–time
• Construct multiple motion graphs from verbal or numerical data.
• Solve quantitative problems involving uniformly accelerated motion (UAM).
• Demonstrate and describe the motion of a free-falling body.

Activity 1: “Who is the Fastest?”

• Line graph provided: horizontal axis = time (s); vertical axis = distance (m)
• Three runners: Bob, Nikki, Leroy.
• Guide-question cues (leads for analysis):
  • Identify runner with smallest completion time (steepest curve to 100 m).
  • Determine who started the fastest (greatest initial slope).
  • Interpret Leroy’s flat region from $t=8\text{ s}$ to $t\approx12.5\text{ s}$ (no change in $x$ ⇒ at rest).
  • Straight-line segment ⇒ constant speed; curved segment ⇒ changing speed (acceleration or deceleration).
  • Slope of graph = instantaneous speed (or average speed for linear segments).
• Real-world relevance: same analysis applies to 100-m dash data, car travel logs, fitness-tracker plots.

Concept of Motion

• Motion = change of an object’s position relative to surroundings as time progresses.
• Can be described via:
  • Words (qualitative narrative)
  • Tables (data sets)
  • Numerical data (explicit $t$ and $x$ pairs)
  • Equations (functional relationships)
  • Graphs (visual summaries)

Basic Quantities Describing Motion

• Distance
  • Scalar; “how much ground” covered irrespective of direction.
• Displacement
  • Vector; overall change in position.
  • Symbol: $\Delta x = x<em>f - x</em>i$ in 1-D, $\vec{\Delta r}$ in 2-D/3-D.
• Speed
  • Scalar; rate of covering distance.
  • Average speed $= \dfrac{\text{distance}}{\text{time}}$.
• Velocity
  • Vector counterpart of speed; rate of change of displacement.
  • Average velocity $= \dfrac{\Delta x}{\Delta t}$.
• Acceleration
  • Vector; rate of change of velocity.
  • $\vec a = \dfrac{\Delta \vec v}{\Delta t}$.

Graphical Analysis of 1-D Motion

• Example scenario (Page 13): three objects A, B, C on position–time graph.
  • A: starts at origin, moves slowly forward.
  • B: stationary (horizontal line).
  • C: reverses direction quickly, passes origin (negative slope then positive? or vice versa depending on sign conventions).
• Second scenario (Page 14) with labels A → D illustrates:
  • Gradual speed increases, passing the origin, slowing, reversing, and returning near start.

Concavity and Direction on Position–Time Graphs

• Concave-up curve ⇒ a>0 (positive acceleration).
• Concave-down curve ⇒ a<0 (negative acceleration / deceleration relative to chosen positive axis).
• Increasing $x(t)$ (upward slope) ⇒ motion in the positive direction.
• Decreasing $x(t)$ (downward slope) ⇒ motion in the negative direction.

Slope Interpretation: Velocity

• Slope of a displacement–time graph equals instantaneous or average velocity.
• Using two arbitrary points $(x<em>1,t</em>1)$ and $(x<em>2,t</em>2)$:
  $\text{slope} = \dfrac{\Delta x}{\Delta t} = \dfrac{x<em>2 - x</em>1}{t<em>2 - t</em>1} = v$
• Graphical construction (Page 19) illustrates linear segments whose slopes directly yield velocities of Bob, Nikki, Leroy.

Time–Position–Velocity Relationship (Mnemonic Table)

• Constant velocity: straight, non-zero slope on $x$–$t$; horizontal line on $v$–$t$; zero line on $a$–$t$.
• Speeding up: curve upward on $x$–$t$; slope of $v$–$t$ positive; $a$ positive.
• Slowing down: curve flattens; negative $a$ (assuming velocity positive).

Acceleration and Area under Velocity–Time Graph

• Velocity–time graph: slope = acceleration.
• Area under $v$–$t$ between $t<em>1$ and $t</em>2$ equals displacement $\Delta x$.
• Example integrals (Page 21):
  • Triangular area $= \tfrac12\,bh$ → $\tfrac12\,(\text{base})(\text{height})$ → correct units $\text{(m/s)}·\text{s}=\text{m}$.
• Sign conventions critical: area below time axis ⇒ negative displacement.

Uniformly Accelerated Motion (UAM)

• Definition: motion with constant (uniform) acceleration $a$.
• Acceleration does not vary with time: $a(t)=a_0$ = constant numeric value.
• Common real-life examples: objects in free fall (neglecting drag), vehicles undergoing steady throttle, projectile motion (vertical component).
• Ethical/engineering relevance: accurate UAM modeling essential in automotive safety, elevator design, amusement-park ride certification.

Kinematic Equations for UAM

Given constant $a$, initial velocity $v<em>i$, final velocity $v</em>f$, elapsed time $t$, and displacement $\Delta x$:

1. $v<em>f = v</em>i + at$
2. $\Delta x = v_i t + \tfrac12 a t^2$
3. $\Delta x = \left(\dfrac{v<em>f + v</em>i}{2}\right) t$
4. $v<em>f^2 = v</em>i^2 + 2a \Delta x$

• All four are algebraically related; any two known quantities determine the others.
• Use consistent units (SI): $x$ in m, $v$ in m/s, $a$ in m/s², $t$ in s.

Worked Examples (from transcript)

1. Airplane acceleration
   • Given: $v_i = 60\,\text{m/s}$, $a = 0.5\,\text{m/s}^2$, $t = 10\,\text{s}$.
   • Find $v<em>f$ using Eq 1: $v</em>f = 60 + (0.5)(10) = 65\,\text{m/s}$.
2. Airplane distance while accelerating
   • Given: $v_i = 80\,\text{m/s}$, $a = 2.0\,\text{m/s}^2$, $t = 10\,\text{s}$.
   • Use Eq 2:
     $\Delta x = 80(10) + \tfrac12 (2.0)(10)^2 = 800 + 100 = 900\,\text{m}$.

Practice Problem (self-test)

• Car at rest ($v_i = 0$) accelerates at $a = 4\,\text{m/s}^2$ for $t = 7\,\text{s}$.
• Find $v<em>f$: $v</em>f = 0 + (4)(7) = 28\,\text{m/s}$ (≈100 km/h).

Free-Fall Motion

• Special case of UAM in vertical dimension.
• Acceleration due to gravity: $g = 9.8\,\text{m/s}^2$ (downward, often taken as negative when upward is positive).
• Symbolic choices:
  • Upward positive: $a = -g$.
  • Downward positive: $a = +g$.

Conceptual Characteristics of Free Fall

• Magnitude of acceleration constant: $|a| = 9.8\,\text{m/s}^2$ regardless of mass (neglecting air drag).
• If dropped (released from rest): $v_i = 0$.
• At the peak of upward throw: instantaneous $v = 0$ but $a = -9.8\,\text{m/s}^2$ still acts.
• Symmetry: speed at launch equals speed upon returning to same height (opposite direction).
• Practical implications: design of sports trajectories, safety nets, timing of fireworks.

Free-Fall Example

• Rock dropped from rest from a roof $6.0\,\text{m}$ above ground.
• Known: $v_i = 0$, $a = -9.8\,\text{m/s}^2$, $\Delta y = -6.0\,\text{m}$ (downward).
• Use Eq 2 (vertical form):
  $\Delta y = v_i t + \tfrac12 a t^2$ → $-6.0 = 0 + \tfrac12 (-9.8) t^2$
  $t^2 = \dfrac{12.0}{9.8} \approx 1.224$
  $t \approx 1.11\,\text{s}$.
• Interpretation: the rock hits the ground slightly over one second after release.