Motion in a Plane — Class XI Physics Notes | eduPhysics
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Class XI · Chapter 03

Motion in a Plane

Vectors, unlocked: how to add and break down directions properly, and what that machinery buys you — projectile motion and circular motion, solved cleanly instead of by guesswork.

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1Scalars and Vectors

A scalar is fully described by a magnitude alone — mass, temperature, speed, time. A vector needs both magnitude and direction — displacement, velocity, force, acceleration. Everything in Chapter 2 secretly assumed motion along one line, where direction reduced to just a plus or minus sign; motion in a plane is where vectors actually earn their keep.

2Vector Addition — Graphical Method

Vectors add head-to-tail (triangle law), or equivalently as adjacent sides of a parallelogram (parallelogram law) — both give the same result.

Magnitude of the Resultant
|R| = √(A² + B² + 2AB cos θ)
θ = angle between vectors A and B. Direction follows from tan β = B sin θ / (A + B cos θ), measured from A's direction.

3Resolution of Vectors into Components

Any vector can be broken down into perpendicular components — almost always the far more practical way to work with vectors than pure geometry.

Components of a Vector
Ax = A cos θ, Ay = A sin θ
Conversely, given components: |A| = √(Ax² + Ay²), θ = tan⁻¹(Ay/Ax).

4Vector Addition — Analytical Method

Once every vector is broken into components, addition becomes simple arithmetic — add the x-components together, add the y-components together, done.

Component-Wise Addition
R = (Ax + Bx) î + (Ay + By) ĵ
This method scales cleanly to any number of vectors and any angles — no geometric constructions required, which is why it's the standard approach for anything beyond the simplest two-vector problem.

5Motion in a Plane with Constant Acceleration

Every kinematic equation from Chapter 2 carries over unchanged — just written as vectors instead of signed numbers.

Vector Kinematic Equations
v = u + at, r = r₀ + ut + ½at²
The genuinely powerful part: motion along the x and y directions can be treated completely independently of each other whenever acceleration has separate x and y components — this independence is exactly what makes projectile motion solvable.

6Relative Velocity in Two Dimensions

Same idea as Chapter 2's relative velocity, now as a full vector subtraction rather than a single number:

Relative Velocity (Vector Form)
vAB = vA − vB
Classic application: a boat crossing a river needs to account for the river's current as a separate velocity vector added to its own, to find its actual path relative to the ground.

7Projectile Motion

Launch an object at an angle, and — ignoring air resistance — it undergoes two completely independent motions at once: constant-velocity motion horizontally, and constant-acceleration (free fall) motion vertically. Treating these separately is the entire trick to solving projectile problems.

Time of Flight
T = 2u sin θ / g
Maximum Height
H = u² sin² θ / (2g)
Horizontal Range
R = u² sin(2θ) / g
Range is maximum at θ = 45° — since sin(2θ) peaks at 2θ = 90°. Two different launch angles that add up to 90° (like 30° and 60°) always give the same range, though different flight times and heights.
The Trajectory Is a Parabola Eliminating time between the horizontal and vertical equations gives y = x tan θ − gx²/(2u²cos²θ) — a quadratic in x, confirming that every ideal projectile path traces exactly a parabola.

8Uniform Circular Motion

Moving at constant speed along a circle still counts as accelerated motion, because velocity's direction is constantly changing — that continuous change requires a centre-pointing force and acceleration.

Centripetal Acceleration
ac = v² / r = ω² r
Always directed toward the centre of the circle — hence "centripetal" (centre-seeking). ω = angular velocity = v/r = 2π/T, where T is the period of one full revolution.

Formula Summary

Resultant Magnitude
R = √(A²+B²+2ABcosθ)
Vector Components
Ax=Acosθ, Ay=Asinθ
Time of Flight
T = 2u sinθ/g
Maximum Height
H = u²sin²θ/2g
Horizontal Range
R = u²sin2θ/g
Centripetal Acceleration
a_c = v²/r
Angular Velocity
ω = v/r = 2π/T

Solved Examples

Example 1 · Projectile Motion

A ball is thrown with a speed of 20 m/s at 30° above the horizontal. Find the time of flight, maximum height, and range. (g = 10 m/s²)

Solution: T = 2u sinθ/g = (2 × 20 × 0.5) / 10 = 2 s.

H = u²sin²θ/2g = (400 × 0.25) / 20 = 5 m.

R = u²sin2θ/g = (400 × sin60°) / 10 = (400 × 0.866) / 10 ≈ 34.6 m.

Example 2 · Centripetal Acceleration

A stone tied to a 1 m string is whirled in a horizontal circle at a constant speed of 4 m/s. Find its centripetal acceleration.

Solution: ac = v²/r = (4)² / 1 = 16 m/s², directed toward the centre of the circle at every instant.

Quick Check

1. In projectile motion (ignoring air resistance), the horizontal component of velocity:
Increases throughout the flight
Decreases to zero at the peak
✓ Remains constant throughout (correct)
Reverses direction mid-flight
2. In uniform circular motion, centripetal acceleration is directed:
Along the direction of velocity
Away from the centre
✓ Toward the centre (correct)
Tangent to the circle
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Motion in a Straight Line

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Laws of Motion

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