AgentSkillsCN

01 Kinematics 1d

分析旋转运动学、扭矩、转动惯量与旋转能量。 当学生需要解决旋转物体、扭矩、角动量，或滚动运动相关问题时，此方法尤为适用。

SKILL.md
--- frontmatter
# === REQUIRED METADATA ===
id: kinematics-1d
subject: physics
display_name: One-Dimensional Kinematics
description: |
  Analyzes motion in one dimension using position, velocity, and acceleration.
  Use when students need to solve constant-acceleration problems, interpret motion
  graphs, or analyze free-fall scenarios.

# === CATEGORIZATION ===
grade_band: 9-12
khan_tags: [physics, kinematics, motion, ap-physics-1]
standards:
  - NGSS.HS-PS2-1
  - CC.HSN-Q.A.1

# === LEARNING STRUCTURE ===
objectives:
  - Define and distinguish position, displacement, velocity, and acceleration
  - Interpret position-time and velocity-time graphs
  - Apply kinematic equations to constant-acceleration problems
  - Solve free-fall problems using g = 9.8 m/s²
  - Identify when to use each kinematic equation based on known quantities

prerequisites:
  - algebra-linear-equations
  - scientific-notation

estimated_time_minutes: 45

# === VALIDATION CONFIG ===
validator:
  type: numeric_solver
  config:
    unit_library: physics
    default_tolerance: 0.02
    require_units: true

# === SOURCES ===
sources:
  - name: OpenStax College Physics
    chapter: 2
    sections: [2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7]
    url: https://openstax.org/books/college-physics/pages/2-introduction
    license: CC-BY-4.0

One-Dimensional Kinematics

Misconceptions

MisconceptionCorrection
"Zero velocity means zero acceleration"An object at the peak of its trajectory has v=0 but a=g downward. Acceleration is the rate of change of velocity, not velocity itself.
"Heavier objects fall faster"In the absence of air resistance, all objects fall at the same rate (g ≈ 9.8 m/s²). Galileo demonstrated this at the Leaning Tower of Pisa.
"Acceleration means speeding up"Acceleration is any change in velocity—it can mean speeding up, slowing down, or changing direction. Deceleration is just acceleration in the opposite direction of motion.
"Distance and displacement are the same"Distance is the total path length (always positive), while displacement is the change in position (can be positive, negative, or zero).
"Average velocity equals average speed"Average speed is total distance / time, while average velocity is displacement / time. They're equal only for motion in one direction.

Key Concepts

Position and Displacement

Position (x) describes where an object is located relative to a chosen origin point.

Displacement (Δx) is the change in position:

code
Δx = x_final - x_initial = x₂ - x₁

Key distinction:

  • •Distance is the total path length traveled (scalar, always ≥ 0)
  • •Displacement is the straight-line change in position (vector, can be +, -, or 0)

Example: If you walk 3 m east then 3 m west, your distance is 6 m but your displacement is 0 m.

Velocity

Average velocity is displacement divided by time:

code
v_avg = Δx / Δt = (x₂ - x₁) / (t₂ - t₁)

Instantaneous velocity is the slope of the position-time graph at a specific moment:

code
v = lim(Δt→0) Δx/Δt = dx/dt

Key insight: On an x-t graph:

  • •Steep slope = fast motion
  • •Positive slope = moving in positive direction
  • •Negative slope = moving in negative direction
  • •Zero slope = at rest

Acceleration

Average acceleration is the rate of change of velocity:

code
a_avg = Δv / Δt = (v₂ - v₁) / (t₂ - t₁)

Instantaneous acceleration is the slope of the velocity-time graph:

code
a = lim(Δt→0) Δv/Δt = dv/dt

Key insight: On a v-t graph:

  • •Area under the curve = displacement
  • •Slope = acceleration
  • •Horizontal line = constant velocity (zero acceleration)

Equations

The four kinematic equations for constant acceleration:

code
[1] v = v₀ + at                    (no x)
[2] x = x₀ + v₀t + ½at²            (no v)
[3] v² = v₀² + 2a(x - x₀)          (no t)
[4] x = x₀ + ½(v₀ + v)t            (no a)

Equation Selection Strategy:

If you don't have...Use equation...
x (displacement)[1]
v (final velocity)[2]
t (time)[3]
a (acceleration)[4]

Free Fall

Objects in free fall near Earth's surface experience constant acceleration:

code
g = 9.8 m/s² ≈ 10 m/s² (downward)

Sign convention (standard):

  • •Upward = positive
  • •Downward = negative
  • •Therefore: a = -g = -9.8 m/s²

Worked Examples

Example 1: Basic Constant Acceleration

Problem: A car accelerates from rest at 3.0 m/s² for 8.0 seconds. How far does it travel?

Solution:

  1. •

    Identify knowns and unknowns:

    • •v₀ = 0 m/s (starts from rest)
    • •a = 3.0 m/s²
    • •t = 8.0 s
    • •x₀ = 0 m (choose starting position as origin)
    • •Find: x = ?
  2. •

    Select equation: Need x, have v₀, a, t, don't need v → Use equation [2]

  3. •

    Solve:

    code
    x = x₀ + v₀t + ½at²
x = 0 + (0)(8.0) + ½(3.0)(8.0)²
x = ½(3.0)(64)
x = 96 m
  4. •

    Check: Units are meters ✓, positive displacement for positive acceleration ✓, reasonable magnitude ✓

Answer: 96 m

Example 2: Free Fall - Time to Fall

Problem: A ball is dropped from a height of 45 m. How long does it take to hit the ground?

Solution:

  1. •

    Identify knowns and unknowns:

    • •y₀ = 45 m (using y for vertical)
    • •y = 0 m (ground level)
    • •v₀ = 0 m/s (dropped, not thrown)
    • •a = -9.8 m/s² (downward)
    • •Find: t = ?
  2. •

    Select equation: Need t, have y, y₀, v₀, a → Use equation [2]

  3. •

    Solve:

    code
    y = y₀ + v₀t + ½at²
0 = 45 + (0)t + ½(-9.8)t²
-45 = -4.9t²
t² = 45/4.9 = 9.18
t = 3.0 s
  4. •

    Check: Units are seconds ✓, positive time ✓, reasonable for ~45 m drop ✓

Answer: 3.0 s

Example 3: Finding Initial Velocity

Problem: A car traveling at unknown speed brakes with deceleration 5.0 m/s² and stops in 40 m. What was its initial speed?

Solution:

  1. •

    Identify knowns and unknowns:

    • •v = 0 m/s (stops)
    • •a = -5.0 m/s² (deceleration = negative acceleration)
    • •Δx = 40 m
    • •Find: v₀ = ?
  2. •

    Select equation: Need v₀, have v, a, Δx, don't need t → Use equation [3]

  3. •

    Solve:

    code
    v² = v₀² + 2aΔx
0² = v₀² + 2(-5.0)(40)
0 = v₀² - 400
v₀² = 400
v₀ = 20 m/s
  4. •

    Check: Positive initial velocity ✓, reasonable car speed (~45 mph) ✓

Answer: 20 m/s

Explanation Patterns

When teaching kinematics, follow this sequence:

For Conceptual Questions

  1. •Start with physical intuition (real-world example)
  2. •Connect to the mathematical representation
  3. •Verify with limiting cases (what happens when v→0, t→∞, etc.)
  4. •Address common misconceptions explicitly

For Problem Solving

  1. •Draw a diagram with coordinate system and sign conventions
  2. •List knowns and unknowns explicitly (including units)
  3. •Identify which kinematic equation to use based on what's missing
  4. •Solve algebraically first - keep symbols until the last step
  5. •Substitute numbers with units
  6. •Check:
    • •Are units correct?
    • •Is the sign reasonable?
    • •Is the magnitude physically sensible?

For Graph Interpretation

  1. •Identify what's on each axis (x-t, v-t, or a-t)
  2. •For x-t: slope = velocity, curvature = acceleration
  3. •For v-t: slope = acceleration, area = displacement
  4. •For a-t: area = change in velocity

When to Use This Skill

Use kinematics when:

  • •Motion is in one dimension (straight line)
  • •Acceleration is constant (or zero)
  • •You need to find position, velocity, time, or acceleration
  • •Problems involve free fall or projectile motion (vertical component)

Do NOT use kinematics when:

  • •Acceleration changes over time (need calculus or numerical methods)
  • •Forces are the focus (use Newton's laws first, then kinematics)
  • •Motion is circular (use circular motion equations)

Advanced Topics

For deeper exploration:

  • •[references/calculus-kinematics.md] - Calculus-based derivations of kinematic equations
  • •[references/relative-motion.md] - Reference frames and relative velocity
  • •[references/motion-graphs.md] - Detailed graph interpretation strategies

Resources