Motion in a Plane

2.1 Core concepts

Physical quantities split into two classes. A scalar has only magnitude with the proper unit (distance, mass, temperature, time, density, work, energy) — its value is specified by a single number. A vector has both magnitude and direction and obeys the triangle/parallelogram law of addition (displacement, velocity, acceleration, force, momentum, electric field) (NCERT §3.2, pp. 27–28). The position vector OP = r locates a particle from a chosen origin O; the displacement vector PP′ joins the initial to the final position and is path-independent — its magnitude is always less than or equal to the actual path length traversed (NCERT §3.2.1, p. 28, Fig. 3.1). Two vectors are equal if and only if they have the same magnitude and the same direction; vectors of equal length but different directions are unequal (NCERT §3.2.2, p. 28).

Scalar multiplication scales a vector A by a real number λ: a positive λ keeps the direction unchanged and multiplies the magnitude by λ; a negative λ reverses the direction; multiplying by zero gives the null/zero vector 0, which has zero magnitude and unspecified direction (NCERT §3.3, p. 29). Graphical vector addition uses either the head-to-tail (triangle) method (NCERT Fig. 3.4, p. 29) or the equivalent parallelogram method (NCERT Fig. 3.6, p. 31). Vector addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)) (NCERT §3.4, pp. 29–31). The null vector 0 satisfies A + 0 = A, λ·0 = 0, and 0·A = 0; it arises naturally when an object returns to its starting point so that the net displacement is zero (NCERT §3.4, p. 30). Subtraction is defined as A − B = A + (−B) (NCERT §3.4, p. 30).

A vector A in a plane can be resolved into components along any two non-collinear directions a, b: A = λa + μb. In rectangular Cartesian axes A = Ax î + Ay ĵ, where Ax = A cos θ and Ay = A sin θ (NCERT §3.5, pp. 31–32). The unit vectors î, ĵ, k̂ are vectors of magnitude 1, dimensionless and unit-less, mutually perpendicular along the x-, y-, z-axes; their sole role is to specify direction (NCERT §3.5, p. 32). The magnitude is recovered as A = √(Ax² + Ay²), and the direction by tan θ = Ay/Ax. Analytical addition is then straightforward: if R = A + B, then Rx = Ax + Bx, Ry = Ay + By. For two vectors A, B with angle θ between them, the law of cosines gives R = √(A² + B² + 2AB cos θ), and the direction by the law of sines R/sin β = A/sin γ = B/sin α (NCERT §3.6, pp. 33–34).

For motion in a plane, the position vector r = x î + y ĵ depends on time. The average velocity is v̄ = Δr/Δt; the instantaneous velocity v = dr/dt = vx î + vy ĵ is always tangential to the path (NCERT §3.7, pp. 35–36, Fig. 3.13). The average acceleration is ā = Δv/Δt; the instantaneous acceleration a = dv/dt = ax î + ay ĵ. A key qualitative difference from 1-D motion: in a plane, v and a can have any angle between 0° and 180° (NCERT §3.7, pp. 36–37). For constant acceleration in a plane, the vector kinematic equations are v = v₀ + at and r = r₀ + v₀t + ½ at²; the motions along x- and y-axes are independent simultaneous 1-D motions — a deep simplification that NCERT proves component-wise (NCERT §3.8, pp. 37–38).

The canonical application is projectile motion. A projectile fired with initial speed v₀ at angle θ₀ to the horizontal experiences ax = 0 (no horizontal force in the absence of air resistance) and ay = −g. The component equations are: x = (v₀ cos θ₀) t, y = (v₀ sin θ₀) t − ½ g t², vx = v₀ cos θ₀ (constant), vy = v₀ sin θ₀ − g t (NCERT §3.9, pp. 38–39, Fig. 3.16). Eliminating t between the first two yields the equation of trajectory y = (tan θ₀) x − g x²/(2 v₀² cos² θ₀), a parabola (NCERT §3.9, p. 39, Fig. 3.17). The time to reach maximum height is tm = v₀ sin θ₀ /g; the total time of flight is Tf = 2 v₀ sin θ₀ /g; the maximum height is hm = (v₀ sin θ₀)²/(2g); the horizontal range is R = v₀² sin 2θ₀ /g (NCERT §3.9, pp. 39–40). The range is maximum when θ₀ = 45° (sin 2θ₀ = 1), with Rm = v₀²/g. For elevations exceeding or falling short of 45° by equal amounts, the range is the same — because sin(90° + 2α) = sin(90° − 2α) = cos 2α (NCERT §3.9, Example 3.6, p. 40).

Uniform circular motion (UCM). The speed v is constant but the direction changes continuously, so the particle is accelerated — even though its speed is unchanging. The acceleration always points toward the centre of the circle and is called centripetal acceleration, with magnitude a_c = v²/R (NCERT §3.10, pp. 40–42). The angular speed ω = Δθ/Δt connects to the linear speed by v = ωR, and hence a_c = ω²R. The time period is T = 2πR/v, the frequency ν = 1/T; the connections ω = 2π/T = 2πν, v = 2πRν, a_c = 4π² ν² R follow at once (NCERT §3.10, p. 42). NCERT emphasises that although the magnitude of a_c is constant, its direction changes continuously (always toward the centre), so a_c is not a constant vector — a frequent assertion–reason topic.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

Fig. 3.1 (p. 28) establishes that the displacement vector PQ between two points is the same regardless of whether the particle travels via path PABCQ, PDQ or PBEFQ — displacement is path-independent, distance is not. Fig. 3.4 (p. 29) illustrates the head-to-tail (triangle) law of vector addition and the associative property visually: (A + B) + C = A + (B + C). Fig. 3.6 (p. 31) demonstrates the equivalence of the parallelogram method (bring tails to a common origin, draw diagonal) with the triangle method. Fig. 3.9 (pp. 32–33) shows the unit vectors î, ĵ, k̂ as the directional skeleton of Cartesian coordinates and the resolution A = Ax î + Ay ĵ. Fig. 3.13 (p. 35) explains why instantaneous velocity is tangential: as Δt → 0 the chord Δr aligns with the tangent to the path. Fig. 3.16 (p. 39) plots the projectile trajectory with components — horizontal velocity stays constant, vertical velocity changes uniformly. Fig. 3.17 (p. 39) displays the resulting parabolic path. Fig. 3.18 (p. 41) is the centripetal-acceleration derivation: the vector Δv is perpendicular to Δr, and in the limit Δt → 0 it points toward the centre of the circle.

Two procedural backbones carry most problems: (i) the component method for vector problems — choose axes, resolve each vector into perpendicular components, add components scalar-wise to get Rx and Ry, recover R = √(Rx² + Ry²) and θ = arctan(Ry/Rx); and (ii) the projectile recipe — write x(t) and y(t) separately, eliminate t for trajectory or solve for the required quantity (time of flight from y = 0 at landing, range from x at landing, max height from vy = 0). For UCM the recipe is similarly mechanical: identify R and v (or ω, T, ν), apply a_c = v²/R = ω²R = 4π² ν² R, remember that a_c always points to the centre and that v is always tangential — they are mutually perpendicular at every instant. CUET routinely tests students on whether they can convert between the various forms (v, ω, T, ν, a_c) of UCM parameters with a single substitution.

2.4 Common confusions / NTA trap points

• Treating displacement and path length as the same — they coincide only when the object moves in one direction without reversing (NCERT Points to Ponder 1, p. 46).
• Forgetting that a component Ax of a vector is itself a scalar number, while Ax î is a vector (NCERT §3.5, p. 32).
• Using the formula v = v₀ + at for uniform circular motion — the kinematic equations of uniform acceleration do not apply because in UCM the direction of a changes (NCERT Points to Ponder 4, p. 46).
• Confusing the angle of projection that gives maximum height (90°) with the angle that gives maximum range (45°): R is maximum at θ₀ = 45°, hm is maximum at θ₀ = 90° (NCERT §3.9, pp. 39–40).
• Treating centripetal acceleration as a constant vector — its magnitude is constant but its direction continually changes, so it is NOT a constant vector (NCERT §3.10, p. 42; Example 3.9, p. 42).
• Forgetting that in projectile motion the horizontal velocity is unchanged throughout the flight (vx = v₀ cos θ₀) while only vy changes (NCERT §3.9, p. 39).
• The formula R = v₀² sin 2θ₀ /g is valid only when launch and landing are at the same height; using it for a projectile launched from a height is a classic error.
• For UCM, velocity (tangent) and centripetal acceleration (radius) are perpendicular at every instant — so the work done by the centripetal force is zero, hence kinetic energy is constant.
• Two equal-and-opposite vectors give a null vector, not "no vector at all" — the null vector is a legitimate mathematical object with zero magnitude and unspecified direction (NCERT §3.4, p. 30).
• The maximum range Rm = v₀²/g (at θ₀ = 45°) is twice the maximum height attainable at that angle — students sometimes wrongly equate the two.
• The scalar (dot) product is not introduced in this chapter — its first appearance is in Chapter 5 (Work, Energy, Power). Don't import it here without justification.
• Angular speed ω is in rad s⁻¹, not rev s⁻¹; to convert revolutions to radians multiply by 2π. CUET sometimes hands you "5 revolutions per second" and expects ω = 10π rad s⁻¹.

2.5 Key formulas