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E-TRADE TOGETHER GLOBAL ACADEMY

Classical Mechanics

Level 5 · Physics & Physical Modeling · Prerequisites: Courses 1.2 and 2.1

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7 canonical classes · 5 laboratories · 4 module assessments · project · final

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MODULE 1

Kinematics & Physical Modeling

Describe motion using frames, vectors, graphs and calculus.

Dependencies: Geometry, trigonometry, vectors and differential calculus.

Next connection: Forces & Newtonian Dynamics

5.1.1 · v1 · 90 min

Position, velocity and acceleration

Why it matters

Kinematics turns observations of motion into measurable, testable models used in robotics, transportation and sensors.

Prerequisites

Courses 1.2 and 2.1; use SI units unless stated otherwise.

Concept and explanation

Choose a reference frame and positive axis before calculating. Position x(t) locates the body; displacement Δx=x_f−x_i is a vector component; distance is path length. Velocity v=dx/dt and acceleration a=dv/dt. Speed is |v|, not velocity.

Key terms

reference frame; position; displacement; distance; velocity; speed; acceleration; trajectory

Physical model, notation and representation

System: one modeled body. Frame: ground-fixed x-axis. Known: x(t) or motion data with seconds and meters. Unknown: displacement, velocity or acceleration. Assume the reported trajectory and clock are adequate; do not infer causes of motion from kinematics alone.

Mathematical development

For constant acceleration, v=v₀+at and x=x₀+v₀t+½at². These equations require constant a. On x–t graphs slope is velocity; on v–t graphs slope is acceleration and signed area is displacement.

Learning objectives

  • Distinguish position, displacement, distance, velocity, speed and acceleration.
  • Connect slopes and signed areas across motion graphs.

Worked examples

Worked example 1

Physical situation: A robot moves from x=−2 m to x=7 m, then to x=3 m.

Given, goal, system and frame: Find displacement and distance; positive x is right.

Diagram/text description and physical model: Δx=3−(−2)=5 m; distance=9+4=13 m.

Development, calculation, result and units: Vector displacement uses endpoints; scalar distance uses the path.

Verification, interpretation and limitations: Units and path reconstruction verify the distinction; one-dimensional idealization is a limitation.

Worked example 2

Physical situation: A cart has x(t)=2+3t−t² meters.

Given, goal, system and frame: Find v and a at t=2 s.

Diagram/text description and physical model: v=3−2t, a=−2; therefore v(2)=−1 m/s and a=−2 m/s².

Development, calculation, result and units: Differentiate with units; negative v means left in the chosen frame.

Verification, interpretation and limitations: Finite-difference slopes near 2 s confirm approximately −1 m/s; the model may not describe real friction.

Common mistake and counterexample

Incorrect: distance equals displacement. It seems plausible on a one-way trip, but fails after reversal. Correct by defining the path and endpoints separately, then checking distance≥|displacement|.

Knowledge check, feedback and mastery

  1. Position depends on reference frame.
  2. Velocity is the derivative of position.
  3. Signed v–t area gives displacement.
  4. Constant-a formulas require constant acceleration.

Mastery criterion: 4/4 correct with system, vector/sign convention, units and validation evidence. Correct a failed step, explain it, then complete a fresh equivalent check.

Related laboratory and next class

Projectile Motion → Newton’s laws