KINEMATICS – Motion Along a Straight Line (Week 3)

Overview: Mechanics & Kinematics

• Mechanics is the broad study of motion and the forces that cause it.

  • Kinematics → mathematical description of motion (what happens).

  • Dynamics → investigates causes of motion (why it happens).

• Focus of Week 3: 1-D or straight-line motion (no curves, no vectors in other axes).

Distance & Displacement

• Distance (d)

  • Scalar; total path length.

  • Example: D = 500\,\text{m} + 350\,\text{m} = 850\,\text{m}.

• Displacement ((\Delta x) or (\vec{d}))

  • Vector; straight-line segment from initial position to final position.

  • Indicates "how far AND in what direction" from the start.

Speed & Velocity

• Speed (v)

  • Scalar; “how fast.”

  • Average speed \bar v = \frac{\text{distance}}{\text{time}}.

• Velocity ((\vec v))

  • Vector; speed with direction.

  • When direction matters, use +/– signs (e.g., right = +, left = –).

• Example (Lydia de Vega, 100-m dash, Seoul 1986):

  • d = 100\,\text{m},\; t = 11.53\,\text{s} \Rightarrow \bar v = \frac{100}{11.53} = 8.67\,\text{m/s}.

• Practice Q’s (slide 14):

  1. SLEX 51 km @ 50 km/h ⇒ t = \frac{51}{50} \approx 1.02\,\text{h} = 61\,\text{min}.

  2. Three marathoners after 5 s: d = v t, then compare Xiao–Diluc, Venti–Xiao.

Acceleration

• Measures rate of change of velocity.

  • a = \frac{\Delta v}{\Delta t} = \frac{vf - v0}{tf - t0}.

  • If (a) ≠ 0, object is speeding up, slowing down, or changing direction.

  • Positive/negative sign denotes direction.

• If acceleration varies → formula above gives average acceleration.

• Lydia de Vega example recap:

  • (a) From rest to 6 m/s in 2 s: a = \frac{6-0}{2} = 3\,\text{m/s}^2.

  • (b) 6 → 10 m/s from 2 s → 8 s: a = \frac{10-6}{6} = 0.67\,\text{m/s}^2.

  • (c) 10 → 7 m/s in 2 s (deceleration): a = \frac{7-10}{2} = -1.5\,\text{m/s}^2.

  • Note the negative sign indicates slowing down.

Graphical Description of Motion

• Position–time (d–t) graph

  • Slope = average speed/velocity: \text{slope} = \frac{\Delta d}{\Delta t}.

  • Straight line → constant velocity.

  • Tangent slope on curve → instantaneous velocity.

• Velocity–time (v–t) graph

  • Area under curve = displacement (or distance if velocity ≥ 0).

  • Slope = acceleration.

• Acceleration–time (a–t) graph

  • For constant a: horizontal line.

  • Area under curve = change in velocity.

• Tri-graph relations for constant +a (activity slides 41-44):

  • a–t: flat positive line.

  • v–t: straight line with positive slope.

  • d–t: upward-opening parabola (increasing curvature).

Uniformly Accelerated Motion (UAM)

• Assumption: acceleration is constant (many real cases: falling objects, skidding cars).

• Four core kinematic equations (initial time & position set to zero):

  1. vf = v0 + a t

  2. d = \tfrac{1}{2}(vf + v0)t

  3. vf^2 = v0^2 + 2 a d

  4. d = v_0 t + \tfrac{1}{2} a t^2

• Variable coverage table (slide 51) indicates which equation omits which variable.

• Runway design example:

  • Need 30 m/s, (a = 2.00\,\text{m/s}^2).
    A. Time: t = \frac{vf - v0}{a} = \frac{30}{2} = 15\,\text{s}.
    B. Length: d = \frac{vf^2 - v0^2}{2a} = \frac{30^2}{2\times 2} = 225\,\text{m}.

Free-Fall Motion in One Dimension

• Only force: gravity ⇒ constant (g = 9.8\,\text{m/s}^2) (downward).

• Sign convention: upward + ; downward –.

• CASE 1 – Dropped from rest ((v_0 = 0))

  • Velocity after (t): v = -gt.

  • Displacement: y = -\tfrac{1}{2} g t^2.

  • Table (slides 74-80):

  • t = 1 s → v = –9.8 m/s, y = –4.9 m

  • 2 s → v = –19.6 m/s, y = –19.6 m

  • 3 s → v = –29.4 m/s, y = –44.1 m

  • 4 s → v = –39.2 m/s, y = –78.4 m

  • 5 s → v = –49.0 m/s, y = –122.5 m

• CASE 2 – Thrown upward ((v_0 = +29.4\,\text{m/s}))

  • Use v = v_0 - g t.

  • Table (slides 88-90):

  • 0 s: +29.4 m/s

  • 1 s: +19.6 m/s

  • 2 s: +9.8 m/s

  • 3 s: 0 m/s (peak)

  • 4 s: –9.8 m/s (downward)

  • 5 s: –19.6 m/s

  • 6 s: –29.4 m/s

  • 7 s: –39.2 m/s

  • Displacement on way up can be found via Eq. 4; motion on the way down mirrors CASE 1.

Worked & Practice Problems Recap

• SLEX travel time, marathoner spacing, Lydia de Vega acceleration intervals, airplane runway design, stone drop, ball throw—all illustrate applying the four UAM equations appropriately.

• Key strategy: Identify knowns/unknowns, pick the equation missing the unwanted variable, apply correct sign convention.

Connections & Real-World Relevance

• Transportation safety (speed limits, braking distance, runway lengths).

• Athletic performance analysis (sprinter speeds, marathon pacing).

• Engineering design (elevator acceleration comfort limits, roller-coaster drops).

• Scientific instrumentation (drop-tower micro-gravity, timing free-fall to measure (g)).

Ethical / Practical Implications

• Ensuring runways & roadways meet safety margins protects lives.

• Proper sign convention & units avoid catastrophic miscalculations (e.g., Mars Climate Orbiter loss due to unit error).

• Education on motion fosters critical assessment of stunt challenges or viral “gravity” videos.

Key Takeaways

• Distinguish scalar vs vector forms (distance vs displacement, speed vs velocity).

• Graph slopes & areas convert between position, velocity, and acceleration views.

• Four UAM equations are cornerstones; memorize along with variable-is-missing logic.

• Gravity provides a natural, nearly constant acceleration useful for simple models; air resistance introduces complexity beyond this course segment.

• Clear documentation (units, signs, significant figures) is as crucial as numeric answers.

Use these organized notes as a stand-alone reference for core kinematic principles, example calculations, and conceptual insights vital for examinations or practical applications.

The table you've provided is a very helpful tool for solving problems involving Uniformly Accelerated Motion (UAM). It lists the four core kinematic equations and indicates which variables are included or excluded in each equation. This helps you select the most appropriate equation for a given problem.

Here's how to use it step-by-step:

1. Understand the Variables:

   • v_0: Initial velocity (velocity at t=0)

   • d: Displacement (change in position)

   • a: Acceleration (constant acceleration)

   • v_f: Final velocity (velocity at some time t)

   • t: Time interval

2. Understand the Symbols:

   • A checkmark (✓) in a column means that variable is present in that equation.

   • A dash (—) in a column means that variable is not present in that equation. This is the key insight for problem-solving.

3. Steps to Use the Table for Problem Solving:

   • Step 1: List your Knowns and Unknowns. When you're faced with a UAM problem, first identify and write down all the information you are given (your 'knowns') and what you are asked to find (your 'unknown'). For example, you might be given v0, a and t, and asked to find vf and d.

   • Step 2: Identify the "Missing" Variable (the variable you don't care about). Look at your list of knowns and the single unknown you want to find. There will typically be one variable that is neither given nor asked for. This is the variable you want to exclude from your equation choice.

   • Step 3: Find the Equation (Row) that Excludes that Variable. Go to the table and look across each row. Find the row (equation) that has a dash (—) under the column of the variable you identified in Step 2 (the one you don't know and don't need).

   • Step 4: Use that Equation. Once you've found the correct row, that's the equation you should use to solve your problem. You'll have all the other variables in that equation as knowns (or the one unknown you are trying to find), allowing you to solve for it.

Example from the notes (Runway Design):

Let's revisit the runway design example to illustrate:

Scenario: An airplane starts from rest (v0 = 0), needs to reach a takeoff speed of vf = 30 ext{ m/s} with acceleration a = 2.00 ext{ m/s}^2.

• Problem A: Find the time (t) required.

  • Knowns: v0 = 0 ext{ m/s}, vf = 30 ext{ m/s}, a = 2.00 ext{ m/s}^2

  • Unknown to find: t

  • Variable not known and not needed: d (displacement/length of runway)

  • Using the table: Look for the row with a dash (—) under 'd'. That is the first equation: vf = v0 + at

  • Solve: Rearrange to t = rac{vf - v0}{a} = rac{30 - 0}{2.00} = 15 ext{ s}.

• Problem B: Find the minimum length (d) of the runway.

  • Knowns: v0 = 0 ext{ m/s}, vf = 30 ext{ m/s}, a = 2.00 ext{ m/s}^2

  • Unknown to find: d

  • Variable not known and not needed: t (time)

  Overview: Mechanics & Kinematics

  • Mechanics is the broad study of motion and the forces that cause it.

    • Kinematics → mathematical description of motion (what happens).

    • Dynamics → investigates causes of motion (why it happens).

  • Focus of Week 3: 1-D or straight-line motion (no curves, no vectors in other axes).

  Distance & Displacement

  • Distance (d)

    • Scalar; total path length.

    • Example: D = 500\,\text{m} + 350\,\text{m} = 850\,\text{m}.

  • Displacement (\Delta x or \vec{d})

    • Vector; straight-line segment from initial position to final position.

    • Indicates "how far AND in what direction" from the start.

  Speed & Velocity

  • Speed (v)

    • Scalar; “how fast.”

    • Average speed \bar v = \frac{\text{distance}}{\text{time}}.

  • Velocity (\vec v)

    • Vector; speed with direction.

    • When direction matters, use +/– signs (e.g., right = +, left = –).

  • Example (Lydia de Vega, 100-m dash, Seoul 1986):

    • d = 100\,\text{m},\; t = 11.53\,\text{s} \Rightarrow \bar v = \frac{100}{11.53} = 8.67\,\text{m/s}.

  • Practice Q’s (slide 14):

    1. SLEX 51 km @ 50 km/h \Rightarrow t = \frac{51}{50} \approx 1.02\,\text{h} = 61\,\text{min}.

    2. Three marathoners after 5 s: d = v t,
       then compare Xiao–Diluc, Venti–Xiao.

  Acceleration

  • Measures rate of change of velocity.

    • a = \frac{\Delta v}{\Delta t} = \frac{vf - v0}{tf - t0}.

    • If (a) \neq 0,
      object is speeding up, slowing down, or changing direction.

    • Positive/negative sign denotes direction.

  • If acceleration varies → formula above gives average acceleration.

  • Lydia de Vega example recap:

    • (a) From rest to 6 m/s in 2 s: a = \frac{6-0}{2} = 3\,\text{m/s}^2.

    • (b) 6 → 10 m/s from 2 s → 8 s: a = \frac{10-6}{6} = 0.67\,\text{m/s}^2.

    • (c) 10 → 7 m/s in 2 s (deceleration): a = \frac{7-10}{2} = -1.5\,\text{m/s}^2.

    • Note the negative sign indicates slowing down.

  Graphical Description of Motion

  • Position–time (d–t) graph

    • Slope = average speed/velocity: \text{slope} = \frac{\Delta d}{\Delta t}.

    • Straight line → constant velocity.

    • Tangent slope on curve → instantaneous velocity.

  • Velocity–time (v–t) graph

    • Area under curve = displacement (or distance if velocity \ge 0).

    • Slope = acceleration.

  • Acceleration–time (a–t) graph

    • For constant a: horizontal line.

    • Area under curve = change in velocity.

  • Tri-graph relations for constant +a (activity slides 41-44):

    • a–t: flat positive line.

    • v–t: straight line with positive slope.

    • d–t: upward-opening parabola (increasing curvature).

  Uniformly Accelerated Motion (UAM)

  • Assumption: acceleration is constant (many real cases: falling objects, skidding cars).

  • Four core kinematic equations (initial time & position set to zero):

    1. vf = v0 + at

    2. d = \frac{1}{2}(vf + v0)t

    3. vf^2 = v0^2 + 2ad

    4. d = v_0 t + \frac{1}{2} at^2

  • Variable coverage table (slide 51) indicates which equation omits which variable.

  • Runway design example:

    • Need 30 m/s, (a = 2.00\,\text{m/s}^2).
      

      A. Time: t = \frac{vf - v0}{a} = \frac{30}{2} = 15\,\text{s}.

      B. Length: d = \frac{vf^2 - v0^2}{2a} = \frac{30^2}{2\times 2} = 225\,\text{m}.

  Free-Fall Motion in One Dimension

  • Only force: gravity \Rightarrow constant (g = 9.8\,\text{m/s}^2) (downward).

  • Sign convention: upward + ; downward –.

  • CASE 1 – Dropped from rest ((v_0 = 0))

    • Velocity after (t): v = -gt.

    • Displacement: y = -\frac{1}{2} g t^2.

    • Table (slides 74-80):

    • t = 1 s → v = –9.8 m/s, y = –4.9 m

    • 2 s → v = –19.6 m/s, y = –19.6 m

    • 3 s → v = –29.4 m/s, y = –44.1 m

    • 4 s → v = –39.2 m/s, y = –78.4 m

    • 5 s → v = –49.0 m/s, y = –122.5 m

  • CASE 2 – Thrown upward ((v_0 = +29.4\,\text{m/s}))

    • Use v = v_0 - g t.

    • Table (slides 88-90):

    • 0 s: +29.4 m/s

    • 1 s: +19.6 m/s

    • 2 s: +9.8 m/s

    • 3 s: 0 m/s (peak)

    • 4 s: –9.8 m/s (downward)

    • 5 s: –19.6 m/s

    • 6 s: –29.4 m/s

    • 7 s: –39.2 m/s

    • Displacement on way up can be found via Eq. 4; motion on the way down mirrors CASE 1.

  Worked & Practice Problems Recap

  • SLEX travel time, marathoner spacing, Lydia de Vega acceleration intervals, airplane runway design, stone drop, ball throw—all illustrate applying the four UAM equations appropriately.

  • Key strategy: Identify knowns/unknowns, pick the equation missing the unwanted variable, apply correct sign convention.

  Connections & Real-World Relevance

  • Transportation safety (speed limits, braking distance, runway lengths).

  • Athletic performance analysis (sprinter speeds, marathon pacing).

  • Engineering design (elevator acceleration comfort limits, roller-coaster drops).

  • Scientific instrumentation (drop-tower micro-gravity, timing free-fall to measure (g)).

  Ethical / Practical Implications

  • Ensuring runways & roadways meet safety margins protects lives.

  • Proper sign convention & units avoid catastrophic miscalculations (e.g., Mars Climate Orbiter loss due to unit error).

  • Education on motion fosters critical assessment of stunt challenges or viral “gravity” videos.

  Key Takeaways

  • Distinguish scalar vs vector forms (distance vs displacement, speed vs velocity).

  • Graph slopes & areas convert between position, velocity, and acceleration views.

  • Four UAM equations are cornerstones; memorize along with variable-is-missing logic.

  • Gravity provides a natural, nearly constant acceleration useful for simple models; air resistance introduces complexity beyond this course segment.

  • Clear documentation (units, signs, significant figures) is as crucial as numeric answers.

  Use these organized notes as a stand-alone reference for core kinematic principles, example calculations, and conceptual insights vital for examinations or practical applications.

  The table you've provided is a very helpful tool for solving problems involving Uniformly Accelerated Motion (UAM). It lists the four core kinematic equations and indicates which variables are included or excluded in each equation. This helps you select the most appropriate equation for a given problem.

  Here's how to use it step-by-step:

  1. Understand the Variables:

     • v_0:

  Initial velocity (velocity at t=0)

  - d: 
  

  Displacement (change in position)

  - a: 
  

  Acceleration (constant acceleration)

  - v_f: 
  

  Final velocity (velocity at some time t)

  - t: 
  

  Time interval

  2. Understand the Symbols:

     • A checkmark (✓) in a column means that variable is present in that equation.

     • A dash (—) in a column means that variable is not present in that equation. This is the key insight for problem-solving.

  4. Steps to Use the Table for Problem Solving:

     • Step 1: List your Knowns and Unknowns. When you're faced with a UAM problem, first identify and write down all the information you are given (your 'knowns') and what you are asked to find (your 'unknown'). For example, you might be given v0, a and t, and asked to find vf and d.

     • Step 2: Identify the "Missing" Variable (the variable you don't care about). Look at your list of knowns and the single unknown you want to find. There will typically be one variable that is neither given nor asked for. This is the variable you want to exclude from your equation choice.

     • Step 3: Find the Equation (Row) that Excludes that Variable. Go to the table and look across each row. Find the row (equation) that has a dash (—) under the column of the variable you identified in Step 2 (the one you don't know and don't need).

     • Step 4: Use that Equation. Once you've found the correct row, that's the equation you should use to solve your problem. You'll have all the other variables in that equation as knowns (or the one unknown you are trying to find), allowing you to solve for it.

  Example from the notes (Runway Design):

  Let's revisit the runway design example to illustrate:

  Scenario: An airplane starts from rest (v0 = 0), needs to reach a takeoff speed of vf = 30 \text{ m/s} with acceleration a = 2.00 \text{ m/s}^2.

  • Problem A: Find the time (t) required.

    • Knowns: v0 = 0 \text{ m/s}, vf = 30 \text{ m/s}, a = 2.00 \text{ m/s}^2

    • Unknown to find: t

    • Variable not known and not needed: d (displacement/length of runway)

    • Using the table: Look for the row with a dash (—) under 'd'. That is the first equation: vf = v0 + at

    • Solve: Rearrange to t = \frac{vf - v0}{a} = \frac{30 - 0}{2.00} = 15 \text{ s}.

  • Problem B: Find the minimum length (d) of the runway.

    • Knowns: v0 = 0 \text{ m/s}, vf = 30 \text{ m/s}, a = 2.00 \text{ m/s}^2

    • Unknown to find: d

    • Variable not known and not needed: t (time)

    • Using the table: Look for the row with a dash (—) under 't'. That is the fourth equation: vf^2 = v0^2 + 2ad

    • Solve: Rearrange to d = \frac{vf^2 - v0^2}{2a} = \frac{(30)^2 - (0)^2}{2 \times 2.00} = \frac{900}{4} = 225 \text{ m}.

  This table streamlines the process of choosing the correct kinematic equation by visually indicating which variable is absent from each formula, making problem-solving more efficient.

  • Solve: Rearrange to d = rac{vf^2 - v0^2}{2a} = rac{(30)^2 - (0)^2}{2 imes 2.00} = rac{900}{4} = 225 ext{ m}.

This table streamlines the process of choosing the correct kinematic equation by visually indicating which variable is absent from each formula, making problem-solving more efficient.