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.

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.

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)

Distance
- Scalar; “how much ground” covered irrespective of direction.
Displacement
- Vector; overall change in position.
- Symbol: \Delta x = xf - xi 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}.

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.

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 of a displacement–time graph equals instantaneous or average velocity.
Using two arbitrary points (x1,t1) and (x2,t2):
\text{slope} = \dfrac{\Delta x}{\Delta t} = \dfrac{x2 - x1}{t2 - t1} = v
Graphical construction (Page 19) illustrates linear segments whose slopes directly yield velocities of Bob, Nikki, Leroy.

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).

Velocity–time graph: slope = acceleration.
Area under v–t between t1 and t2 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.

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.

Given constant a, initial velocity vi, final velocity vf, elapsed time t, and displacement \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.

Airplane acceleration
- Given: v_i = 60\,\text{m/s}, a = 0.5\,\text{m/s}^2, t = 10\,\text{s}.
- Find vf using Eq 1: vf = 60 + (0.5)(10) = 65\,\text{m/s}.
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}.

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.

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.

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.