Motion in 1D/2D (calc-based)

Motion in one and two dimensions forms the foundation of classical mechanics, utilizing calculus to describe position, velocity, and acceleration.

Position Vector: r⃗(t) describes an object's location in space as a function of time. In 1D: x(t), in 2D: r⃗(t) = x(t)î + y(t)ĵ.

Velocity is the instantaneous rate of change of position: v⃗ = dr⃗/dt. In component form: v⃗ = (dx/dt)î + (dy/dt)ĵ = vₓî + vᵧĵ. The magnitude gives speed: v = √(vₓ² + vᵧ²).

Acceleration is the instantaneous rate of change of velocity: a⃗ = dv⃗/dt = d²r⃗/dt². In components: a⃗ = (dvₓ/dt)î + (dvᵧ/dt)ĵ = aₓî + aᵧĵ.

Key Kinematic Equations (constant acceleration only):

• v = v₀ + at
• x = x₀ + v₀t + ½at²
• v² = v₀² + 2a(x - x₀)

Calculus Integration: From acceleration to velocity: v(t) = v₀ + ∫a dt. From velocity to position: x(t) = x₀ + ∫v dt.

Projectile Motion: A prime 2D example where aₓ = 0 and aᵧ = -g. Horizontal and vertical motions are independent and analyzed separately.

Memory Aid (DIVA): Displacement → Integrate → Velocity → Integrate → Acceleration (reverse process: differentiate twice from position to acceleration).

Vector Independence Principle: In 2D motion, x and y components are solved completely independently, each following 1D kinematic rules.

Understanding motion through calculus reveals the continuous nature of physical change, unlike algebraic approaches that give only snapshots.

Real-World Application 1: GPS Navigation Your phone's GPS continuously tracks dr⃗/dt to calculate velocity, then integrates to predict your future position. When you accelerate in a car, the system detects d²r⃗/dt² and recalculates arrival times. This real-time calculus happens millions of times per second.

Real-World Application 2: Sports Ballistics A cricket ball's trajectory is pure 2D projectile motion. The bowler imparts initial velocity v₀ at angle θ. Horizontally: vₓ = v₀cos(θ) remains constant (ignoring air resistance). Vertically: vᵧ = v₀sin(θ) - gt decreases linearly. The ball's height y(t) = (v₀sin(θ))t - ½gt² follows a parabolic path. Fielders intuitively integrate these equations to position themselves!

Analogy: The Speedometer vs. Odometer Think of velocity as your speedometer (instantaneous rate) and position as your odometer (accumulated distance). Acceleration is how quickly your speedometer needle moves. Differentiation asks: "How fast is the odometer changing right now?" Integration asks: "If I know my speedometer reading over time, what's my total distance?"

The Calculus Connection When a car's braking system applies force, acceleration becomes negative (deceleration). The rate of velocity change dv/dt determines stopping distance through integration: Δx = ∫v dt. Modern anti-lock brakes optimize this integral by preventing wheel lock-up, maintaining maximum friction coefficient and minimizing stopping distance—a literal life-saving calculus computation.

Example 1: Calculus-Based 1D Motion A particle's position is given by x(t) = 2t³ - 6t² + 4t + 1 (meters). Find: (a) velocity at t = 2s, (b) when acceleration is zero.

Solution: (a) v(t) = dx/dt = 6t² - 12t + 4 At t = 2s: v(2) = 6(4) - 12(2) + 4 = 24 - 24 + 4 = 4 m/s*

(b) a(t) = dv/dt = 12t - 12 Set a = 0: 12t - 12 = 0 → t = 1 s*

Examiner Note: Always show differentiation steps explicitly; don't skip to the answer.

Example 2: 2D Projectile Motion A ball is launched at 20 m/s at 37° above horizontal. Find: (a) maximum height, (b) range.

Solution: (a) v₀ᵧ = 20sin(37°) = 12 m/s, vᵧ = 0 at peak Using vᵧ² = v₀ᵧ² - 2gh: 0 = 144 - 2(10)h h = 7.2 m*

(b) Flight time: vᵧ = v₀ᵧ - gt → 0 = 12 - 10t → t = 1.2 s (to peak) Total time = 2.4 s v₀ₓ = 20cos(37°) = 16 m/s Range: R = v₀ₓ × t = 16 × 2.4 = 38.4 m*

Example 3: Integration Problem Given a(t) = 4t, v₀ = 2 m/s, x₀ = 0, find x(3).

v(t) = ∫4t dt = 2t² + C; using v(0) = 2: C = 2, so v(t) = 2t² + 2 x(t) = ∫(2t² + 2)dt = ⅔t³ + 2t + C; using x(0) = 0: C = 0 x(3) = ⅔(27) + 6 = 24 m*