Graph/parametric descriptions

Graphical and parametric descriptions form the visual and mathematical language of motion in kinematics. In Physics C: Mechanics, understanding these representations is fundamental to solving complex motion problems.

Position-Time Graphs (x-t): Plot position x versus time t. The slope represents instantaneous velocity: v = dx/dt. Curved lines indicate changing velocity (acceleration). A horizontal line means the object is stationary; a straight line indicates constant velocity.

Velocity-Time Graphs (v-t): Plot velocity v versus time t. The slope gives instantaneous acceleration: a = dv/dt. The area under the curve represents displacement: Δx = ∫v dt. This is crucial for solving problems where acceleration varies with time.

Acceleration-Time Graphs (a-t): Plot acceleration a versus time t. The area under the curve yields the change in velocity: Δv = ∫a dt.

Parametric Equations: Express position as functions of time for both coordinates: x(t) and y(t). For projectile motion: x(t) = x₀ + v₀ₓt and y(t) = y₀ + v₀ᵧt - ½gt². Velocity components are derivatives: vₓ = dx/dt and vᵧ = dy/dt.

Key Relationships:

• Position → derivative → Velocity → derivative → Acceleration
• Acceleration → integral → Velocity → integral → Position

Remember: Graphs encode calculus operations visually. Slopes are derivatives; areas are integrals.

Mnemonic: "Slopes Show Speed" (position graphs), "Areas Are Displacements" (velocity graphs).

Real-World Application: Car Motion on a Highway

Imagine analyzing a car's journey using a velocity-time graph. When the car accelerates from rest at a traffic light, the v-t graph shows a positive slope (positive acceleration). The steeper the slope, the more powerful the acceleration. When cruising at constant highway speed, the graph becomes horizontal (zero acceleration). Braking creates a negative slope.

The area under the v-t curve tells you how far the car traveled. If the car maintained 30 m/s for 10 seconds, the rectangular area (30 × 10 = 300 m) gives the displacement. This is why speed cameras integrate your velocity over time—they're calculating areas under graphs!

Parametric Motion: Football Trajectory

When a quarterback throws a football, the motion is described parametrically:

• Horizontal: x(t) = v₀cos(θ)·t (constant velocity, no air resistance)
• Vertical: y(t) = v₀sin(θ)·t - ½gt² (parabolic path due to gravity)

Each coordinate evolves independently with time as the parameter. At t = 2s, you can find the exact position by plugging into both equations. This parametric approach is essential for analyzing projectile motion, where horizontal and vertical components behave differently.

Analogy: Think of parametric equations as GPS coordinates changing with time. The parameter t acts like a timeline slider—move it forward, and both x and y update according to their individual rules. The object's path is the collection of all (x(t), y(t)) pairs.

Key Insight: Graphs convert complex motion into visual patterns your brain recognizes instantly, while parametric equations provide precise mathematical control.

Example 1: Velocity-Time Graph Analysis

Question: A particle's velocity is given by the v-t graph showing v = 10 m/s for 0 ≤ t ≤ 4s, then linearly decreasing to v = 0 at t = 8s. Find (a) acceleration during 4-8s, (b) total displacement.

Solution: (a) Acceleration = slope of v-t graph during 4-8s a = Δv/Δt = (0 - 10)/(8 - 4) = -2.5 m/s²

(b) Total displacement = area under v-t curve

• Rectangle (0-4s): A₁ = 10 × 4 = 40 m
• Triangle (4-8s): A₂ = ½ × 4 × 10 = 20 m
• Total: 60 m

Examiner Note: Always label areas and show calculation steps separately.

Example 2: Parametric Projectile Motion

Question: A ball is thrown with x(t) = 15t and y(t) = 20t - 5t² (SI units). Find velocity at t = 2s.

Solution: Velocity components are derivatives:

• vₓ = dx/dt = 15 m/s (constant)
• vᵧ = dy/dt = 20 - 10t

At t = 2s:

• vₓ = 15 m/s
• vᵧ = 20 - 10(2) = 0 m/s

Magnitude: |v| = √(15² + 0²) = 15 m/s Direction: θ = tan⁻¹(0/15) = 0° (horizontal)

Examiner Note: At maximum height, vᵧ = 0, making this the peak of trajectory.

Example 3: Graph to Equation

Question: Given a-t graph showing constant a = 3 m/s² from t = 0-5s, starting from rest, find v(t) and x(t).

Solution: v(t) = v₀ + ∫a dt = 0 + 3t = 3t m/s x(t) = x₀ + ∫v dt = 0 + ∫3t dt = 1.5t² m