Unit 2: The Art of Observation

Lecture 2.2: The Rules of Falling

Unit 2: The Art of Observation

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Today's Mission

  • Bridge Back: Yesterday you learned to read the story of motion from graphs. The rocket's velocity-time graph showed constant acceleration during engine burn. Today, we'll develop the mathematical tools to predict exactly where and when objects reach their destinations.

  • Thinking Lens: We will use Mathematical Modeling to create equations that describe constant acceleration motion.

  • Essential Questions:

    • How do we derive the kinematic equations from first principles?
    • What makes freefall a special case of constant acceleration?
    • How can we systematically solve motion problems?
Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

The Foundation: Constant Acceleration

When acceleration is constant, we can create precise mathematical relationships.

  • Definition of acceleration:

    a⃗=Δv⃗Δt=v⃗f−v⃗it\vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_f - \vec{v}_i}{t}

  • Rearranging gives us our first kinematic equation:

    v⃗f=v⃗i+a⃗t(1)\vec{v}_f = \vec{v}_i + \vec{a}t \quad \text{(1)}

  • This tells us the final velocity if we know initial velocity, acceleration, and time.

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Deriving Position from Velocity

  • For constant acceleration, the average velocity is:

    v⃗avg=v⃗i+v⃗f2\vec{v}_{avg} = \frac{\vec{v}_i + \vec{v}_f}{2}

  • Since displacement equals average velocity times time:

    Δx⃗=v⃗avg⋅t=v⃗i+v⃗f2⋅t\Delta \vec{x} = \vec{v}_{avg} \cdot t = \frac{\vec{v}_i + \vec{v}_f}{2} \cdot t

  • Substituting equation (1), v⃗f=v⃗i+a⃗t\vec{v}_f = \vec{v}_i + \vec{a}t:

    Δx⃗=v⃗it+12a⃗t2(2)\Delta \vec{x} = \vec{v}_i t + \frac{1}{2}\vec{a}t^2 \quad \text{(2)}

  • This is our second kinematic equation.

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

The Time-Independent Equation

Sometimes we need to eliminate time from our calculations.

  • Starting with:

    • Equation (1): v⃗f=v⃗i+a⃗t\vec{v}_f = \vec{v}_i + \vec{a}t → t=v⃗f−v⃗ia⃗t = \frac{\vec{v}_f - \vec{v}_i}{\vec{a}}
    • Equation (2): Δx⃗=v⃗it+12a⃗t2\Delta \vec{x} = \vec{v}_i t + \frac{1}{2}\vec{a}t^2

  • Substituting the expression for tt:

    Δx⃗=v⃗i(v⃗f−v⃗ia⃗)+12a⃗(v⃗f−v⃗ia⃗)2\Delta \vec{x} = \vec{v}_i \left(\frac{\vec{v}_f - \vec{v}_i}{\vec{a}}\right) + \frac{1}{2}\vec{a}\left(\frac{\vec{v}_f - \vec{v}_i}{\vec{a}}\right)^2

  • After algebraic manipulation:

    v⃗f2=v⃗i2+2a⃗Δx⃗(3)\vec{v}_f^2 = \vec{v}_i^2 + 2\vec{a}\Delta \vec{x} \quad \text{(3)}

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

The Three Kinematic Equations

The Complete Set:

  • Equation (1): v⃗f=v⃗i+a⃗t\vec{v}_f = \vec{v}_i + \vec{a}t

  • Equation (2): Δx⃗=v⃗it+12a⃗t2\Delta \vec{x} = \vec{v}_i t + \frac{1}{2}\vec{a}t^2

  • Equation (3): v⃗f2=v⃗i2+2a⃗Δx⃗\vec{v}_f^2 = \vec{v}_i^2 + 2\vec{a}\Delta \vec{x}

How to Choose:

  • Each equation contains four of the five variables: v⃗i\vec{v}_i, v⃗f\vec{v}_f, a⃗\vec{a}, tt, Δx⃗\Delta \vec{x}

  • Select the equation that contains your three known variables and one unknown

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Freefall: Galileo's Revolutionary Discovery

  • Key Insight: All objects fall with the same acceleration (ignoring air resistance).

  • On Earth's surface:

    ∣a⃗g∣=9.80 m/s2|\vec{a}_g| = 9.80 \text{ m/s}^2

  • Convention: Choose positive direction as upward, so:

    a⃗=−a⃗g=−9.80 m/s2\vec{a} = -\vec{a}_g = -9.80 \text{ m/s}^2

  • Freefall is simply constant acceleration motion with a⃗=−9.80 m/s2\vec{a} = -9.80 \text{ m/s}^2!

  • Watch: Galileo's Discovery of Freefall

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Problem-Solving Strategy

Steps 1-3:

  • Step 1: Identify what you know and what you want to find
  • Step 2: Choose coordinate system (positive direction)
  • Step 3: Select the kinematic equation that contains your known and unknown variables

Steps 4-6:

  • Step 4: Substitute values with proper units and significant figures
  • Step 5: Solve algebraically, then substitute numerical values
  • Step 6: Check your answer for reasonableness
Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Worked Example: Stone Drop

Problem: A stone is dropped from rest from a height of 45.0 m45.0 \text{ m}. How long does it take to reach the ground, and what is its final velocity?

  • Given:

    • v⃗i=0.0 m/s\vec{v}_i = 0.0 \text{ m/s} (dropped from rest)
    • Δx⃗=−45.0 m\Delta \vec{x} = -45.0 \text{ m} (downward displacement)
    • a⃗=−9.80 m/s2\vec{a} = -9.80 \text{ m/s}^2

  • Find: tt and v⃗f\vec{v}_f

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Stone Drop Solution: Finding Time

Use equation (2): Δx⃗=v⃗it+12a⃗t2\Delta \vec{x} = \vec{v}_i t + \frac{1}{2}\vec{a}t^2

  • Since the stone is dropped from rest, v⃗i=0.0 m/s\vec{v}_i = 0.0 \text{ m/s}:
    Δx⃗=12a⃗t2\Delta \vec{x} = \frac{1}{2}\vec{a}t^2

  • Solve algebraically for tt:
    t=2Δx⃗a⃗t = \sqrt{\frac{2\Delta \vec{x}}{\vec{a}}}

  • Now substitute numerical values:
    t=2(−45.0 m)−9.80 m/s2=9.18 s2=3.03 st = \sqrt{\frac{2(-45.0 \text{ m})}{-9.80 \text{ m/s}^2}} = \sqrt{9.18 \text{ s}^2} = 3.03 \text{ s}

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Stone Drop Solution: Finding Velocity

Use equation (1): v⃗f=v⃗i+a⃗t\vec{v}_f = \vec{v}_i + \vec{a}t

  • Solve algebraically (with v⃗i=0\vec{v}_i = 0):
    v⃗f=a⃗t\vec{v}_f = \vec{a}t

  • Substitute numerical values:
    v⃗f=(−9.80 m/s2)(3.03 s)=−29.7 m/s\vec{v}_f = (-9.80 \text{ m/s}^2)(3.03 \text{ s}) = -29.7 \text{ m/s}

The negative sign indicates downward motion.

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Worked Example: Rocket at Apogee

Problem: A model rocket is launched vertically upward with initial velocity v⃗i=125 m/s\vec{v}_i = 125 \text{ m/s}. Find the maximum height and time to reach it.

  • Given:

    • v⃗i=125 m/s\vec{v}_i = 125 \text{ m/s} (upward)
    • v⃗f=0.0 m/s\vec{v}_f = 0.0 \text{ m/s} (at maximum height)
    • a⃗=−9.80 m/s2\vec{a} = -9.80 \text{ m/s}^2 (gravitational acceleration)

  • Find: tt and Δx⃗\Delta \vec{x}

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Rocket Solution: Finding Time

Use equation (1): v⃗f=v⃗i+a⃗t\vec{v}_f = \vec{v}_i + \vec{a}t

  • Solve algebraically for tt:
    t=v⃗f−v⃗ia⃗t = \frac{\vec{v}_f - \vec{v}_i}{\vec{a}}

  • Substitute numerical values:
    t=0.0 m/s−125 m/s−9.80 m/s2=12.8 st = \frac{0.0 \text{ m/s} - 125 \text{ m/s}}{-9.80 \text{ m/s}^2} = 12.8 \text{ s}

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Rocket Solution: Finding Height

Use equation (3): v⃗f2=v⃗i2+2a⃗Δx⃗\vec{v}_f^2 = \vec{v}_i^2 + 2\vec{a}\Delta \vec{x}

  • Solve algebraically for Δx⃗\Delta \vec{x}:
    Δx⃗=v⃗f2−v⃗i22a⃗\Delta \vec{x} = \frac{\vec{v}_f^2 - \vec{v}_i^2}{2\vec{a}}

  • Substitute numerical values:
    Δx⃗=(0.0 m/s)2−(125 m/s)22(−9.80 m/s2)=797 m\Delta \vec{x} = \frac{(0.0 \text{ m/s})^2 - (125 \text{ m/s})^2}{2(-9.80 \text{ m/s}^2)} = 797 \text{ m}

The rocket reaches a maximum height of 797 m.

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Check for Understanding

A ball is thrown upward with initial velocity 15.0 m/s15.0 \text{ m/s}.

1. Which equation would you use to find how high it goes?

  • Use equation (3), v⃗f2=v⃗i2+2a⃗Δx⃗\vec{v}_f^2 = \vec{v}_i^2 + 2\vec{a}\Delta \vec{x} with v⃗f=0.0 m/s\vec{v}_f = 0.0 \text{ m/s} at maximum height

  • Solve algebraically: Δx⃗=v⃗f2−v⃗i22a⃗\Delta \vec{x} = \frac{\vec{v}_f^2 - \vec{v}_i^2}{2\vec{a}}

  • Substitute values: Δx⃗=(0.0 m/s)2−(15.0 m/s)22(−9.80 m/s2)=−225 (m/s)2−19.6 m/s2=11.5 m\Delta \vec{x} = \frac{(0.0 \text{ m/s})^2 - (15.0 \text{ m/s})^2}{2(-9.80 \text{ m/s}^2)} = \frac{-225 \text{ (m/s)}^2}{-19.6 \text{ m/s}^2} = 11.5 \text{ m}

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Check for Understanding (continued)

2. What is the acceleration throughout the motion?

  • a⃗=−9.80 m/s2\vec{a} = -9.80 \text{ m/s}^2 (constant throughout entire flight)

3. How long does it take to return to the starting point?

  • First find time to reach maximum height: tup=v⃗f−v⃗ia⃗=0.0 m/s−v⃗ia⃗t_{up} = \frac{\vec{v}_f - \vec{v}_i}{\vec{a}} = \frac{0.0 \text{ m/s} - \vec{v}_i}{\vec{a}}

  • By symmetry, total flight time = 2tup=2(15.0 m/s)9.80 m/s2=3.06 s2t_{up} = \frac{2(15.0 \text{ m/s})}{9.80 \text{ m/s}^2} = 3.06 \text{ s}

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Connection to Our Rocket Investigation

The Power of Equations:

  • We can now predict when the rocket engine cuts off
  • We can calculate exactly when it reaches maximum altitude
  • We can determine its velocity at any point in the trajectory
  • These same equations govern satellites, projectiles, and all falling objects

From graphs to equations: Yesterday's slopes and areas are now precise mathematical relationships!

Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Answering Our Essential Questions

1. How do we derive the kinematic equations from first principles?

  • Start with definition of acceleration: a⃗=Δv⃗Δt\vec{a} = \frac{\Delta \vec{v}}{\Delta t}
  • Apply algebraic manipulation to eliminate variables

2. What makes freefall a special case of constant acceleration?

  • Freefall has constant acceleration ag⃗=−9.80 m/s2\vec{a_g} = -9.80 \text{ m/s}^2
  • Same kinematic equations apply with this specific acceleration value (on Earth)

3. How can we systematically solve motion problems?

  • Identify knowns and unknowns • Choose appropriate equation • Solve algebraically first
Lecture 2.2: The Rules of Falling
Unit 2: The Art of Observation

Summary and Next Steps

The Kinematic Toolbox:

  • Equation (1): v⃗f=v⃗i+a⃗t\vec{v}_f = \vec{v}_i + \vec{a}t
  • Equation (2): Δx⃗=v⃗it+12a⃗t2\Delta \vec{x} = \vec{v}_i t + \frac{1}{2}\vec{a}t^2
  • Equation (3): v⃗f2=v⃗i2+2a⃗Δx⃗\vec{v}_f^2 = \vec{v}_i^2 + 2\vec{a}\Delta \vec{x}

Key Principles:

  • Always define coordinate system clearly
  • Use proper vector notation and units throughout
  • Freefall is constant acceleration with a⃗=−9.80 m/s2\vec{a} = -9.80 \text{ m/s}^2
Lecture 2.2: The Rules of Falling