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Class 11 Physics Chapter 3 Motion in a Plane — Formulas & Key Points

Motion in a Plane extends kinematics into two dimensions, requiring vector treatment of displacement, velocity and acceleration. CBSE Class 11 Physics Chapter 3 introduces vector algebra, applies it to projectile motion (a stone thrown at an angle, a football kick) and analyses uniform circular motion. This formula sheet organizes every equation from the NCERT syllabus into quick-reference tables, highlights sign conventions and units, and provides three solved mini-examples so you can confidently tackle numericals in board exams and competitive tests.

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Key takeaways

  • Vectors obey triangle and parallelogram laws; scalars add algebraically while vectors require magnitude and direction consideration.
  • Projectile motion splits into two independent motions: uniform motion horizontally (vₓ = u cos θ) and uniformly accelerated motion vertically (a = –g).
  • Maximum range occurs at 45° projection angle; same range achieved at complementary angles (θ and 90° – θ).
  • Centripetal acceleration in uniform circular motion always points towards the centre with magnitude aᴄ = v²/r = ω²r.
  • Dot product yields a scalar (A·B = AB cos θ); cross product yields a vector perpendicular to both (A×B = AB sin θ n̂).
  • Time of flight T = 2u sin θ / g; maximum height H = u² sin² θ / 2g; horizontal range R = u² sin 2θ / g for projectile motion.
  • Angular displacement θ, angular velocity ω and angular acceleration α relate to linear quantities via radius: v = rω, aₜ = rα.

Vector Algebra — Core Formulas & Definitions

Vectors possess both magnitude and direction; scalars have magnitude only. Addition follows the triangle law (place tail of second vector at head of first) or parallelogram law (diagonals represent sum and difference). Resolution breaks a vector into perpendicular components. The dot product (scalar product) measures projection of one vector onto another; the cross product (vector product) finds a vector perpendicular to both. These operations form the backbone of two-dimensional motion analysis in NCERT Class 11 Physics Chapter 3.
  • Position vector: r = x î + y ĵ (in Cartesian coordinates)
  • Magnitude of vector A: |A| = √(Aₓ² + Aᵧ²) in 2D, √(Aₓ² + Aᵧ² + Aᴢ²) in 3D
  • Unit vector: â = A / |A| (dimensionless, magnitude = 1)
  • Null vector has zero magnitude and arbitrary direction
  • Equal vectors have same magnitude and direction; parallel vectors differ only in magnitude or sense

Projectile Motion — Essential Equations

Projectile motion treats horizontal and vertical motions independently. Horizontally, no acceleration exists (air resistance neglected), so velocity remains constant at u cos θ. Vertically, gravitational acceleration –g acts downward, making it uniformly accelerated motion. CBSE Class 11 Physics Chapter 3 derives range, maximum height, time of flight and trajectory equation from these principles. All formulas assume projection from ground level and return to same level unless stated otherwise.
  • Initial velocity components: uₓ = u cos θ (horizontal), uᵧ = u sin θ (vertical)
  • Velocity at time t: vₓ = u cos θ (constant), vᵧ = u sin θ – gt
  • Displacement: x = (u cos θ)t, y = (u sin θ)t – ½gt²
  • At maximum height, vertical velocity becomes zero: vᵧ = 0
  • Symmetry: time to reach maximum height = time to fall back = T/2

Uniform Circular Motion — Angular & Linear Relations

When a particle moves in a circle at constant speed, its velocity direction changes continuously, producing centripetal acceleration towards the centre. Angular displacement θ (in radians), angular velocity ω (rad/s) and angular acceleration α (rad/s²) link to linear quantities through radius r. NCERT Class 11 Physics emphasises uniform circular motion where speed is constant but velocity is not, because direction keeps changing. Period T and frequency ν describe the time for one complete revolution.
  • One complete revolution = 2π radians = 360°
  • Centripetal force required: Fᴄ = mv²/r = mω²r (directed towards centre)
  • In non-uniform circular motion, tangential acceleration aₜ = rα also exists
  • Total acceleration: a = √(aᴄ² + aₜ²) when both centripetal and tangential components present
  • Centripetal acceleration changes direction of velocity; tangential acceleration changes magnitude

Key Definitions & Terminology (Class 11 Physics Chapter 3)

Understanding precise definitions prevents conceptual errors in Motion in a Plane. A scalar has magnitude only (speed, mass, temperature); a vector requires magnitude and direction (velocity, force, momentum). Displacement is the straight-line vector from initial to final position; distance is the total path length (scalar). Average velocity = total displacement / total time; average speed = total distance / total time. Instantaneous velocity and acceleration are derivatives. Projectile is any object moving under gravity alone after initial projection. Centripetal means 'centre-seeking' and describes the inward acceleration in circular motion. These definitions appear frequently in CBSE Class 11 Physics Chapter 3 NCERT solutions and board exam questions.
  • Scalar: Physical quantity with magnitude only (distance, speed, mass, energy, temperature)
  • Vector: Physical quantity with magnitude and direction (displacement, velocity, acceleration, force, momentum)
  • Position vector: Vector from origin to the position of the particle (r = x î + y ĵ + z k̂)
  • Displacement: Change in position vector (Δr = r₂ – r₁); shortest distance between two points
  • Average velocity: v̄ = Δr / Δt (vector); instantaneous velocity: v = dr / dt
  • Average acceleration: ā = Δv / Δt; instantaneous acceleration: a = dv / dt
  • Projectile: Object in flight under gravity only (after initial projection, no propulsion)
  • Trajectory: Path followed by a projectile (parabola in absence of air resistance)
  • Centripetal acceleration: Acceleration towards centre in circular motion (changes direction of velocity)
  • Angular displacement: Angle swept by position vector (measured in radians)
  • Uniform circular motion: Circular motion with constant speed (magnitude of velocity constant, direction changing)

Common Mistakes, Sign Conventions & Unit Pitfalls

Many marks are lost in CBSE Class 11 Physics Chapter 3 numericals due to sign errors and unit mismatches. Always take vertically upward as positive y-direction, making g = –9.8 m/s² (negative because it acts downward). Angles in projectile motion are measured from the horizontal unless stated otherwise. Radians are dimensionless but must be used in ω = v/r formulas; degrees will give wrong answers. When a projectile is thrown downward or from a height, adjust the trajectory and range equations accordingly. Confusing speed (scalar, always positive) with velocity (vector, can be negative) is a frequent error. Double-check whether the question asks for distance or displacement, speed or velocity.
  • Sign convention: Take upward as +y, downward as –y; acceleration due to gravity a = –g = –9.8 m/s² (or –10 m/s² for simplicity)
  • Angle measurement: θ is angle of projection from horizontal (not vertical) unless specified
  • Units for angular quantities: θ in radians (not degrees) when using ω = v/r or α = aₜ/r
  • Range formula R = u² sin 2θ / g valid only when projectile lands at same level as launch; modify for projection from height
  • Dot product A·B gives scalar (no direction); cross product A×B gives vector (has direction via right-hand rule)
  • Centripetal acceleration aᴄ = v²/r always points inward; do not confuse with tangential acceleration aₜ which changes speed
  • Time of flight formula T = 2u sin θ / g assumes symmetric flight; for projection from height h, use full kinematic equations
  • Velocity at highest point is u cos θ (not zero); only vertical component becomes zero
  • Maximum height H = u² sin² θ / (2g) uses sin² θ, not sin 2θ
  • Complementary angles θ and (90° – θ) give same range because sin 2θ = sin(180° – 2θ)

Memory Tricks & Mnemonics for Quick Recall

Mnemonics help lock formulas into long-term memory for CBSE Class 11 Physics exams. For projectile motion, remember 'T-H-R': Time of flight has 2u sin θ, Height has u² sin² θ / 2g, Range has u² sin 2θ / g. The factor of 2 moves from numerator (T) to denominator (H) to the angle function (R). For maximum range, think '45° is halfway between 0° and 90°' — symmetry gives maximum. In circular motion, 'C-V-R' means Centripetal acceleration = V² / R. Dot product is 'Direction cosine' (cos θ), cross product is 'Direction sine' (sin θ). Vector addition in triangle law: 'Head to Tail, result from free Tail to free Head.'
  • T-H-R mnemonic: Time = 2u sin θ / g, Height = u² sin²θ / 2g, Range = u² sin 2θ / g
  • Maximum range at 45°: sin 2θ maximum when 2θ = 90°, so θ = 45°
  • Complementary angles: Remember sin 2θ = sin(180° – 2θ), so R(30°) = R(60°)
  • Dot product for Work: W = F·d (both cosine and Work start with consonants)
  • Cross product for Torque: τ = r × F (both have direction, use right-hand rule)
  • C-V-R: Centripetal = V² / R (sounds like 'see-v-r')
  • Period and frequency: 'Periodတိုင်းတ frequency' — they are reciprocals, T = 1/ν
  • Radians reminder: 'Radian Relates Radius' — θ = s/r, arc length over radius
  • Horizontal velocity constant: 'No Horizontal Acceleration' in projectile motion (neglecting air resistance)
  • Vertical motion: 'G Always Down' — acceleration due to gravity always acts downward

Three Solved Mini-Examples Applying the Formulas

Worked examples cement formula application and reveal the step-by-step logic expected in CBSE Class 11 Physics board answers. Example 1 demonstrates vector addition using the parallelogram law and finding resultant magnitude and direction. Example 2 applies projectile motion formulas to calculate range, maximum height and time of flight for a given initial velocity and angle. Example 3 analyses uniform circular motion, linking linear speed, angular velocity, centripetal acceleration and period. These mirror typical NCERT Class 11 Physics Chapter 3 numerical problems and show correct unit handling, sign conventions and rounding practices.

Last-Minute Revision Box — One-Glance Summary

This condensed box contains the absolute essentials for a final review 24 hours before your CBSE Class 11 Physics exam. Read it once in the morning, once before entering the exam hall. It covers the top-10 formulas by frequency in board papers, critical sign rules, standard values and the most common conceptual checkpoints. Pair this with NCERT exemplar problems and previous years' question papers for maximum confidence. For personalised doubt-solving and step-by-step solutions to any Class 11 Physics Chapter 3 problem (even from a photo of your homework), explore CBSETUTOR.ai — one flat fee of ₹999/month for 24×7 AI tutoring across all subjects and classes 6 to 12, with a 3-day free trial to experience the platform risk-free.
  • **Vector addition magnitude:** R = √(A² + B² + 2AB cos θ); if perpendicular, R = √(A² + B²)
  • **Dot product:** A·B = AB cos θ = AₓBₓ + AᵧBᵧ + AᴢBᴢ (scalar output)
  • **Cross product magnitude:** |A×B| = AB sin θ (vector output, right-hand rule for direction)
  • **Projectile time of flight:** T = 2u sin θ / g
  • **Projectile max height:** H = u² sin² θ / (2g)
  • **Projectile range:** R = u² sin 2θ / g; Rₘₐₓ at θ = 45°
  • **Centripetal acceleration:** aᴄ = v²/r = ω²r = 4π²r/T²
  • **Angular & linear relation:** v = rω, aₜ = rα (tangential acceleration)
  • **Period & frequency:** T = 2π/ω = 1/ν; ν = ω/(2π)
  • **Sign rule:** Take g = –10 m/s² if upward is +y; angle θ measured from horizontal in projectile motion; radians not degrees in ω = v/r

How CBSETUTOR.ai Helps Master Motion in a Plane

Motion in a Plane numericals often trip students because they require choosing the right formula, applying correct signs and visualising vector diagrams. CBSETUTOR.ai offers a 24×7 AI tutor that accepts photo uploads of your textbook problem or homework question and delivers step-by-step worked solutions within seconds. Whether you are stuck on a tricky projectile-motion derivation at 11 p.m. or need clarity on centripetal versus tangential acceleration, the platform provides instant, NCERT-aligned explanations. At a single price of ₹999/month for every class (6 to 12) and every subject, it eliminates the need for multiple tutors or expensive coaching. A 3-day free trial lets you test the AI tutor with real Class 11 Physics Chapter 3 questions before committing. Thousands of CBSE students already rely on CBSETUTOR.ai for rapid doubt resolution, formula recall practice and exam-focused question banks, making it an indispensable companion for Physics mastery.
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Frequently asked questions

What is the difference between distance and displacement in Motion in a Plane?+
Distance is the total path length travelled (scalar, always positive). Displacement is the straight-line vector from initial to final position (can be zero even if distance is non-zero, e.g. completing a circular lap). Class 11 Physics Chapter 3 emphasises vector treatment of displacement.
Why is centripetal acceleration present even when speed is constant in circular motion?+
Acceleration is the rate of change of velocity (a vector). Even if speed (magnitude) is constant, the direction of velocity changes continuously in circular motion, producing centripetal acceleration aᴄ = v²/r directed towards the centre. This concept is central to NCERT Class 11 Physics Chapter 3.
How do I remember whether to use sin θ or sin 2θ in projectile formulas?+
Use the T-H-R mnemonic: Time of flight has sin θ, Height has sin² θ, Range has sin 2θ. The doubling of angle (2θ) appears only in the range formula R = u² sin 2θ / g.
At what angle should a projectile be launched for maximum range?+
45° gives maximum range for a given initial speed on level ground. This is because sin 2θ reaches its maximum value of 1 when 2θ = 90°, i.e. θ = 45°. Complementary angles (e.g. 30° and 60°) yield equal but smaller ranges.
What is the velocity of a projectile at the highest point of its trajectory?+
At maximum height, the vertical component of velocity becomes zero, but the horizontal component u cos θ remains unchanged (no horizontal acceleration). So velocity at highest point is v = u cos θ (horizontal direction only).
How is dot product different from cross product in vector algebra?+
Dot product A·B = AB cos θ yields a scalar (used in work, power). Cross product A×B = AB sin θ n̂ yields a vector perpendicular to both A and B (used in torque, angular momentum). Dot product is commutative; cross product is anti-commutative.
Can the range formula R = u² sin 2θ / g be used if the projectile is launched from a height?+
No. The standard range formula assumes the projectile lands at the same level as the launch point. For projection from height h, you must use the full trajectory equation or kinematic equations to find where y returns to ground level (y = –h).
Why must angles be in radians when using ω = v / r?+
The formula ω = v/r is derived from θ = s/r (arc length definition of radian). If θ is in radians, ω comes out in rad/s. Using degrees will give numerically incorrect results. Always convert degrees to radians (multiply by π/180) before substituting.
What is the trajectory shape of a projectile in Motion in a Plane?+
A parabola, described by the equation y = x tan θ – (gx²) / (2u² cos² θ). The path is symmetric about the vertical line through the maximum height point when air resistance is neglected.
How does CBSETUTOR.ai help with Class 11 Physics Chapter 3 numericals?+
CBSETUTOR.ai provides 24×7 AI tutoring where you upload a photo of your problem and receive instant step-by-step solutions. It covers all formulas, sign conventions and common mistakes for Motion in a Plane. At ₹999/month for all subjects and classes, with a 3-day free trial, it is an affordable, always-available tutor for mastering projectile motion, vectors and circular motion.

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