NEET Physics · NCERT Class 11

Motion in a Plane notes

Scalars & Vectors, Vector Operations, Resolution & Relative Velocity, Motion in a Plane, Projectile Motion and Uniform Circular Motion — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 7 topics
  • 67 min read
  • 37 formulas
  • 26 PYQs
1

Chapter Overview

7 min

Motion in a plane extends straight-line motion to two dimensions, usually described using x and y coordinates. Many real situations such as projectile motion, circular motion, boat crossing a river, rain seen by a moving person and airplane motion in wind require two-dimensional analysis. The chapter begins with scalar and vector quantities, then explains vector addition, subtraction, dot product, cross product and resolution into components. The key idea is that motion along perpendicular axes can often be treated independently. For NEET, this chapter is highly important because it combines conceptual vector questions, formula-based projectile numericals, relative velocity problems and circular motion applications.

Key points6

  • In plane motion, x and y components are solved separately and then combined vectorially.
  • A scalar has magnitude only; a vector has magnitude and direction.
  • Vector diagrams are essential for NEET relative velocity and projectile questions.
  • Projectile range is maximum at 45 degrees on level ground.
  • Complementary projection angles give the same horizontal range for equal speed.
  • Velocity in circular motion is tangential while centripetal acceleration is radial inward.

Memory tricks2

Chapter Flow: Remember V-R-P-C: Vectors, Resolution, Projectile, Circular motion.

Component Method: Break first, solve separately, combine finally.

Examples2

Daily Life Example

A basketball shot, a water stream from a hose, a boat crossing a river and a car turning on a circular track are all examples of motion in a plane.

Quick Check

If a ball is projected horizontally, its horizontal velocity remains constant while vertical velocity increases due to gravity.

Common mistakes3

Treating Vector Magnitudes Like Algebraic Numbers

Vectors must be added using direction. Two vectors of 5 N and 5 N can have resultant from 0 to 10 N depending on angle.

Mixing Components

Do not use vertical acceleration in horizontal projectile equations. In ideal projectile motion, ax = 0 and ay = -g.

Forgetting Direction in Relative Velocity

Relative velocity is vector subtraction, not always simple subtraction of speeds.

Reference tables2

Quick revision8

Formulas3

Vector Magnitude in 2D
∣A∣=√(Ax2+Ay2)|A| = √(Ax^2 + Ay^2)

Magnitude of a vector from its rectangular components.

|A|
Magnitude of vector A
Ax
x-component of vector A
Ay
y-component of vector A
Projectile Range
R=u2sin2θ/gR = u^2 sin 2θ / g

Horizontal range of an oblique projectile on level ground.

R
Horizontal range
u
Initial speed
θ
Angle of projection
g
Acceleration due to gravity
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Diagrams3

2

Scalars & Vectors

8 min

Physical quantities are measurable properties used to describe nature. Some quantities, such as mass, time, temperature, distance and speed, need only magnitude and are called scalars. Others, such as displacement, velocity, acceleration, force and momentum, need both magnitude and direction and are called vectors. In two-dimensional motion, vectors are represented by arrows or by components along coordinate axes. Equal vectors have the same magnitude and direction, while negative vectors have the same magnitude but opposite direction. Unit vectors show direction with magnitude one. Understanding zero, parallel, anti-parallel and position vectors is essential before solving vector operations, projectile motion and relative velocity problems.

Key points6

  • A vector is unchanged if shifted parallel to itself without changing magnitude and direction.
  • In 2D, position vector is r = xi + yj.
  • Unit vectors i and j represent positive x and positive y directions.
  • Magnitude of a vector is always non-negative.
  • Negative sign before a vector reverses its direction, not its magnitude.
  • Types of vectors include equal, negative, zero, unit, position, parallel, anti-parallel, coinitial and coplanar vectors.

Memory tricks2

Scalar vs Vector: Scalar says 'how much'; vector says 'how much and where'.

Unit Vector: Unit vector is a direction tag: magnitude one, direction same.

Examples2

Real-Life Vector

A wind blowing at 20 km/h east is a vector because both speed and direction are required.

Position Vector Example

A point at (3 m, 4 m) has position vector r = 3i + 4j and magnitude 5 m.

Common mistakes3

Calling Speed a Vector

Speed has no direction. Velocity is the vector quantity.

Thinking Negative Vector Has Negative Magnitude

Magnitude is never negative. The negative sign reverses direction.

Confusing Equal and Parallel Vectors

Parallel vectors need not have equal magnitude; equal vectors must have same magnitude and same direction.

Reference tables2

Quick revision8

Formulas3

Position Vector in 2D
r=xi+yjr = xi + yj

Vector from origin to a point with coordinates (x, y).

r
Position vector
x
Coordinate along x-axis
y
Coordinate along y-axis
i, j
Unit vectors along x and y axes
Magnitude of a 2D Vector
∣A∣=√(Ax2+Ay2)|A| = √(Ax^2 + Ay^2)

Length of a vector from its rectangular components.

|A|
Magnitude of vector A
Ax
x-component
Ay
y-component
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Diagrams3

3

Vector Operations

10 min

Vector operations allow us to combine physical quantities that have direction. Vector addition can be performed by triangle law, parallelogram law or polygon law. In triangle law, the tail of the second vector is placed at the head of the first; the resultant goes from the first tail to the last head. Subtraction means adding the negative vector. Multiplication by a scalar changes magnitude and may reverse direction if the scalar is negative. Dot product gives a scalar and is used in work done and projections. Cross product gives a vector perpendicular to the plane of the two vectors and is used in torque, angular momentum and magnetic force.

Key points6

  • Vector addition is commutative: A + B = B + A.
  • Vector subtraction is not commutative: A - B is generally not equal to B - A.
  • Dot product is maximum when vectors are parallel and zero when perpendicular.
  • Cross product is zero when vectors are parallel and maximum when perpendicular.
  • Work done W = F · s depends on component of force along displacement.
  • Right hand rule: curl fingers from first vector to second; thumb gives A × B direction.

Memory tricks3

Dot vs Cross: Dot gives how much along; cross gives how much perpendicular.

Maximum Rules: Dot loves parallel, cross loves perpendicular.

Subtraction Trick: Never subtract directly. Reverse the second vector and add.

Examples3

Solved Resultant

Two vectors of 3 units and 4 units at 90 degrees have resultant R = √(9 + 16) = 5 units.

Work Example

A force of 10 N acts at 60 degrees to displacement 5 m. Work = 10 × 5 × cos60 = 25 J.

NEET Trick

For two equal vectors A at angle 120 degrees, resultant is A because R = √(A^2 + A^2 + 2A^2 cos120) = A.

Common mistakes3

Adding Magnitudes Directly

A + B is not always A + B numerically. It depends on the angle between vectors.

Forgetting Cross Product Direction

A × B and B × A have opposite directions.

Using sin Instead of cos in Work

Work done is F · s = Fs cosθ, not Fs sinθ.

Reference tables2

Quick revision8

Formulas6

Resultant of Two Vectors
R=√(A2+B2+2AB cosθ)R = √(A^2 + B^2 + 2\text{AB cos}θ)

Magnitude of resultant of two vectors A and B with angle θ between them.

R
Magnitude of resultant vector
A, B
Magnitudes of two vectors
θ
Angle between vectors
Direction of Resultant
tanα=Bsinθ/(A+Bcosθ)tanα = B sinθ / (A + B cosθ)

Angle α made by resultant with vector A.

α
Angle of resultant with A
A, B
Magnitudes of vectors
θ
Angle between A and B
Vector Subtraction
A−B=A+(−B)A - B = A + (-B)

Subtracting a vector means adding its opposite vector.

A, B
Vectors
-B
Vector equal in magnitude and opposite in direction to B
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Diagrams4

4

Resolution & Relative Velocity

11 min

Resolution of vectors means splitting a vector into components along chosen axes. Rectangular components along x and y axes are most common because perpendicular components are independent. If a vector A makes angle θ with x-axis, Ax = A cosθ and Ay = A sinθ. Components can also be taken along an inclined plane or arbitrary axes if they simplify the problem. Relative velocity describes velocity of one object as observed from another and is found by vector subtraction. Boat-river, rain-man and airplane-wind problems are all relative velocity applications. The safest NEET method is to draw vectors, resolve into components and then apply the required condition.

Key points6

  • Always attach cos to the component adjacent to the angle and sin to the opposite component.
  • Along an inclined plane, weight components are mg sinθ down the plane and mg cosθ perpendicular to plane.
  • For shortest river crossing time, boat should head perpendicular to river bank.
  • For zero drift while crossing river, boat must aim upstream with suitable component cancelling river velocity.
  • Rain appears vertical to a moving person when horizontal component of relative rain velocity is zero.
  • Airplane ground velocity is the vector sum of airplane velocity relative to air and wind velocity.

Memory tricks3

Component Rule: Cos is close to the angle; sin is away from the angle.

Relative Velocity: Velocity of A with respect to B means remove B: vAB = vA - vB.

Boat Shortcut: Shortest time: aim straight across. No drift: aim upstream.

Examples3

Component Example

A 10 N force at 30 degrees to x-axis has Fx = 10 cos30 = 5√3 N and Fy = 10 sin30 = 5 N.

Boat Example

River width is 100 m and boat speed in still water is 5 m/s. For shortest time, t = 100/5 = 20 s, independent of river speed.

Rain-Man NEET Style

If rain falls vertically at 10 m/s and man runs horizontally at 10 m/s, relative rain makes 45 degrees with vertical.

Common mistakes3

Using sin and cos in Wrong Places

Check where the angle is measured. The component adjacent to the angle gets cos.

Treating River Velocity as Acceleration

River current adds velocity, not acceleration.

Wrong Umbrella Direction

The umbrella must be aligned opposite to relative rain velocity, not actual rain velocity alone.

Reference tables2

Quick revision8

Formulas6

Rectangular Components
Ax=AcosθAx = A cosθ
Ay=AsinθAy = A sinθ

Components of vector A when θ is measured from positive x-axis.

A
Magnitude of vector
Ax
Component along x-axis
Ay
Component along y-axis
θ
Angle with x-axis
Component Reconstruction
A=Axi+AyjA = Axi + Ayj

A vector can be written as sum of its component vectors.

A
Original vector
i, j
Unit vectors along x and y axes
Magnitude and Direction
∣A∣=√(Ax2+Ay2)|A| = √(Ax^2 + Ay^2)
tanθ=Ay/Axtanθ = Ay/Ax

Find vector magnitude and direction from components.

|A|
Magnitude of vector
θ
Direction angle from x-axis
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Diagrams3

5

Motion in a Plane

9 min

Motion in a plane is described by position, displacement, velocity and acceleration vectors that change with time. A particle at coordinates (x, y) has position vector r = xi + yj. Its displacement is the change in position vector. Velocity is the rate of change of position vector, and acceleration is the rate of change of velocity vector. When acceleration is constant, the equations of motion can be applied separately along x and y directions. Parametric equations x(t) and y(t) describe the path of the particle by eliminating time if required. This component method is the foundation for projectile motion, relative motion and many NEET numerical problems.

Key points6

  • Two-dimensional motion can be treated as two simultaneous one-dimensional motions.
  • Components along perpendicular axes are independent.
  • The velocity vector is tangent to the path at every point.
  • Acceleration may change speed, direction or both.
  • Constant acceleration in a plane does not always mean straight-line motion.
  • Projectile motion is a special case with ax = 0 and ay = -g.

Memory tricks2

Vector Kinematics Chain: r changes to v, v changes to a: position → velocity → acceleration.

Component Independence: x never asks y for permission; y never asks x. Solve separately.

Examples3

Displacement Vector Example

A particle moves from (2, 3) m to (8, 6) m. Displacement = 6i + 3j and magnitude = √45 = 3√5 m.

Velocity Components Example

If vx = 3 m/s and vy = 4 m/s, speed = √(3^2 + 4^2) = 5 m/s.

Parametric Example

If x = 2t and y = 5t^2, the motion is two-dimensional because both coordinates vary with time.

Common mistakes3

Assuming Constant Acceleration Means Straight Path

Projectile motion has constant acceleration but curved parabolic path.

Confusing Position Vector and Displacement Vector

Position is from origin; displacement is from initial point to final point.

Forgetting Velocity Direction

Velocity is tangent to the path, not necessarily in the direction of acceleration.

Reference tables2

Quick revision8

Formulas5

Position Vector
r=xi+yjr = xi + yj

Locates the particle in a two-dimensional coordinate system.

r
Position vector
x, y
Coordinates of the particle
i, j
Unit vectors along x and y axes
Displacement Vector
Δr=(x2−x1)i+(y2−y1)jΔr = (x2 - x1)i + (y2 - y1)j

Change in position vector between two points.

Δr
Displacement vector
x1, y1
Initial coordinates
x2, y2
Final coordinates
Velocity Vector
v=dr/dt=vxi+vyjv = dr/dt = vxi + vyj

Rate of change of position vector with time.

v
Velocity vector
vx, vy
Velocity components along x and y axes
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Diagrams3

6

Projectile Motion

12 min

Projectile motion is the motion of an object projected into air and then moving under the effect of gravity alone, neglecting air resistance. It is a two-dimensional motion with horizontal and vertical components. Horizontally, acceleration is zero, so horizontal velocity remains constant. Vertically, acceleration is g downward, so vertical velocity changes uniformly. In oblique projection, initial velocity is resolved as ux = u cosθ and uy = u sinθ. Important results include time of flight, maximum height, horizontal range, trajectory equation and velocity components at any instant. For level ground, range is maximum at 45 degrees and complementary angles give the same range.

Key points6

  • At highest point, vertical velocity is zero but horizontal velocity remains u cosθ.
  • Acceleration is always g downward throughout projectile motion.
  • Trajectory is parabolic when air resistance is neglected.
  • For horizontal projection, initial vertical velocity is zero.
  • Complementary angles θ and 90° - θ have equal range for same speed on level ground.
  • Time of ascent equals time of descent only when landing level equals projection level.

Memory tricks3

Projectile Split: Horizontal is steady, vertical is gravity-ready.

Maximum Range: Range loves sin2θ; sin2θ is maximum at 90°, so θ = 45°.

Complementary Angles: θ and 90° - θ share the same sin2θ, so they share the same range.

Examples3

Solved Example

A projectile is fired at 20 m/s at 30 degrees. Time of flight = 2×20×sin30/10 = 2 s. Range = 20^2×sin60/10 = 20√3 m.

Maximum Height Example

For u = 10 m/s and θ = 30 degrees, H = 100×(1/2)^2/(2×10) = 1.25 m.

Horizontal Projection Example

A ball rolls off a table with horizontal speed 5 m/s and falls for 2 s. Horizontal distance = 5×2 = 10 m.

Common mistakes3

Using g in Horizontal Motion

Gravity acts vertically, so horizontal acceleration is zero in ideal projectile motion.

Taking Velocity Zero at Highest Point

Only vertical velocity is zero at the highest point. Horizontal velocity remains u cosθ.

Using Range Formula for Unequal Heights

R = u^2 sin2θ/g is valid only when projection and landing points are at the same level.

Reference tables2

Quick revision8

Formulas7

Velocity Components
ux=ucosθux = u cosθ
uy=usinθuy = u sinθ

Initial velocity of oblique projectile resolved into horizontal and vertical parts.

u
Initial speed
θ
Angle of projection with horizontal
ux
Initial horizontal velocity
uy
Initial vertical velocity
Projectile Equations
x=ucosθtx = u cosθ t
y=usinθt−(1/2)gt2y = u sinθ t - (1/2)gt^2

Parametric equations of an oblique projectile from origin.

x, y
Horizontal and vertical coordinates
t
Time after projection
g
Acceleration due to gravity
Time of Flight
T=2usinθ/gT = 2u sinθ / g

Total time for projectile to return to same vertical level.

T
Time of flight
u sinθ
Initial vertical velocity
Maximum Height
H=u2sin2θ/2gH = u^2 sin^2θ / 2g

Greatest vertical height reached by an oblique projectile.

H
Maximum height
u
Initial speed
θ
Projection angle
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Diagrams4

7

Uniform Circular Motion

10 min

Uniform circular motion is motion of a body along a circular path with constant speed. Although speed remains constant, velocity continuously changes direction, so the motion is accelerated. This acceleration is centripetal acceleration and always points toward the centre of the circle. The force responsible for this inward acceleration is centripetal force, which may be tension, friction, gravity or normal force depending on the situation. Angular displacement measures the angle swept by the radius vector, angular velocity measures its rate of change, and angular acceleration measures change in angular velocity. Period and frequency describe repetition. NEET often asks formulas for centripetal acceleration, centripetal force, v-r-ω relation and real applications.

Key points7

  • Centripetal force is not a new force; it is the name of the required inward net force.
  • In uniform circular motion, tangential acceleration is zero but radial acceleration is non-zero.
  • Angular displacement is measured in radians.
  • One full revolution corresponds to 2π radians.
  • For a stone tied to a string, tension provides centripetal force.
  • For satellite motion, gravity provides centripetal force.
  • For a car on a flat circular track, friction provides centripetal force.

Memory tricks3

Velocity and Acceleration: Velocity touches the circle; acceleration targets the centre.

Centripetal Formula: Two faces of inward acceleration: v²/r and ω²r.

Force Provider: Centripetal force is a role, not a separate force; tension, gravity or friction may play that role.

Examples4

Stone Tied to String

When a stone is whirled in a circle, string tension acts toward the centre and provides centripetal force.

Numerical Example

A body of mass 2 kg moves in a circle of radius 1 m with speed 4 m/s. Centripetal force = mv²/r = 2×16/1 = 32 N.

Satellite Example

A satellite stays in orbit because gravitational force continuously bends its velocity toward Earth, acting as centripetal force.

Previous NEET-Style Question

If speed is doubled at the same radius, centripetal acceleration becomes four times because ac ∝ v².

Common mistakes3

Thinking Uniform Circular Motion Has Zero Acceleration

Speed is constant, but velocity direction changes continuously, so acceleration is non-zero.

Drawing Centripetal Acceleration Tangentially

Centripetal acceleration always points toward the centre, not along the tangent.

Calling Centripetal Force a New Force

Centripetal force is the net inward force required for circular motion, provided by real forces.

Reference tables3

Quick revision8

Formulas7

Angular Displacement
θ=s/rθ = s/r

Angle swept by radius vector in radians.

θ
Angular displacement in radians
s
Arc length
r
Radius
Angular Velocity
ω=dθ/dt=2π/T=2πfω = dθ/dt = 2π/T = 2πf

Rate of change of angular displacement.

ω
Angular velocity
θ
Angular displacement
T
Time period
f
Frequency
Angular Acceleration
α=dω/dtα = dω/dt

Rate of change of angular velocity.

α
Angular acceleration
ω
Angular velocity
t
Time
Linear and Angular Velocity
v=rωv = rω

Relation between tangential speed and angular velocity.

v
Linear or tangential speed
r
Radius
ω
Angular velocity
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Diagrams3

Videos for Motion in a Plane

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Watch with a free account

NEET PYQs from Motion in a Plane

The latest 3 of 26 previous-year questions from this chapter.

  1. NEET 2025Set 45HardQ1

    Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of service and the speed (assumed constant) of the buses.

    • A9 min, 40 km/h
    • B25 min, 100 km/h
    • C10 min, 90 km/h
    • D15 min, 120 km/h
  2. NEET 2024Set T1EasyQ2

    A particle moving with uniform speed in a circular path maintains:

    • Aconstant velocity but varying acceleration.
    • Bvarying velocity and varying acceleration.
    • Cconstant velocity.
    • Dconstant acceleration.
  3. NEET 2023Set E1MediumQ3

    A bullet is fired from a gun at the speed of 280 m s⁻¹ in the direction 30° above the horizontal. The maximum height attained by the bullet is : (g = 9.8 m s⁻², sin 30° = 0.5)

    • A2800 m
    • B2000 m
    • C1000 m
    • D3000 m

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