Module 2: One-Dimensional Kinematics Notes

One-Dimensional Kinematics: Overview

• Today and next class focus on one-dimensional motion along a single axis (x). We’ll build intuition for how position, velocity, and acceleration relate, and how to solve problems with and without constant acceleration.
• Module plan reminder:
  • We’re spending today and next Wednesday on module two.
  • Module three opens next Friday.
  • Labor Day break: no class Monday; next meeting is Wednesday.
  • No video notes due for Wednesday; math assessment due Sunday; about 70% of students already completed it.
  • After reviewing assessment results, the session may adjust in-class practice on Wednesday.
• Important due dates for TA-based work:
  • Expert TA 0 (tutorial) and Expert TA 1 due Tuesday at 23:59.
  • Recitation next Tuesday evening (06:30–08:00, location in the shed).
• Recap: We’ll use the calculus relationships when acceleration is not constant, but we’ll also rely on the four kinematics equations derived for constant acceleration.
• Goals for today:
  • Review calculus relationships between position, velocity, and acceleration.
  • Interpret motion graphically (x(t), v(t), a(t)) and relate areas/slope to physical quantities.
  • Reinforce problem-solving strategies: identify knowns/unknowns, pick the right equation, and minimize the number of equations used.
  • Practice with an example from the course packet and discuss a real in-class activity on reaction time.

Key Concepts

• One-dimensional motion: define motion along a single axis (x). Sign conventions matter (positive to the right, negative to the left).
• Instantaneous vs average quantities:
  • Instantaneous velocity v(t) is the slope (tangent) of the x(t) curve at time t.
  • Instantaneous acceleration a(t) is the slope of the v(t) curve (or the derivative dv/dt).
  • Average velocity over an interval is the change in position divided by the time interval.
• Graphical interpretations:
  • For x as a function of t, the slope at a point gives v(t).
  • If x increases with time, the displacement is positive; if it decreases, the displacement is negative.
  • For a as a function of t, the area under the a(t) curve between t1 and t2 equals the change in velocity Δv over that interval.
  • For v as a function of t, the slope equals a; a negative slope implies negative acceleration (deceleration if velocity is positive, etc.).
• Constant acceleration:
  • Acceleration is time-independent (a is a constant).
  • With constant a, we can derive and use the four kinematics equations.
  • If acceleration is not constant, rely on calculus relations v = dx/dt, a = dv/dt, x = ∫ v dt.
• Free fall example:
  • Constant acceleration due to gravity: a = -g (g ≈ 9.8 m/s²).
  • The kinematics equations apply when a is constant (e.g., free fall near the Earth's surface).
• Problem-solving strategy (unknowns vs knowns):
  • Write down knowns and unknowns first.
  • Prefer using a single equation when possible; if you have two unknowns, you’ll typically need a system of two equations.
  • Signs matter: include direction in velocity and acceleration terms; positive/negative directions must be consistent.

Calculus Relationships and Constant Acceleration

• Core relationships (in one dimension):
  • Position and velocity: $$v = rac{dx}{dt} \