Motion in a Plane

In the previous chapter you described motion along a straight line, where + or − signs were enough to fix direction. Motion in a plane needs a full direction, not just a sign, so this chapter opens by sorting physical quantities into two families.

A scalar quantity has only magnitude: mass, distance, speed, temperature, work. It obeys the ordinary rules of algebra — you can add, subtract, multiply and divide scalars directly, but only if they have the same physical dimensions.

A vector quantity has both magnitude and direction, and it obeys a special rule of addition (the triangle/parallelogram law) that you will study in this chapter: displacement, velocity, acceleration, force, momentum. A vector is written as A and drawn as an arrow whose length (to scale) gives the magnitude |A| and whose arrowhead gives the direction. Two vectors are equal only if they have the same magnitude and the same direction — position does not matter, so a vector can be freely moved parallel to itself.

NCERT is careful to distinguish a vector from anything that merely has a direction: current in a wire has a direction but is a scalar, because it does not add by the vector law. Keep this distinction in mind — it is a favourite one-mark trap.

Multiplying a vector A by a positive real number λ gives a new vector λA whose magnitude is λ|A| and whose direction is the same as A. Multiplying by a negative number λ reverses the direction while scaling the magnitude by |λ|. Multiplying by zero gives the null vector 0, which has zero magnitude and no specified direction.

This simple rule underlies two ideas you use constantly later: (i) a unit vector, obtained by dividing a vector by its own magnitude, Â = A/|A|, which has magnitude 1 and points in the direction of A; and (ii) writing any vector in a plane in terms of the standard unit vectors î (along x) and ĵ (along y) as A = A_x î + A_y ĵ.

If a physical quantity like mass (a positive scalar) multiplies a vector — for example, momentum p = mv — the result is still a vector, in the same direction as v (since mass is always positive) but with different physical dimensions and units.

Two vectors are added by the triangle law: draw A, then draw B starting from the head of A; the resultant R = A + B is the vector from the tail of A to the head of B. The equivalent parallelogram law places A and B tail to tail; R is the diagonal of the parallelogram they form. Both give the same resultant, and vector addition is commutative, A + B = B + A, and associative, (A + B) + C = A + (B + C).

Subtraction is addition of the negative: A − B = A + (−B), where −B has the same magnitude as B but points opposite to it. Draw −B and add it to A by the triangle law.

For n vectors in sequence, the polygon law applies: draw each vector head to tail in turn; the resultant is the single vector from the start point to the final head. If the polygon closes on itself, the resultant is zero — this is a common way to test whether a set of forces or displacements is balanced.

Resolution is the reverse of addition: splitting one vector into two (or more) component vectors along chosen directions, usually mutually perpendicular x and y axes. If a vector A makes angle θ with the x-axis, its rectangular components are A_x = A cosθ and A_y = A sinθ, and the vector is written A = A_x î + A_y ĵ. The magnitude and direction are recovered from A = √(A_x² + A_y²) and tanθ = A_y/A_x.

Resolution is the single most useful tool in this chapter: every projectile and circular-motion problem is solved by resolving velocity or acceleration along two convenient perpendicular directions and treating each direction as independent one-dimensional motion.

In three dimensions the idea extends with a third unit vector k̂ along z, giving A = A_x î + A_y ĵ + A_z k̂ and magnitude A = √(A_x² + A_y² + A_z²), though Class 11 numericals mostly stay in a plane.

The analytical method gives an exact formula instead of a scale drawing. For two vectors A and B with angle θ between them, complete the parallelogram: dropping a perpendicular from the head of B onto the extension of A and applying Pythagoras gives the magnitude of the resultant, and the tangent rule gives its direction β measured from A.

Derivation (asked for 3–5 marks): Let OP represent A and OQ represent B, with angle θ between them; complete the parallelogram OPSQ so OS represents R = A + B. Drop a perpendicular from S to meet OP extended at N. In right triangle ONS, SN = B sinθ and PN = B cosθ. Then ON = OP + PN = A + B cosθ. By Pythagoras, OS² = ON² + SN² = (A + B cosθ)² + (B sinθ)², which expands and simplifies to R² = A² + B² + 2AB cosθ, so R = √(A² + B² + 2AB cosθ). For direction, tanβ = SN/ON = B sinθ/(A + B cosθ).

Two convenient special cases are worth memorising: when θ = 90°, R = √(A² + B²); when the vectors are equal in magnitude and θ = 120°, R = A (a favourite MCQ trap).

The position of a particle in the x-y plane at time t is given by the position vector r(t) = x(t) î + y(t) ĵ, drawn from the chosen origin. If the particle moves from P₁ (position r₁) to P₂ (position r₂) in time Δt, the displacement is Δr = r₂ − r₁, a vector along the straight line P₁P₂ regardless of the actual path taken.

The average velocity is v_avg = Δr/Δt, which is parallel to the displacement. The instantaneous velocity is the limit as Δt → 0, v = dr/dt; it is always tangent to the actual path at that point, and its magnitude is the instantaneous speed. In component form v = v_x î + v_y ĵ, with v_x = dx/dt and v_y = dy/dt, so the x and y motions can be treated completely independently — this independence is what makes projectile motion tractable.

When acceleration a is constant in both magnitude and direction, the x and y components of motion each obey the familiar straight-line kinematic equations independently, because a = a_x î + a_y ĵ with both components constant.

In vector form: v = v₀ + at and r = r₀ + v₀t + ½at². Written out in components this is exactly two copies of the 1-D equations of motion running side by side, one along x and one along y, joined only by the common time variable t. This is the key idea the examiner is testing when a two-mark question asks you to 'justify that motion in a plane can be treated as two independent one-dimensional motions'.

Constant acceleration in 2-D covers two important special motions in this chapter: projectile motion (acceleration = g, vertically down, constant) and, as a contrast, uniform circular motion (acceleration constant in magnitude but continuously changing in direction, so it is not covered by this section's equations).

The velocity of object A relative to object B is the vector v_AB = v_A − v_B — it is what an observer moving with B would measure A's velocity to be. This extends the 1-D idea from the previous chapter into a full vector subtraction, so it must be evaluated by the triangle or component method, not by ordinary subtraction of magnitudes.

The two classic setups are: (i) a person walking in the rain, where the rain's velocity relative to the person, v_rain,person = v_rain − v_person, decides the angle at which the person should hold the umbrella; and (ii) a boat crossing a river, where the boat's velocity relative to the ground is the vector sum of its velocity relative to the water and the water's velocity relative to the ground, v_boat,ground = v_boat,water + v_water,ground. Both are solved by drawing the vector triangle and applying resolution or the analytical formula from section 3.6.

Note the sign convention: v_AB = −v_BA. If two bodies move with equal velocity, their relative velocity is zero even though each has a non-zero velocity with respect to the ground.

A projectile is a body given an initial velocity and then left to move under gravity alone (air resistance neglected). Take the point of projection as origin, with the projectile launched at speed u at angle θ above the horizontal. Resolve the initial velocity: u_x = u cosθ (constant throughout, since gravity acts only vertically) and u_y = u sinθ (changes under a = −g).

Trajectory: x = (u cosθ)t and y = (u sinθ)t − ½gt². Eliminating t gives the path equation y = x tanθ − gx²/(2u²cos²θ), which is of the form y = ax − bx², a parabola — this is the standard 'derive the equation of trajectory and show it is a parabola' question.

Time of flight: total time to return to the same horizontal level is when y = 0 (other than t = 0), giving T = 2u sinθ/g. Maximum height: reached when v_y = 0, at t = T/2, giving H = u²sin²θ/2g. Horizontal range: R = (horizontal velocity) × (time of flight) = u²sin2θ/g, which is maximum at θ = 45°, and R is the same for complementary angles θ and (90° − θ).

When a body moves in a circle of radius R at constant speed v, its velocity direction keeps changing even though its magnitude does not — so it is still accelerating. This acceleration is the centripetal acceleration, directed always towards the centre of the circle, magnitude a_c = v²/R.

Derivation (asked for 3 marks): Consider the particle at P at time t (velocity v) and at P′ at time t + Δt (velocity v + Δv), the radius turning through a small angle Δθ = Δt·v/R. Since |v| is constant, the velocity vectors form an isosceles triangle with angle Δθ between them, so |Δv| ≈ v Δθ for small Δθ. Then a = |Δv|/Δt = v(Δθ/Δt) = v·(v/R) = v²/R, and in the limit Δt → 0 the direction of Δv, and hence of a, becomes perpendicular to v, i.e. along the radius towards the centre.

The angular speed ω = v/R = 2π/T (T = period), so a_c can also be written a_c = ω²R = ωv. Because the acceleration is always perpendicular to the velocity, it changes only the direction of motion, never the speed — this is why the motion is called uniform circular motion.