Motion in a Plane— Notes, Formulas & Revision

Triangle law: place tail of B at head of A; resultant R goes from tail of A to head of B.

Parallelogram law: place both tails together; R is the diagonal of the parallelogram.

R² = A² + B² + 2AB cosθ, where θ is the angle between them.

Vectors commute: A + B = B + A.

If θ = 0: R = A + B (max). If θ = 180°: R = |A − B| (min). If θ = 90°: R = √(A² + B²).

Any 2D vector can be resolved into perpendicular components: Vₓ = V cosθ, Vᵧ = V sinθ.

Magnitude: |V| = √(Vₓ² + Vᵧ²). Direction: tanθ = Vᵧ/Vₓ.

Components allow us to treat x and y motions independently — core of 2D kinematics.

A·B = |A||B| cosθ — a scalar.

A·B = Aₓ Bₓ + Aᵧ Bᵧ + A_z B_z.

Perpendicular vectors: A·B = 0. Parallel: A·B = AB. Anti-parallel: A·B = −AB.

Work = F·d; Power = F·v — both are dot products.

|A × B| = |A||B| sinθ — a vector.

Direction: right-hand rule, perpendicular to both A and B.

|A × B| equals the area of the parallelogram formed by A and B.

A × B = −(B × A). A × A = 0.

Torque τ = r × F; angular momentum L = r × p — both are cross products.

Projectile motion is a combination of two independent motions: uniform horizontal motion and uniformly accelerated vertical motion (under gravity).

The horizontal component of velocity (uₓ = u cos θ) remains constant throughout the flight (no air resistance).

The vertical component of velocity (uᵧ = u sin θ) changes due to gravitational acceleration g = 9.8 m/s².

At the highest point, the vertical velocity is zero, but horizontal velocity persists — the object is NOT at rest.

The trajectory of a projectile is a parabola described by: y = x tan θ − (gx²)/(2u²cos²θ).

Time of flight, maximum height, and range depend only on initial speed and launch angle (in vacuum).

The motion is symmetric about the highest point — time of ascent equals time of descent.

Projectile motion analysis is a cornerstone of JEE Mechanics and connects to relative motion, energy, and circular motion concepts.

Projectile motion = uniform horizontal motion + uniformly accelerated vertical motion.

Range R = u² sin(2θ)/g, max at θ = 45°.

Max height H = u² sin²θ/(2g). Flight time T = 2u sinθ/g.

Complementary angles (θ and 90°−θ) give the same range.

R vs θ is a symmetric curve peaking at θ = 45°.

R(θ) = R(90°−θ) — the range is symmetric about 45°.

Maximum range: Rmax = u²/g.

From height h with initial velocity (v₀ₓ, v₀ᵧ), time of flight: T = (v₀ᵧ + √(v₀ᵧ² + 2gh))/g.

Range R = v₀ₓ · T.

Impact speed: v = √(v₀ₓ² + v₀ᵧ² + 2gh) (from energy conservation).

The trajectory is asymmetric — ascent time ≠ descent time.

v_resultant = v_boat + v_river (vector addition).

To cross river straight: point upstream at angle φ = sin⁻¹(v_river/v_boat) — only works if v_boat > v_river.

Min crossing time: point boat perpendicular to current (ignore drift).

Min drift: use angle cos⁻¹(v_river/v_boat) from perpendicular — requires v_boat > v_river.

Rain falls vertically with speed v_r. Man walks with speed v_m horizontally.

Rain relative to man: v_rain/man = v_rain − v_man.

Umbrella tilt angle (from vertical) = tan⁻¹(v_m/v_r).

Speed of rain relative to man: √(v_r² + v_m²).

Speed is constant but velocity direction changes → centripetal acceleration.

a_c = v²/r = ω²r, directed toward the center.

Angular velocity ω = 2π/T = 2πf; linear velocity v = ωr.

Centripetal force F_c = mv²/r = mω²r (always needed, provided by string/gravity/friction/etc.)

Motion is always described relative to a frame of reference. There is no concept of 'absolute motion' in classical mechanics.

Relative velocity of A with respect to B: v_AB = v_A − v_B (vector subtraction).

In 1D, if objects move in the same direction, relative velocity is the difference; if in opposite directions, it is the sum.

The relative velocity determines how fast two objects approach or separate from each other.

In a moving frame, an observer sees the other object's velocity as the relative velocity — this simplifies many problems.

Rain-umbrella problems use relative velocity: tilt angle = tan⁻¹(v_man / v_rain).

River-boat problems: to cross in shortest time, row perpendicular; to cross shortest path, adjust angle upstream.

Relative motion analysis converts two-body problems into single-body problems — a powerful JEE problem-solving technique.