Motion in a Plane extends one-dimensional kinematics to two dimensions using vectors, covering vector addition, dot and cross products, and motion under constant acceleration. It builds up to the chapter’s most tested ideas, projectile motion, relative velocity, and uniform circular motion. For NEET it is a high-yield foundation chapter whose vector tools and projectile and circular-motion formulas reappear throughout mechanics.
Table of Contents
- Key Concepts — Scalars & vectors, vector addition, resolution, projectile & circular motion
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Scalars and Vectors
A scalar is a quantity that has only magnitude — like mass, time, speed, distance, energy, or temperature. You can add scalars with ordinary arithmetic.
A vector has both magnitude and direction, and it follows the triangle (or parallelogram) law of addition. Displacement, velocity, acceleration, force, and momentum are vectors. A vector is written in bold (A) or with an arrow, and its magnitude is written |A| or simply A.
Types of Vectors
- Equal vectors: same magnitude and same direction.
- Unit vector: a vector of magnitude 1 used to show direction; Â = A/|A|.
- Zero (null) vector: magnitude zero, arbitrary direction; result of adding a vector to its negative.
- Negative vector: same magnitude, opposite direction (−A).
- Collinear vectors: act along the same or parallel lines.
2. Unit Vectors
A unit vector has a magnitude of exactly one and only specifies direction. The standard unit vectors along the x, y and z axes are î, ĵ, k̂ respectively.
Any vector can be written in terms of unit vectors: A = Aₓ î + A_y ĵ + A_z k̂
- î, ĵ, k̂ are mutually perpendicular (orthogonal).
- Each has magnitude 1 and no units.
- The unit vector of any vector is  = A/|A|.
3. Addition of Vectors — Triangle Law
The triangle law states that if two vectors are represented by two sides of a triangle taken in order, their resultant is given by the third side taken in the opposite order.
[DIAGRAM: Vector A drawn head-to-tail with vector B; the closing side from the tail of A to the head of B is the resultant R = A + B.]
Vector addition is commutative (A + B = B + A) and associative (A + (B + C) = (A + B) + C).
4. Addition of Vectors — Parallelogram Law
The parallelogram law states that if two vectors acting at a point are represented by two adjacent sides of a parallelogram, their resultant is represented by the diagonal passing through that point.
For two vectors A and B with angle θ between them, the magnitude of the resultant R is:
R = √(A² + B² + 2AB cos θ)
The direction of the resultant (angle β with A) is given by:
tan β = (B sin θ)/(A + B cos θ)
- When θ = 0°: R = A + B (maximum).
- When θ = 180°: R = |A − B| (minimum).
- When θ = 90°: R = √(A² + B²).
5. Subtraction of Vectors
Subtracting a vector means adding its negative: A − B = A + (−B). So to subtract B, reverse its direction and add it to A using the triangle law.
Magnitude: |A − B| = √(A² + B² − 2AB cos θ)
6. Resolution of Vectors
Splitting a vector into two or more components is called resolution. The most useful is resolving a vector into two mutually perpendicular components (rectangular components).
If a vector A makes an angle θ with the x-axis:
- x-component: Aₓ = A cos θ
- y-component: A_y = A sin θ
Then A = Aₓ î + A_y ĵ, with magnitude A = √(Aₓ² + A_y²) and direction tan θ = A_y/Aₓ.
7. Analytical Method of Vector Addition
To add several vectors accurately, resolve each into components, add the components separately, then recombine.
If R = A + B, then:
- Rₓ = Aₓ + Bₓ
- R_y = A_y + B_y
Magnitude: R = √(Rₓ² + R_y²), and direction tan α = R_y/Rₓ. This method extends easily to any number of vectors and to three dimensions.
8. Motion in a Plane with Constant Acceleration
In two dimensions, position, velocity and acceleration are all vectors, and motion along x and y is independent. The equations of motion apply separately to each axis.
- v = u + at → vₓ = uₓ + aₓt, v_y = u_y + a_yt
- r = r₀ + ut + ½at²
- v² = u² + 2a·s (along each axis)
Key idea: The horizontal and vertical motions are independent — this is exactly what makes projectile motion solvable.
9. Projectile Motion
A projectile is any object thrown into the air that moves under gravity alone (air resistance neglected). Its path is a parabola. The horizontal velocity stays constant; the vertical velocity changes due to g.
For a projectile launched with speed u at angle θ to the horizontal:
- Horizontal velocity: uₓ = u cos θ (constant)
- Vertical velocity: u_y = u sin θ (decreases, then increases)
Key Projectile Equations
| Quantity | Formula |
|---|---|
| Time of flight (T) | T = (2u sin θ)/g |
| Maximum height (H) | H = (u² sin²θ)/(2g) |
| Horizontal range (R) | R = (u² sin 2θ)/g |
| Equation of path | y = x tan θ − (gx²)/(2u²cos²θ) |
Maximum range occurs at θ = 45°, giving R_max = u²/g. Two complementary angles (θ and 90° − θ) give the same range.
Horizontal Projectile
For a body projected horizontally with speed u from height h: time of fall t = √(2h/g), and horizontal range = u√(2h/g).
10. Uniform Circular Motion
When a body moves along a circle at constant speed, it is in uniform circular motion. The speed is constant, but the velocity keeps changing direction, so the body is accelerating.
Angular Velocity
Angular velocity (ω) is the rate of change of angular displacement. It links to linear speed by v = ωr.
- ω = θ/t = 2π/T = 2πn, where T is the time period and n is frequency.
- SI unit: radian per second (rad/s).
- Linear speed: v = ωr
Centripetal Acceleration
The acceleration in uniform circular motion always points towards the centre and is called centripetal acceleration.
a_c = v²/r = ω²r = (4π²r)/T²
Important: In uniform circular motion the speed is constant, so there is no tangential acceleration — only the centre-directed centripetal acceleration that changes direction.
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Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 6–8 marks | Vector addition/resolution, projectile equations, circular motion |
| JEE Main / Advanced | 1–3 questions | Projectile range/height, relative velocity, circular motion |
| NEET | 2–3 questions | Resolution of vectors, projectile motion, centripetal acceleration |
[TABLE: Question-type split — VSA (1 mark): scalar/vector, unit vectors, definitions; SA (2–3 marks): resolution, projectile numericals, ω–v relation; LA (5 marks): derivation of time of flight, max height and range; centripetal acceleration derivation.]
Important Definitions
| Term | Definition |
|---|---|
| Scalar | A quantity with magnitude only, e.g. mass, speed, energy |
| Vector | A quantity with both magnitude and direction obeying the triangle law |
| Unit vector | A vector of magnitude one showing direction: Â = A/|A| |
| Resolution of a vector | Splitting a vector into rectangular components Aₓ = A cos θ, A_y = A sin θ |
| Resultant vector | Single vector with the same effect as two or more vectors combined |
| Projectile | A body moving under gravity alone after being thrown; path is parabolic |
| Time of flight | Total time a projectile stays in air: T = 2u sin θ/g |
| Horizontal range | Horizontal distance covered by a projectile: R = u² sin 2θ/g |
| Angular velocity | Rate of change of angular displacement: ω = θ/t = 2π/T |
| Centripetal acceleration | Centre-directed acceleration in circular motion: a_c = v²/r = ω²r |
Solved Examples
Example 1
Two vectors of magnitude 3 units and 4 units act at right angles to each other. Find the magnitude of their resultant.
Answer: R = √(A² + B² + 2AB cos 90°) = √(3² + 4² + 0) = √(9 + 16) = √25 = 5 units.
Example 2
A force of 10 N makes an angle of 30° with the x-axis. Find its rectangular components.
Answer: Fₓ = F cos 30° = 10 × (√3/2) = 8.66 N; F_y = F sin 30° = 10 × 0.5 = 5 N.
Example 3
A ball is thrown with a speed of 20 m/s at an angle of 30° to the horizontal. Find the time of flight. (g = 10 m/s²)
Answer: T = (2u sin θ)/g = (2 × 20 × sin 30°)/10 = (2 × 20 × 0.5)/10 = 20/10 = 2 s.
Example 4
For the ball in Example 3, find the maximum height and horizontal range.
Answer: H = (u² sin²θ)/(2g) = (400 × 0.25)/20 = 100/20 = 5 m. R = (u² sin 2θ)/g = (400 × sin 60°)/10 = (400 × 0.866)/10 = 34.64 m.
Example 5
A particle moves in a circle of radius 2 m at a constant speed of 4 m/s. Find its centripetal acceleration.
Answer: a_c = v²/r = 4²/2 = 16/2 = 8 m/s², directed towards the centre.
Example 6
A stone tied to a string of length 0.5 m is whirled in a horizontal circle, completing 2 revolutions per second. Find its angular velocity and linear speed.
Answer: ω = 2πn = 2π × 2 = 4π ≈ 12.57 rad/s. v = ωr = 12.57 × 0.5 = 6.28 m/s.
Important Questions for Board Exams
1-Mark Questions (VSA)
- Define a unit vector. What is its use?
- Can the resultant of two vectors of unequal magnitude be zero? Justify.
- What is the angle of projection for maximum horizontal range?
- Is uniform circular motion accelerated motion? Why?
- Give one example each of a scalar and a vector quantity.
2–3-Mark Questions (SA)
- State the parallelogram law of vector addition and write the expression for the magnitude of the resultant.
- Show that the horizontal and vertical motions of a projectile are independent of each other.
- Derive the expression for the time of flight of a projectile.
- Define angular velocity and derive the relation v = ωr.
5-Mark Questions (LA)
- Define projectile motion. Derive expressions for time of flight, maximum height, and horizontal range of a projectile.
- Show that the path of a projectile is a parabola and find the angle for maximum range.
- Derive an expression for the centripetal acceleration of a body in uniform circular motion.
Quick Revision Points
- Scalar = magnitude only; vector = magnitude + direction (obeys triangle law)
- Unit vector  = A/|A|; standard unit vectors î, ĵ, k̂ are mutually perpendicular
- Parallelogram law: R = √(A² + B² + 2AB cos θ); max at θ = 0°, min at θ = 180°
- Resolution: Aₓ = A cos θ, A_y = A sin θ; A = √(Aₓ² + A_y²)
- Analytical addition: Rₓ = ΣAₓ, R_y = ΣA_y, R = √(Rₓ² + R_y²)
- 2D motion: x and y are independent; apply v = u + at and r = r₀ + ut + ½at² per axis
- Projectile path is a parabola; horizontal velocity u cos θ is constant
- Time of flight T = 2u sin θ/g; max height H = u²sin²θ/2g; range R = u²sin 2θ/g
- Maximum range at θ = 45°, R_max = u²/g; complementary angles give equal range
- Uniform circular motion: v = ωr; ω = 2π/T = 2πn
- Centripetal acceleration a_c = v²/r = ω²r, always towards the centre
Next Chapter: Chapter 4 — Laws of Motion
Class 11 Physics · Chapter 4 – swipe through all 9 cards to understand the whole chapter.
Vectors and Components
A vector has both magnitude and direction, so it adds tip-to-tail, never by plain arithmetic.
Resolve into ⊥ components first, add component-wise, then recombine.
- Scalar = size only (mass, speed); vector = size + direction (velocity, force)
- A component can never exceed the vector (cos, sin ≤ 1)
- Unit vector n̂ has magnitude 1 — direction only, no size
Resultant of Two Vectors
Combine two vectors at angle θ using the parallelogram law.
α is the direction of R measured from vector A.
- Max resultant = A + B at θ = 0° (parallel)
- Min resultant = |A − B| at θ = 180° (antiparallel)
- At θ = 90°: R = √(A2 + B2)
Dot and Cross Product
Two ways to multiply vectors: dot gives a number, cross gives a vector.
Dot ⊥ → 0 (work F·d); cross ∥ → 0 (torque r×F).
- Dot = ‘how parallel’ (cos); cross = ‘how perpendicular’ (sin)
- Components: A·B = AₓBₓ + AᵧBᵧ + A_zB_z
- |A×B| = area of parallelogram; A×B = −(B×A)
2D Kinematics
Motion along x and y is independent — each axis obeys 1D equations sharing the same time t.
Apply each equation separately to x and y; time links the two axes.
- Perpendicular motions don’t affect each other
- A bullet fired horizontally and one dropped land at the same time
- Avg velocity = displacement/time (vector); avg speed = path/time (scalar)
Projectile Motion
Launched at speed u and angle θ: horizontal speed u cos θ is constant, vertical u sin θ changes with g.
Path is a parabola: y = x tan θ − gx2/(2u2 cos2θ).
- Range is max at θ = 45°: R_max = u2/g
- θ and (90° − θ) give the same range (e.g. 30° and 60°)
- At the top vertical velocity = 0, but speed ≠ 0 (u cos θ survives)
Relative Velocity
Velocity of A as seen by B is the vector difference of their velocities.
Note the minus sign — relative velocity subtracts (unlike the resultant law).
- River boat, shortest time: aim straight across, t = width/v_b, you drift
- Land directly opposite: aim upstream, sin θ = v_r/v_b, across speed = √(v_b2 − v_r2)
- Rain-man: tilt umbrella along v_rain − v_man (forward, into your motion)
Uniform Circular Motion
Constant speed in a circle is still accelerated motion because direction keeps changing.
a_c always points to the centre, perpendicular to velocity.
- Speed is constant but velocity is not — so it accelerates
- Centripetal force F = mv2/r = mω2r, directed inward
- ‘Centrifugal’ is only a pseudo-force in the rotating frame
Non-Uniform Circular Motion
When the speed also changes, a tangential acceleration adds to the centripetal one.
a_c is radial (changes direction), a_t is tangential (changes speed).
- a_c = v2/r points to the centre; a_t is along the path
- The two are perpendicular, so combine by Pythagoras
- Remove the inward force and the object flies off along the tangent
Quick Wins for NEET
High-yield shortcuts and traps examiners reuse from this chapter.
Three is the fewest unequal coplanar vectors that can sum to zero.
- Zero dot product ⇒ vectors are perpendicular (A·A = A2, A×A = 0)
- Range needs sin 2θ, not sin θ; horizontal velocity never shrinks
- Always write the vector equation first, then resolve into components
📝 Practice Motion in a Plane — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Chapter Navigation
Previous: Motion in a Straight Line Class 11 Notes
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Related Chapters in Class 11 Physics
- Laws of Motion Class 11 Notes
- Work, Energy and Power Class 11 Notes
- Systems of Particles and Rotational Motion Class 11 Notes
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Frequently Asked Questions
A scalar has only magnitude, such as mass, speed or distance, while a vector has both magnitude and direction, such as displacement, velocity or force. Vectors must be added tip-to-tail or by components, never by simple arithmetic.
For a projectile launched at speed u and angle theta, the time of flight is T = 2u sin(theta)/g, the maximum height is H = u squared sin squared(theta)/2g, and the horizontal range is R = u squared sin(2 theta)/g. The horizontal velocity u cos(theta) stays constant while the vertical velocity changes due to gravity.
Even at constant speed the direction of the velocity keeps changing, and any change in velocity is an acceleration. This centripetal acceleration, a = v squared/r = omega squared r, always points toward the centre, perpendicular to the velocity.
Yes, it is a high-weightage Class 11 mechanics chapter and questions on projectile motion, vectors and circular motion appear almost every year. Its vector methods are also reused in later chapters like Laws of Motion and Work, Energy and Power.
The dot product A.B = AB cos(theta) gives a scalar and is zero when the vectors are perpendicular, as in work F.d. The cross product has magnitude AB sin(theta), gives a vector perpendicular to both, and is zero when the vectors are parallel, as in torque r cross F.