Kinematics 1D Study Guide – White to Black Belt Mastery

by | Aug 10, 2025

🥋 Kinematics in One Dimension – White to Black Belt Mastery

“Even the longest journey begins with a single step — but in physics, every step must be measured.”

White Belt – Fundamentals

What it is: Kinematics in one dimension is the study of how objects move along a straight line, without considering what causes the motion.

Core Principles

  1. Displacement (s⃗) – change in position.
  2. Velocity (v⃗) – rate of change of displacement.
  3. Acceleration (a⃗) – rate of change of velocity.
  4. Motion can be uniform (constant velocity) or accelerated (velocity changes).

Key Equations (SI Units)

  • v⃗ = Δs⃗ / Δt
  • For constant acceleration:
    • v⃗ = v0 + at
    • s⃗ = s0 + v0t + 0.5at²
    • v⃗² = v0² + 2aΔs⃗

Common Mistakes

  • Mixing up distance (scalar) and displacement (vector).
  • Forgetting to include signs (direction) in velocity and acceleration.
  • Using kinematics formulas when acceleration is not constant.

Sensei’s Shortcuts

  • Draw a motion diagram before solving to prevent sign errors.
  • Identify knowns and unknowns, then choose the equation that fits.
  • For free fall problems, set a = −9.81 m/s² (downward).

Worked Example – Step by Step

Problem: A car starts from rest and accelerates at 3.00 m/s² for 5.00 s. Find its final velocity and displacement.

Solution:

Final velocity: v = 0 + (3.00)(5.00) = 15.0 m/s

Displacement: s = 0 + 0 + 0.5(3.00)(5.00²) = 37.5 m

Final Answer: v = 15.0 m/s, s = 37.5 m

Practice Drill

  1. A sprinter accelerates from rest at 4.00 m/s² for 3.00 s. Find v and s.
  2. A ball is thrown straight up at 12.0 m/s. How long until it reaches the top?
  3. A train moving at 20.0 m/s slows uniformly to rest over 200 m. Find its acceleration.

Answers:

  1. v = 12.0 m/s, s = 18.0 m
  2. t = 1.22 s
  3. a = −1.00 m/s²

Yellow Belt – Deeper Skills

Example: Solving multi-step motion problems

Problem: A car accelerates from 10.0 m/s to 25.0 m/s in 4.00 s, then continues at that speed for 6.00 s. Find the total displacement.

Solution Outline:

  1. a = (25.0 − 10.0) / 4.00 = 3.75 m/s²
  2. s₁ = (10.0)(4.00) + 0.5(3.75)(4.00²) = 70.0 m
  3. s₂ = (25.0)(6.00) = 150 m
  4. Total displacement = 220 m

Black Belt – Exam Strategy and Challenge

Challenge: A rock is thrown straight upward from the edge of a 45.0 m cliff with an initial velocity of 15.0 m/s.

  1. How high above the cliff edge does it rise?
  2. How long until it hits the ground below?
  3. What is its impact velocity?

Sensei Strategy Notes:

  • Break motion into upward and downward phases.
  • Use v = 0 at the highest point to find height above the cliff.
  • For total time, sum the time up and time down.
  • Impact velocity will be downward; include sign in your final answer.

Sensei’s Final Words

“The path may be straight, but the lessons curve deep into your mind.”