Displacement: Physics of Motion & Kinematics
Master the fundamental concepts of displacement, velocity, acceleration, and motion analysis!
What is Displacement and Why is it Important?
Displacement is a vector quantity that represents the change in position of an object. Unlike distance, which is the total path traveled, displacement only considers the straight-line distance between the starting and ending points, along with the direction.
Why displacement matters: Understanding displacement is fundamental to kinematics and physics. It helps us analyze motion, predict future positions, calculate velocities and accelerations, and solve complex engineering problems involving moving objects.
Real-World Displacement Examples:
- Car Journey: Drive 100km east, then 60km north → displacement = 117km northeast
- Elevator Motion: Start at ground floor, go to 10th floor → displacement = 30m upward
- Pendulum Swing: Swings left and returns to center → displacement = 0m
- Airplane Flight: Flies from NYC to LA → displacement = 3,944km west-southwest
- Satellite Orbit: Completes one orbit → displacement = 0km (returns to start)
Key Characteristics of Displacement:
- Vector Quantity: Has both magnitude and direction
- Path Independent: Only depends on start and end positions
- Can be Zero: When object returns to starting position
- Can be Negative: When moving in opposite direction to reference
- SI Unit: Meters (m), but can be expressed in any length unit
The Fundamental Displacement Equations
Displacement calculations use kinematic equations that relate position, velocity, acceleration, and time. These equations form the foundation of motion analysis in physics.
The Complete Set of Kinematic Equations:
First Equation of Motion:
v = u + at
Final velocity = Initial velocity + (acceleration × time)
Second Equation of Motion:
s = ut + ½at²
Displacement = (initial velocity × time) + ½(acceleration × time²)
Third Equation of Motion:
v² = u² + 2as
Final velocity² = Initial velocity² + 2(acceleration × displacement)
Key Variables Explained:
- s: Displacement (change in position)
- u: Initial velocity (velocity at start)
- v: Final velocity (velocity at end)
- a: Acceleration (rate of change of velocity)
- t: Time duration of motion
Additional Useful Formulas:
- Average Velocity: v̄ = s/t = (u + v)/2
- Position Function: x(t) = x₀ + ut + ½at²
- Vector Displacement: s = √(Δx² + Δy²)
- Direction Angle: θ = tan⁻¹(Δy/Δx)
- Relative Displacement: s₁₂ = s₁ – s₂
Displacement vs Distance: Understand the Difference
Displacement and distance are often confused, but they represent fundamentally different concepts in physics. Understanding this distinction is crucial for solving motion problems correctly.
Displacement vs Distance Comparison:
| Property | Displacement | Distance | Example |
|---|---|---|---|
| Type | Vector quantity | Scalar quantity | Vector has direction, scalar doesn’t |
| Path Dependence | Path independent | Path dependent | Straight line vs actual route |
| Can be Zero? | Yes (round trip) | No (always positive) | Return to starting point |
| Can be Negative? | Yes (direction dependent) | No (always positive) | Moving backward from reference |
| Magnitude | ≤ Distance | ≥ Displacement | Shortest path vs total path |
| Symbol | s or Δx | d | Mathematical notation |
Practical Examples of Displacement vs Distance:
- Walking in a Square: Walk 400m around a square → Distance = 400m, Displacement = 0m
- Straight Line Motion: Walk 100m east → Distance = 100m, Displacement = 100m east
- Zigzag Path: Walk 50m north, then 30m south → Distance = 80m, Displacement = 20m north
- Circular Motion: Half circle with radius 10m → Distance = 31.4m, Displacement = 20m
Types of Motion and Displacement Analysis
Different types of motion require different approaches to displacement calculation. Understanding these motion types helps choose the correct formula and method for analysis.
Motion Types and Displacement Characteristics:
| Motion Type | Acceleration | Displacement Formula | Key Features | Examples |
|---|---|---|---|---|
| Uniform Motion | a = 0 | s = ut | Constant velocity | Car on cruise control |
| Uniformly Accelerated | a = constant | s = ut + ½at² | Constant acceleration | Free fall, car accelerating |
| Variable Acceleration | a = f(t) | s = ∫∫a dt dt | Changing acceleration | Rocket launch, pendulum |
| Circular Motion | Centripetal | s = rθ | Curved path | Satellite orbit, wheel rotation |
| Oscillatory Motion | a ∝ -x | s = A sin(ωt + φ) | Periodic motion | Spring, pendulum, waves |
Special Cases in Displacement Analysis:
- Free Fall: a = g = 9.81 m/s², s = ut + ½gt² (downward positive)
- Projectile Motion: Horizontal: s = ut, Vertical: s = ut + ½gt²
- Simple Harmonic Motion: s = A cos(ωt + φ), where A is amplitude
- Uniformly Retarded Motion: Negative acceleration, object slowing down
- Multi-dimensional Motion: s = √(sx² + sy² + sz²)
Vector Displacement and 2D Motion
Vector displacement becomes essential when dealing with motion in two or three dimensions. Understanding vector addition and component analysis is crucial for accurate displacement calculations.
Vector Displacement Analysis:
Component Form:
s⃗ = sₓî + sᵧĵ + sᵤk̂
Displacement vector in component form
Magnitude:
|s⃗| = √(sₓ² + sᵧ² + sᵤ²)
Magnitude of displacement vector
Direction (2D):
θ = tan⁻¹(sᵧ/sₓ)
Angle from positive x-axis
Vector Addition Rules:
- Component Addition: s⃗ₜₒₜₐₗ = s⃗₁ + s⃗₂ + s⃗₃ + …
- Head-to-Tail Method: Connect vectors end-to-end
- Parallelogram Method: For two vectors only
- Unit Vector Notation: Express in î, ĵ, k̂ components
- Resultant Calculation: Use Pythagorean theorem for magnitude
Practice Problems and Worked Solutions
Problem 1: Basic Uniformly Accelerated Motion
Question: A car starts from rest and accelerates at 2 m/s² for 10 seconds. Calculate its displacement.
Click to see detailed solution
Given: u = 0 m/s (starts from rest), a = 2 m/s², t = 10 s
Formula: s = ut + ½at²
Substitution: s = (0)(10) + ½(2)(10)²
Calculation: s = 0 + ½(2)(100) = 0 + 100 = 100 m
Final velocity: v = u + at = 0 + 2(10) = 20 m/s
Answer: Displacement = 100 m, Final velocity = 20 m/s
Problem 2: Motion with Initial Velocity
Question: A ball is thrown upward with initial velocity 30 m/s. Find its displacement after 4 seconds. (g = 9.81 m/s²)
Click to see detailed solution
Given: u = 30 m/s (upward), a = -9.81 m/s² (gravity downward), t = 4 s
Formula: s = ut + ½at²
Substitution: s = (30)(4) + ½(-9.81)(4)²
Calculation: s = 120 + ½(-9.81)(16) = 120 – 78.48 = 41.52 m
Final velocity: v = 30 + (-9.81)(4) = 30 – 39.24 = -9.24 m/s
Answer: Displacement = 41.52 m upward, Final velocity = 9.24 m/s downward
Problem 3: Vector Displacement in 2D
Question: A person walks 100 m east, then 75 m north. Calculate the total displacement and direction.
Click to see detailed solution
Given: East displacement = 100 m, North displacement = 75 m
Components: sₓ = 100 m, sᵧ = 75 m
Magnitude: |s| = √(sₓ² + sᵧ²) = √(100² + 75²) = √(10000 + 5625) = √15625 = 125 m
Direction: θ = tan⁻¹(sᵧ/sₓ) = tan⁻¹(75/100) = tan⁻¹(0.75) = 36.87°
Answer: Displacement = 125 m at 36.87° north of east
Problem 4: Using Third Equation of Motion
Question: A car accelerates from 10 m/s to 25 m/s with acceleration 3 m/s². Find the displacement during this acceleration.
Click to see detailed solution
Given: u = 10 m/s, v = 25 m/s, a = 3 m/s²
Formula: v² = u² + 2as → s = (v² – u²)/(2a)
Substitution: s = (25² – 10²)/(2 × 3) = (625 – 100)/6 = 525/6
Calculation: s = 87.5 m
Time taken: t = (v – u)/a = (25 – 10)/3 = 5 s
Verification: s = ut + ½at² = 10(5) + ½(3)(25) = 50 + 37.5 = 87.5 m ✓
Answer: Displacement = 87.5 m
Advanced Topics and Special Cases
Relative Motion and Displacement:
- Relative Displacement: s₁₂ = s₁ – s₂ (displacement of object 1 relative to object 2)
- Reference Frame: Displacement depends on chosen coordinate system
- Moving Observer: Displacement appears different to moving vs stationary observer
- Galilean Transformation: s’ = s – vt for moving reference frame
- Vector Subtraction: Use component method for relative displacement
Non-Uniform Motion Analysis:
- Variable Acceleration: Use calculus: s = ∫∫a(t) dt dt
- Graphical Analysis: Area under velocity-time graph gives displacement
- Numerical Methods: Approximate solutions for complex motion
- Piecewise Motion: Break complex motion into simpler segments
- Parametric Equations: x(t) and y(t) for 2D motion analysis
Common Mistakes and How to Avoid Them:
- Confusing Distance and Displacement: Remember displacement is vector, distance is scalar
- Sign Conventions: Establish positive direction and stick to it consistently
- Unit Consistency: Convert all quantities to same unit system before calculation
- Vector Components: Don’t forget to consider direction in 2D/3D problems
- Initial Conditions: Clearly identify initial position, velocity, and time
