Lesson 2: Kinematics — Motion in 1D and 2D

Weeks 2–3 — Describing how things move

🎯 Learning Objectives

Distinguish between distance/displacement and speed/velocity
Derive and apply the kinematic equations for constant acceleration
Solve projectile motion problems by separating x and y components

🎯 Intuition First: How Do We Describe Motion?

Imagine you're in a car. You can describe your motion in several ways:

Where are you? → Position (x)
How fast are you going? → Speed (scalar) or Velocity (vector)
Are you speeding up or slowing down? → Acceleration

These are the building blocks of kinematics — the branch of physics that describes motion without worrying about what causes it.

📐 Position, Velocity, Acceleration

Distance vs. Displacement

Concept	Definition	Type
Distance	Total path length traveled	Scalar (no direction)
Displacement	Change in position (final − initial)	Vector (has direction)

Speed vs. Velocity

Concept	Definition	Type
Speed	How fast (rate of distance)	Scalar
Velocity	How fast + direction (rate of displacement)	Vector

Average velocity: v̄ = Δx / Δt

Average speed = total distance / total time

Acceleration

Acceleration tells you how fast velocity is changing. If velocity is constant, acceleration is zero.

Average acceleration: ā = Δv / Δt

Units: m/s² (meters per second per second)

💡 Key Insight:

A car going 60 mph at constant speed has zero acceleration. A car braking at 60 mph is accelerating (negatively, or decelerating). A car turning the corner at 30 mph is accelerating — because its direction is changing, even if its speed is constant!

📋 The Kinematic Equations (Constant Acceleration)

When acceleration is constant (which is a good approximation for many everyday situations), we can derive three essential equations:

1) v = v₀ + at

2) x = x₀ + v₀t + ½at²

3) v² = v₀² + 2a(x − x₀)

Where:

v₀ = initial velocity
v = final velocity
a = constant acceleration
t = time elapsed
x₀ = initial position
x = final position

📝 Derivation (Quick Overview):

Equation 1: From a = Δv/Δt, rearrange: v = v₀ + at

Equation 2: Average velocity = (v₀ + v)/2 = (v₀ + v₀ + at)/2 = v₀ + ½at

Then: Δx = v̄t = (v₀ + ½at)t = v₀t + ½at²

Equation 3: Eliminate t from (1) and (2): substitute t = (v − v₀)/a into (2)

This gives: v² = v₀² + 2aΔx ✦

🍎 Worked Example: Free Fall

A ball is thrown straight upward at 20 m/s. How high does it go?

📝 Step-by-Step Solution:

Given:

v₀ = +20 m/s (up is positive)
a = −g = −9.8 m/s² (gravity points down)
At the peak: v = 0 m/s

Find: Δx = ?

Choose Equation 3 (relates velocities, acceleration, and displacement):

v² = v₀² + 2aΔx

0² = (20)² + 2(−9.8)Δx

0 = 400 − 19.6Δx

Δx = 400/19.6 = 20.4 m

✅ The ball rises 20.4 meters before stopping and falling.

🎯 Projectile Motion

Projectile motion is the most elegant application of kinematics. The key insight:

💡 Core Principle:

Horizontal and vertical motions are independent. The horizontal motion doesn't affect the vertical motion and vice versa.

Direction	Acceleration	Velocity	Motion
Horizontal (x)	aₓ = 0	vₓ = constant	Uniform motion
Vertical (y)	aᵧ = −g = −9.8 m/s²	vᵧ = vᵧ₀ + aᵧt	Constant acceleration

📝 Worked Example: Kicked Ball

A ball is kicked at 15 m/s at an angle of 30° above the horizontal. How far does it travel (range)?

Step 1 — Break into components:

v₀ₓ = 15 cos(30°) = 15 × 0.866 = 13.0 m/s

v₀ᵧ = 15 sin(30°) = 15 × 0.500 = 7.5 m/s

Step 2 — Find time of flight:

Ball lands when y = 0. Use y = v₀ᵧt + ½aᵧt²:

0 = 7.5t + ½(−9.8)t²

0 = 7.5t − 4.9t²

t(7.5 − 4.9t) = 0

t = 0 or t = 7.5/4.9 = 1.53 s

Step 3 — Find horizontal distance:

x = v₀ₓ × t = 13.0 × 1.53 = 19.9 m

✅ Range ≈ 20.0 meters

🏀 Real-World Connection:

When you see a basketball arcing toward the hoop, the ball's path is a parabola — exactly described by the equations above (ignoring air resistance). The horizontal speed stays constant while the vertical speed changes due to gravity.