NEET PhysicsNCERT Class 11Chapter 3

Motion in a Plane: NEET notes

Motion along a line needs only a sign for direction; motion in a plane needs vectors. This chapter builds the vector toolkit (adding, subtracting, scaling and resolving into components) and then uses it for position, velocity and acceleration in two dimensions, with projectile motion and uniform circular motion as the two standard cases.

What NEET asks

NEET asks for the resultant of two vectors at an angle, components and unit vectors, the time of flight, maximum height and range of a projectile, complementary angles giving equal range, and centripetal acceleration in terms of v, ω or frequency. Marks are lost by adding magnitudes instead of vectors, using a cosine where a sine belongs, forgetting that the horizontal velocity of a projectile never changes, and thinking uniform circular motion has zero acceleration.

1. Scalars and vectors

NCERT §3.2

  • A scalar is fully given by a number with its unit: distance, mass, temperature, time, volume. Scalars combine by ordinary algebra.
  • A vector has a magnitude and a direction and combines by the triangle law (equivalently the parallelogram law): displacement, velocity, acceleration and force are vectors. In print a vector is bold (v); by hand it is written with an arrow over it.
  • The position vector r of a point is the arrow from a chosen origin O to that point. If an object moves from P to P′, its displacement is the vector PP′ = r′ − r, which depends only on the start and end points.
  • The path length between two points is never less than the magnitude of the displacement between them: walking 40 m east and then 30 m north covers 70 m, but the displacement is only 50 m.
  • A and B are equal vectors when both their sizes and their directions agree. A vector may be shifted parallel to itself without changing it (a free vector); some applications, like a force acting at a point, care about the line of action as well (a localised vector).
  • The magnitude of a vector is written |A| or simply A; it is never negative.

2. Multiplying a vector by a number

NCERT §3.3

  • Multiplying a vector A by a positive number λ gives λA: same direction, magnitude λ|A|. A 5 m/s velocity east doubled is 10 m/s east.
  • Multiplying by a negative number reverses the direction: −1 × (5 m/s east) is 5 m/s west, and −1.5A points opposite to A with 1.5 times its length.
  • If λ carries a dimension, the product has a new dimension: a constant velocity (m/s) times a time (s) gives a displacement (m).
  • −A has the same magnitude as A and the opposite direction; this is what makes subtraction possible.

3. Adding and subtracting vectors graphically

NCERT §3.4

  • Triangle (head-to-tail) method: place the tail of B at the head of A; the arrow from the tail of A to the head of B is R = A + B.
  • Vector addition is commutative, A + B = B + A, and associative, (A + B) + C = A + (B + C).
  • Adding A and −A gives the null (zero) vector 0, of zero magnitude and no fixed direction. A + 0 = A, λ0 = 0 and 0A = 0. A displacement that returns to its start is the null vector.
  • Subtraction is addition of the reversed vector: A − B = A + (−B).
  • Parallelogram method: draw A and B from a common tail and complete the parallelogram; the diagonal from that tail is A + B. It gives the same result as the triangle method.
  • Rain falling vertically at 8 m/s in a wind blowing from east to west at 6 m/s moves at 10 m/s, at an angle whose tangent is 6/8 (about 37°) from the vertical, drifting west. To meet it, the umbrella is tilted about 37° from the vertical towards the east.

4. Resolving vectors into components

NCERT §3.5

  • Any vector A in a plane can be written as λa + μb along two chosen non-parallel vectors a and b; λa and μb are its components along them.
  • Unit vectors î, ĵ, k̂ have magnitude 1, point along +x, +y and +z, and have no dimension or unit. Any vector A equals |A| n̂, where n̂ is the unit vector along A.
  • A vector of magnitude A at angle θ to the x-axis has components Ax = A cos θ and Ay = A sin θ, so A = Ax î + Ay ĵ. A component can be positive, negative or zero.
  • From the components back to the vector: A = √(Ax² + Ay²) and tan θ = Ay/Ax. A ball thrown at 20 m/s at 30° has components 17.3 m/s and 10.0 m/s.
  • In three dimensions, Ax = A cos α, Ay = A cos β, Az = A cos γ, where α, β, γ are the angles with the axes, and A = √(Ax² + Ay² + Az²).
  • The position vector of a point (x, y, z) is r = x î + y ĵ + z k̂.

5. Adding vectors analytically

NCERT §3.6

  • To add vectors, add their components: if R = A + B, then Rx = Ax + Bx, Ry = Ay + By (and Rz = Az + Bz). The same works for any number of vectors and for subtraction.
  • For two vectors of magnitudes A and B with angle θ between them, R² = A² + B² + 2AB cos θ (the law of cosines).
  • The direction of R from A is given by tan α = B sin θ / (A + B cos θ); the law of sines, R/sin θ = A/sin β = B/sin α, relates the sides and angles of the triangle.
  • The resultant lies between |A − B| (θ = 180°) and A + B (θ = 0°); at θ = 90° it is √(A² + B²).
  • 40 m and 30 m displacements at 60° to each other give R = √(1600 + 900 + 1200) ≈ 60.8 m, at about 25° to the 40 m leg.
  • Worked with components, a boat heading north across a current adds its velocity through the water to the water's velocity; the ground velocity is the vector sum.

6. Velocity and acceleration in a plane

NCERT §3.7

  • The position vector is r = x î + y ĵ; as the object moves from r to r′ in Δt, the displacement is Δr = r′ − r.
  • Average velocity is Δr/Δt, a vector along Δr. Instantaneous velocity is the limit v = dr/dt, with components vx = dx/dt and vy = dy/dt.
  • The instantaneous velocity is always along the tangent to the path, in the direction of motion. Its magnitude is v = √(vx² + vy²) and its angle to x is given by tan θ = vy/vx.
  • Average acceleration is Δv/Δt; instantaneous acceleration is a = dv/dt, with ax = dvx/dt = d²x/dt² and ay = dvy/dt.
  • In one dimension velocity and acceleration lie on the same line; in a plane the angle between v and a can be anything from 0° to 180°.
  • Example: r = 4.0t î + 1.5t² ĵ (metres, t in s) gives v = 4.0 î + 3.0t ĵ and a = 3.0 ĵ m/s²; at t = 1 s the speed is 5.0 m/s at about 37° to the x-axis.

7. Constant acceleration in a plane

NCERT §3.8

  • If a is constant, v = v₀ + a t and r = r₀ + v₀ t + ½ a t², as vectors.
  • In components: x = x₀ + v₀ₓ t + ½ aₓ t² and y = y₀ + v₀ᵧ t + ½ aᵧ t², with vₓ = v₀ₓ + aₓ t and vᵧ = v₀ᵧ + aᵧ t.
  • Motion in a plane with constant acceleration is two independent one-dimensional motions, along x and along y, sharing the same clock t.
  • Example: starting at the origin with v₀ = 4.0 î m/s and a = 3.0 ĵ m/s², after 2 s the object is at 8.0 î + 6.0 ĵ m, 10 m from the start, moving at 4.0 î + 6.0 ĵ m/s (about 7.2 m/s).
  • The velocity turns steadily towards the direction of a, while the component perpendicular to a never changes.

8. Projectile motion

NCERT §3.9

  • A projectile is an object that, once thrown, moves under gravity alone; air resistance is neglected. Its motion is a uniform horizontal motion and a uniformly accelerated vertical motion, independent of each other.
  • Launched at speed v₀ at angle θ₀ above the horizontal: aₓ = 0, aᵧ = −g; x = (v₀ cos θ₀) t and y = (v₀ sin θ₀) t − ½ g t².
  • The horizontal velocity v₀ cos θ₀ never changes; the vertical velocity vᵧ = v₀ sin θ₀ − g t falls by g every second and is zero at the top.
  • The path is y = (tan θ₀) x − g x² / [2 (v₀ cos θ₀)²], a parabola.
  • Time to the top tₘ = v₀ sin θ₀ / g; time of flight T_f = 2 v₀ sin θ₀ / g; maximum height hₘ = (v₀ sin θ₀)² / 2g; horizontal range R = v₀² sin 2θ₀ / g.
  • The range is largest at θ₀ = 45°, Rₘ = v₀²/g. Angles 45° + α and 45° − α (complementary angles such as 30° and 60°) give the same range but different heights and times.
  • A ball thrown at 20 m/s at 30° (g = 9.8 m/s²) rises 5.1 m, stays up 2.04 s and lands 35.3 m away.
  • Thrown horizontally from a height h, a body falls for t = √(2h/g) whatever its horizontal speed: from 4.9 m it takes 1.0 s, and at 10 m/s it lands 10 m away moving at 14 m/s.

9. Uniform circular motion

NCERT §3.10

  • Uniform circular motion is motion on a circle at constant speed. The velocity is along the tangent and keeps changing direction, so the body is accelerated.
  • The acceleration points towards the centre (centripetal acceleration) and has magnitude a_c = v²/R.
  • a_c has constant magnitude but its direction keeps turning, always towards the centre, so it is not a constant vector; the constant-acceleration equations do not apply.
  • Angular speed ω = Δθ/Δt (rad/s); v = R ω and a_c = ω² R.
  • The time period T is the time for one revolution and the frequency ν = 1/T is revolutions per second; v = 2πRν, ω = 2πν and a_c = 4π²ν²R.
  • Jogging at 5.0 m/s on a circle of radius 20 m: ω = 0.25 rad/s, one lap takes 2π/0.25 ≈ 25 s, and a_c = 25/20 = 1.25 m/s² towards the centre.
  • If the speed is changing as well, the acceleration also has a component along the tangent and no longer points at the centre.

Must-know facts

  1. Displacement magnitude ≤ path length; equal only for straight-line motion in one direction.
  2. Unit vectors î, ĵ, k̂ have magnitude 1 and no unit or dimension.
  3. Ax = A cos θ, Ay = A sin θ, where θ is measured from the x-axis.
  4. R² = A² + B² + 2AB cos θ; tan α = B sin θ / (A + B cos θ).
  5. |A − B| ≤ |A + B| ≤ A + B.
  6. A − B = A + (−B); A + (−A) = 0, the null vector.
  7. Instantaneous velocity is tangent to the path.
  8. In a plane the angle between v and a can be anything from 0° to 180°.
  9. Constant a: v = v₀ + at, r = r₀ + v₀t + ½at², component by component.
  10. Projectile: horizontal velocity v₀ cos θ₀ constant; vertical acceleration −g.
  11. T_f = 2v₀ sin θ₀/g; hₘ = v₀² sin² θ₀/2g; R = v₀² sin 2θ₀/g.
  12. Maximum range v₀²/g at 45°; θ and 90° − θ give equal ranges.
  13. The trajectory of a projectile is a parabola: y = x tan θ₀ − gx²/(2v₀² cos² θ₀).
  14. Uniform circular motion: a_c = v²/R = ω²R = 4π²ν²R, towards the centre.
  15. v = Rω; ω = 2πν = 2π/T.
  16. Average speed ≥ magnitude of average velocity.

Common traps

Adding the magnitudes: 40 m and 30 m give 70 m whatever the angle.

Magnitudes add only when the vectors are parallel. At 90° the answer is 50 m; use R² = A² + B² + 2AB cos θ.

Taking Ax = A sin θ out of habit.

The cosine goes with the side next to the angle. If θ is measured from x, Ax = A cos θ; if it is measured from y, the roles swap.

Saying the velocity of a projectile is zero at the highest point.

Only the vertical component is zero there; the body still moves with v₀ cos θ₀ horizontally, and g still acts.

Thinking uniform circular motion has no acceleration because the speed is constant.

The direction of velocity keeps changing, so there is an acceleration v²/R towards the centre.

Using v = u + at along the circle for uniform circular motion.

The acceleration vector keeps turning, so it is not constant; the kinematic equations for constant a do not apply.

Believing a ball thrown horizontally takes longer to fall than one dropped from the same height.

Vertical motion is independent of horizontal motion: both fall for √(2h/g).

Assuming a higher launch angle always sends a projectile farther.

Range is v₀² sin 2θ₀/g, which rises to 45° and then falls; 60° gives the same range as 30°.

Giving a unit vector a unit, such as î metres.

A unit vector is a pure direction, magnitude 1 with no dimension; the unit sits on the number in front of it.

Formulas

Components

Ax = A cos θ, Ay = A sin θ

θ measured from the x-axis.

Magnitude and direction

A = √(Ax² + Ay²), tan θ = Ay/Ax

In 3D add Az² under the root.

Resultant of two vectors

R² = A² + B² + 2AB cos θ

Law of cosines; θ is the angle between A and B.

Direction of the resultant

tan α = B sin θ / (A + B cos θ)

α measured from A.

Law of sines

R/sin θ = A/sin β = B/sin α

Angles of the vector triangle.

Constant acceleration

v = v₀ + at, r = r₀ + v₀t + ½at²

Applies to each component separately.

Projectile position

x = v₀ cos θ₀ t, y = v₀ sin θ₀ t − ½gt²

Launch from the origin; g = 9.8 m/s².

Projectile path

y = x tan θ₀ − gx² / (2v₀² cos² θ₀)

A parabola.

Time of flight

T_f = 2v₀ sin θ₀ / g

Time to the top is half of it.

Maximum height

hₘ = v₀² sin² θ₀ / 2g

Reached when vᵧ = 0.

Horizontal range

R = v₀² sin 2θ₀ / g

Maximum v₀²/g at 45°.

Centripetal acceleration

a_c = v²/R = ω²R = 4π²ν²R

Directed towards the centre.

Angular speed

v = Rω, ω = 2πν = 2π/T

ω in rad/s, ν in s⁻¹ (Hz).

Key terms

Scalar
A quantity fully given by a number and a unit.
Vector
A quantity with magnitude and direction that adds by the triangle law.
Position vector
The arrow from the chosen origin to the point.
Displacement
The change in position vector, from the starting point to the end point.
Null vector
The vector of zero magnitude, the sum of a vector and its reverse.
Unit vector
A dimensionless vector of magnitude 1 that only marks a direction.
Resolution
Splitting a vector into components along chosen directions.
Projectile
A body moving under gravity alone after being launched.
Trajectory
The path a body follows; for a projectile, a parabola.
Horizontal range
The horizontal distance from launch to return to the launch level.
Centripetal acceleration
The acceleration towards the centre of a circle, v²/R in size.
Angular speed
The rate at which the angle swept about the centre changes, in rad/s.
Frequency
Number of revolutions per second, the reciprocal of the time period.

Lumi is not affiliated with or endorsed by NCERT. The official NCERT textbooks are free to read and download from NCERT's own website, ncert.nic.in. These notes and simulations are original work by Lumi (Aikolumi Software Pvt Ltd), © 2026, shared under CC BY-NC 4.0: copy, print, share and adapt them for any non-commercial use, with credit to Lumi and a link to lumineet.com.