Full video lecture by Kanishk Sir covering this entire chapter
Why this chapter matters for AAI ATC? Air Traffic Controllers must visualise aircraft trajectories in 2D/3D airspace. Concepts like velocity vectors, resultant velocities, relative motion, and circular paths (holding patterns!) are the physical basis of ATC operations. This chapter provides the mathematical foundation you need.
Scalar quantities have magnitude only — no direction. Examples: distance, mass, speed, temperature, time. They follow ordinary algebra rules.
Vector quantities have both magnitude and direction. They obey the triangle law (or parallelogram law) of addition. Examples: displacement, velocity, acceleration, force.
Notation
Vector A is written as A (bold) or Ā (arrow). Magnitude = |A| = A
A unit vector has magnitude = 1 and specifies direction only. For Cartesian axes: î (x-axis), ĵ (y-axis), k̂ (z-axis).
⚡ Key Points
Two vectors are equal only if magnitude AND direction are both same.
A null vector has zero magnitude; direction is undefined.
Multiplying vector A by scalar λ: magnitude becomes λ|A|; direction same (if λ > 0) or reversed (if λ < 0).
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[Add image here: Diagram showing scalar vs vector examples, unit vectors î, ĵ, k̂ along axes]
🎯 Practice MCQs — Scalars & Vectors
Q1 An aircraft is flying at 300 km/h towards north. A crosswind blows at 40 km/h towards east. The magnitude of the resultant velocity is approximately:
A 260 km/h
B 302.7 km/h
C 340 km/h
D 310 km/h
✔ Correct Answer: B — 302.7 km/h
The two velocity vectors are perpendicular (north and east).
Resultant = √(300² + 40²) = √(90000 + 1600) = √91600 ≈ 302.7 km/h
This is a direct application of the Pythagorean theorem for perpendicular vectors.
Q2 A vector of magnitude 5 units is multiplied by –3. The resulting vector has:
A Magnitude 5, same direction
B Magnitude 15, same direction
C Magnitude 15, opposite direction
D Magnitude –15, opposite direction
✔ Correct Answer: C — Magnitude 15, opposite direction
When a vector A is multiplied by a negative scalar λ = –3:
• New magnitude = |λ| × |A| = 3 × 5 = 15 units
• Direction is reversed (opposite to original)
Magnitude is always positive; option D is physically meaningless.
Q3 Which of the following is a scalar quantity?
A Angular velocity
B Displacement
C Angular frequency
D Linear momentum
✔ Correct Answer: C — Angular frequency
Angular frequency (ω = 2πν) is a scalar — it has only magnitude (rad/s) with no directional information needed for its definition in scalar context.
• Angular velocity: vector
• Displacement: vector
• Linear momentum: vector
3.4
Vector Addition & Subtraction
Head-to-Tail (Triangle) Method: Place the tail of B at the head of A. The resultant R = A + B joins the tail of A to the head of B.
Parallelogram Method: Place both vectors with tails at a common origin. Complete the parallelogram. The diagonal from the common origin is the resultant.
Magnitude
R = √(A² + B² + 2AB cosθ)
Direction
tan α = (B sinθ) / (A + B cosθ)
Properties: Vector addition is commutative (A+B = B+A) and associative ((A+B)+C = A+(B+C)).
Subtraction: A – B = A + (–B). Vector –B has same magnitude as B but opposite direction.
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[Add image here: Triangle method and Parallelogram method of vector addition diagrams]
🎯 Practice MCQs — Vector Addition
Q1 Two velocity vectors of magnitudes 12 m/s and 5 m/s are perpendicular to each other. What is the magnitude of their resultant?
A 7 m/s
B 13 m/s
C 17 m/s
D 8.5 m/s
✔ Correct Answer: B — 13 m/s
For perpendicular vectors, θ = 90°, cosθ = 0:
R = √(A² + B²) = √(12² + 5²) = √(144 + 25) = √169 = 13 m/s
This is the classic 5-12-13 Pythagorean triple!
Q2 Two equal vectors each of magnitude P are inclined at angle 2θ to each other. Their resultant has magnitude:
Key Insight: Motion in a plane = two independent simultaneous 1D motions along perpendicular axes. x-motion and y-motion are independent of each other.
Equations (const. acceleration)
x = x₀ + v₀ₓt + ½aₓt² | y = y₀ + v₀ᵧt + ½aᵧt²
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[Add image here: Displacement Δr, velocity v as tangent to curved path, acceleration vectors]
🎯 Practice MCQs — Motion in a Plane
Q1 A particle has position r = (3.0t)î + (2.0t²)ĵ metres. Its velocity at t = 2 s is:
A 3î + 4ĵ m/s
B 3î + 8ĵ m/s
C 6î + 8ĵ m/s
D 3î + 16ĵ m/s
✔ Correct Answer: B — 3î + 8ĵ m/s
v = dr/dt = d/dt(3.0t)î + d/dt(2.0t²)ĵ = 3.0î + 4.0tĵ
At t = 2s: v = 3î + 4×2ĵ = 3î + 8ĵ m/s
Magnitude = √(9 + 64) = √73 ≈ 8.54 m/s
Q2 A particle starts from origin with velocity 5î m/s and acceleration (2î + 3ĵ) m/s². Its y-coordinate at t = 3s is:
A 9 m
B 13.5 m
C 27 m
D 4.5 m
✔ Correct Answer: B — 13.5 m
y = y₀ + v₀ᵧt + ½aᵧt²
y₀ = 0, v₀ᵧ = 0 (initial y-velocity = 0), aᵧ = 3 m/s², t = 3s
y = 0 + 0 + ½ × 3 × 9 = 13.5 m
Q3 For motion in a plane with constant acceleration, which statement is correct?
A Velocity and acceleration must be parallel
B x and y motions are independent of each other
C Speed must remain constant
D The path is always circular
✔ Correct Answer: B — x and y motions are independent
This is a key principle: 2D motion = two independent 1D motions along perpendicular axes. The x-equation only involves x-quantities, and the y-equation only involves y-quantities.
Velocity and acceleration can have any angle (0°–180°) between them in 2D.
3.9
Projectile Motion
An object in flight after being projected is a projectile. It has two simultaneous components:
→ Horizontal: Constant velocity (no acceleration, aₓ = 0)
Q2 For maximum horizontal range, a projectile must be launched at:
A 30°
B 60°
C 45°
D 90°
✔ Correct Answer: C — 45°
R = v₀² sin2θ₀ / g. For Rₘₐₓ, sin2θ₀ must be maximum = 1
∴ 2θ₀ = 90° → θ₀ = 45°
At 45°: Rₘₐₓ = v₀²/g
Q3 A stone is thrown horizontally from a cliff 80 m high at 20 m/s. Time to reach the ground is: (g = 10 m/s²)
A 2 s
B 4 s
C 8 s
D 6 s
✔ Correct Answer: B — 4 s
Thrown horizontally → v₀ᵧ = 0, aᵧ = –g
y = ½gt² → 80 = ½ × 10 × t²
t² = 16 → t = 4 s
Horizontal distance = 20 × 4 = 80 m
3.10
Uniform Circular Motion
Uniform Circular Motion (UCM): Object moves in a circle at constant speed (not constant velocity, since direction changes). The object accelerates towards the centre.
Centripetal Accelerationaꜱ = v²/R = ω²R (directed towards centre)
Angular Speedω = Δθ/Δt | v = ωR
Time Period & FrequencyT = 2πR/v = 2π/ω | ν = 1/T
With frequencyω = 2πν | v = 2πνR | aꜱ = 4π²ν²R
✈️ AAI ATC Relevance
Aircraft holding patterns are circular. An aircraft holds at constant speed in a horizontal circle — this is uniform circular motion!
The bank angle in a turn is directly related to centripetal acceleration needed: tan(bank) = v²/(gR).
Centripetal acceleration in UCM is constant in magnitude but not in direction (always pointing to centre).
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[Add image here: Circular path showing velocity tangential at various points, centripetal acceleration pointing to centre, ω, R, T labels]
🎯 Practice MCQs — Uniform Circular Motion
Q1 An aircraft in a holding pattern flies at 300 km/h in a circle of radius 5 km. The centripetal acceleration is:
A 1.39 m/s²
B 0.5 m/s²
C 5 m/s²
D 2.78 m/s²
✔ Correct Answer: A — 1.39 m/s²
v = 300 km/h = 300 × 1000/3600 = 83.33 m/s; R = 5000 m
aꜱ = v²/R = (83.33)²/5000 = 6944.4/5000 = 1.39 m/s²
Q2 An insect completes 10 revolutions in 5 seconds in a circle of radius 0.2 m. Its angular speed is:
A 2 rad/s
B 4π rad/s
C 2π rad/s
D 10π rad/s
✔ Correct Answer: B — 4π rad/s
T = 5/10 = 0.5 s (time for one revolution)
ω = 2π/T = 2π/0.5 = 4π rad/s ≈ 12.57 rad/s
Q3 In uniform circular motion, which quantity remains constant?
A Velocity
B Acceleration
C Speed
D Both velocity and speed
✔ Correct Answer: C — Speed
• Speed (magnitude of velocity) = constant ✓
• Velocity = changes direction continuously ✗
• Acceleration magnitude is constant but direction changes (always centripetal) ✗
UCM: "uniform" refers to speed, not velocity.
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