Core Concepts
Build from fundamentals to advanced theory. Every concept with the reasoning JEE expects.
Position Vector & Displacement in 2D
CBSE asks this in 2-mark questions. JEE uses this as the foundation for ALL 2D motion problems. If you don't understand position vectors deeply, your entire solution framework is weak.
In 2D motion, the position of a particle is described by a position vector from the origin to the particle's location:
Displacement is the change in position vector:
Students confuse distance and displacement. Distance is the total path length (scalar). Displacement is the shortest path from initial to final position (vector). In circular motion, after one complete revolution: distance = 2πr but displacement = 0.
Position vectors r₁ and r₂ from origin; displacement Δr⃗ connects them
Velocity & Acceleration in 2D
Average velocity direction = direction of displacement. Never assume it follows the curved path. This is the key to solving "what is the direction of velocity" questions.
Key property: Speed = |v⃗| = √(vₓ² + vy²) — this is always positive.
In JEE problems with parametric equations [x(t), y(t)]: differentiate to get velocity components, then find speed and direction separately. This 3-step approach solves 80% of such problems.
"Acceleration is opposite to velocity when decelerating" — This is ONLY true in 1D. In 2D, acceleration can be perpendicular to velocity (circular motion) or at any angle. This is a favourite JEE trap.
Projectile Motion — Complete Analysis
A projectile is any object launched with an initial velocity and subject only to gravity (no air resistance). The key insight that unlocks all projectile problems:
The Fundamental Principle
Horizontal and vertical motions are completely independent.
Horizontal: uniform velocity (no force). Vertical: uniform acceleration (gravity = g downward).
↔ Horizontal Motion
Acceleration: aₓ = 0
Velocity: vₓ = v₀cosθ = constant
Position: x = v₀cosθ · t
↕ Vertical Motion
Acceleration: ay = −g
Velocity: vy = v₀sinθ − gt
Position: y = v₀sinθ·t − ½gt²
Key Derived Results
⏱ Time of Flight (T)
▼Derivation: Set y = 0 → v₀sinθ·t − ½gt² = 0 → t(v₀sinθ − ½gt) = 0 → t = 0 or t = 2v₀sinθ/g
T = 2 × (time to reach max height). The projectile takes equal time going up and coming down. Use this to halve your calculation in many problems.
⬆ Maximum Height (H)
▼Derivation: At peak, vy = 0 → v₀sinθ − gt₀ = 0 → t₀ = v₀sinθ/g. Sub into y: H = v₀sinθ · (v₀sinθ/g) − ½g(v₀sinθ/g)² = v₀²sin²θ/2g
Use v² = u² − 2gH directly: at max height, v = 0, so H = u²/(2g) where u = v₀sinθ.
↔ Horizontal Range (R)
▼Derivation: R = v₀cosθ × T = v₀cosθ × (2v₀sinθ/g) = v₀² · 2sinθcosθ / g = v₀²sin2θ/g
• Maximum range at θ = 45° → R_max = v₀²/g
• Complementary angles give same range: θ and (90°−θ)
• R = 4H·cotθ (useful when both R and H are given)
📈 Equation of Trajectory
▼Derivation: From x = v₀cosθ · t → t = x/(v₀cosθ). Sub into y = v₀sinθ·t − ½gt².
The trajectory equation applies only when ground is at y=0 and launch is from origin. If launch height is h, modify the equation accordingly. NEET and JEE regularly test this boundary condition.
Horizontal Projectile from a Height
When a body is thrown horizontally from height h with velocity u:
Time to hit ground
Independent of horizontal velocity!
Horizontal range
Proportional to u and √h
Final velocity
Use energy: v² = u² + vy² = u² + 2gh
A classic JEE/NEET question: "Two balls dropped and thrown horizontally at the same time. Which hits the ground first?" Answer: Both hit simultaneously — time depends only on height, not horizontal velocity. This is the independence of horizontal and vertical motion in action.
Uniform Circular Motion (UCM)
UCM appears in NEET almost every year (1–2 questions). JEE uses it as a subpart in more complex vertical circle, banking, and rotation problems. Master this and you unlock multiple chapters.
In UCM, a particle moves in a circle with constant speed but changing direction — hence velocity is always changing → acceleration exists.
"Centrifugal force is the reaction of centripetal force." FALSE. Centrifugal force is a pseudo-force in a rotating frame. In an inertial frame, only centripetal force exists. Confusing these costs marks every year in NEET and JEE.
Angular Quantities
- θ = arc / radius = s/r
- ω = dθ/dt = v/r
- α = dω/dt (angular acc)
Linear-Angular Relations
- v = ωr
- aₜ = αr (tangential)
- aᶜ = ω²r (centripetal)
Net acceleration = √(aₜ² + aᶜ²). The direction is NOT toward the center but at an angle. JEE tests this exact concept — find the angle between net acceleration and radius vector.
Relative Motion in 2D
🚣 River Boat Problem
▼A boat of speed v_b wants to cross a river of width d with current speed v_r.
Minimum Time
Point boat perpendicular to river.
t_min = d/v_b Drift = v_r × t_min
Minimum Drift (Straight Path)
Only possible if v_b > v_r.
sinα = v_r/v_b Aim upstream at angle α
Minimum time ≠ minimum drift path. These are two different scenarios requiring different boat orientations. Every year, students mix these up.
✈ Rain and Umbrella
▼Rain velocity is v⃗_r, person moves with v⃗_p. Relative velocity of rain w.r.t person:
Draw a vector diagram. The angle of the umbrella from vertical = tan⁻¹(v_p/v_r) when rain falls vertically. This graphical approach avoids sign errors.