Vector & 2D Motion Analysis
The mathematical toolkit for 2D motion. Master vectors and you master this entire chapter.
🔧 Vector Component Visualizer
Vector Fundamentals for 2D Motion
Unit Vectors
î, ĵ, k̂ are unit vectors along x, y, z axes respectively.
î · ĵ = 0 (perpendicular)
î × ĵ = k̂
Vector Operations
|A⃗| = √(Aₓ² + Ay²)
θ = tan⁻¹(Ay/Aₓ)
CBSE asks: "What is the angle made by the resultant with the x-axis?" — Always use tan θ = (sum of y-components)/(sum of x-components). Never guess from the diagram alone.
The Component Method — Master This First
Every 2D problem becomes solvable with this method. It's not a shortcut — it's the only reliable approach at JEE level.
4-Step Component Method
Choose axes: Usually x = horizontal, y = vertical. For inclined planes, rotate axes along and perpendicular to the incline.
Resolve all vectors: Break every vector (velocity, force, etc.) into x and y components.
Apply equations separately: Use kinematics for x-direction, then independently for y-direction.
Combine results: Find resultant magnitude = √(x² + y²) and angle = tan⁻¹(y/x).
If you choose the wrong direction for a component (sign error), all subsequent steps are wrong. Always draw a clear diagram BEFORE writing equations and define positive direction explicitly.
Oblique Projectile — Velocity Analysis
At any time t, the velocity makes angle φ with horizontal:
JEE asks: "At what time does velocity become perpendicular to initial velocity?" Set v⃗ · v⃗₀ = 0 → vₓ·v₀cosθ + vy·v₀sinθ = 0. Solve for t. This is an integer-type question format.
When velocity becomes perpendicular to initial velocity v⃗₀:
v⃗ · v⃗₀ = 0: v₀cosθ·vₓ + v₀sinθ·vy = 0. Since vₓ = v₀cosθ: v₀²cos²θ + v₀sinθ(v₀sinθ − gt) = 0. Simplify: v₀² − gtv₀sinθ = 0 → t = v₀/(gsinθ).
Time when speed equals initial speed again:
At t = T: vₓ is same (v₀cosθ), vy = −v₀sinθ (reversed). Speed = v₀ but downward angle instead of upward. This is why the landing angle equals the launch angle.
Velocity & Acceleration as Vectors in UCM
Velocity Vector in UCM
- Always tangent to the circle
- Magnitude constant (v = ωr)
- Direction changes continuously
- At angle θ: v⃗ = v(−sinθ î + cosθ ĵ)
Acceleration Vector in UCM
- Always toward center (centripetal)
- Perpendicular to velocity
- Magnitude = v²/r = ω²r
- At angle θ: a⃗ = −ω²r(cosθ î + sinθ ĵ)
Velocity (tangent) and centripetal acceleration (toward center) for a particle in UCM
In UCM: v⃗ · a⃗ = 0 always (perpendicular vectors). Their dot product is zero. JEE uses this to test: "Prove that speed is constant in UCM" — because a⃗ ⊥ v⃗ means no component of acceleration along velocity, so no speed change.
Change in Velocity — The Tricky Vector Subtraction
This is one of the most commonly tested vector concepts in NEET and JEE Main.
Change in velocity for circular arc (angle α)
▼v⃗₁ and v⃗₂ both have magnitude v, angle between them = α.
|Δv⃗| = |v⃗₂ − v⃗₁| = √(v₁² + v₂² − 2v₁v₂cosα) = √(2v²(1 − cosα)) = 2v·sin(α/2)
• Quarter circle (α=90°): |Δv⃗| = v√2
• Half circle (α=180°): |Δv⃗| = 2v
• Full circle (α=360°): |Δv⃗| = 0
Direction of Δv⃗ in circular motion
▼Δv⃗ = v⃗₂ − v⃗₁. Graphically: flip v⃗₁ and add to v⃗₂.
For UCM, Δv⃗ always bisects the angle between the two velocity vectors and points toward the center of the circle. This is why centripetal acceleration is always inward.