Motion in a Plane — NEET notes

Scalars & Vectors

Physics quantities split into two camps: scalars (just a number + unit) and vectors (number + unit + direction). Confusing the two is the #1 reason students get vector problems wrong. NEET tests this distinction directly and indirectly in almost every mechanics question.

| A | = A_x^2 + A_y^2 | A = A | A |

Scalar Quantities

A scalar is completely described by a single number (magnitude) and its unit. No direction required.

A scalar quantity has only magnitude (a number with a unit). It obeys ordinary algebra. When you say 'the temperature is 37 °C' or 'the mass is 5 kg,' you've given complete information — no direction needed.

• Examples: mass, temperature, time, speed, distance, energy, work, power, charge, electric potential
• Scalars add, subtract, multiply, and divide using ordinary arithmetic
• A scalar can be positive, negative, or zero (e.g., temperature can be -10 °C, but the negative is NOT a direction)
• Speed is a scalar; velocity is a vector. Distance is a scalar; displacement is a vector

Vector Quantities

A vector needs both magnitude and direction to be fully specified. Vectors follow special addition rules (triangle/parallelogram law), not ordinary arithmetic.

A vector quantity has both magnitude and direction. Saying 'velocity is 5 m/s' is incomplete — you must say '5 m/s eastward.' Vectors are represented by arrows: the length = magnitude, the arrowhead = direction.

• Examples: displacement, velocity, acceleration, force, momentum, torque, electric field
• Notation: A or A (bold) for the vector; | A | or A for its magnitude
• Vectors are equal if and only if they have the same magnitude AND same direction — position doesn't matter
• A vector can be moved parallel to itself without changing it (free vector concept)

Types of Vectors

NCERT defines several special vector types. NEET questions often hinge on knowing the precise definitions — especially null vector and unit vector.

Vectors come in several flavours, each with a precise definition that NEET exploits.

A = A | A | A = | A | A

• The null vector 0 is the identity element for vector addition: A + 0 = A
• The null vector has no fixed direction — NCERT explicitly states this. NEET has asked this as a true/false
• Two vectors are anti-parallel if they point in exactly opposite directions (angle = 180°)
• A unit vector carries only direction information — magnitude is always 1

Vector Addition & Subtraction

You can't just add vector magnitudes — 3 N east + 4 N north does NOT equal 7 N. Vectors follow geometric addition rules (triangle law, parallelogram law). Mastering these laws is essential because every force, velocity, and field problem in NEET requires adding vectors correctly.

| R | = A^2 + B^2 + 2AB

Triangle Law of Addition

The simplest vector addition rule: place the tail of the second vector at the head of the first. The resultant goes from the free tail to the free head.

To add A + B : draw A , then place the tail of B at the head of A . The resultant R goes from the tail of A to the head of B . This forms a triangle — hence the name.

• Head-to-tail rule: tail of B touches head of A
• Resultant R = A + B goes from the free tail to the free head
• Vector addition is commutative: A + B = B + A
• Vector addition is associative: ( A + B ) + C = A + ( B + C )

Parallelogram Law

When two vectors share a common tail, complete the parallelogram. The diagonal from the common point is the resultant. This gives us the master formula for the magnitude and direction of the resultant.

Place A and B with their tails at the same point. Complete the parallelogram. The diagonal from the common tail is the resultant R = A + B .

| R | = A^2 + B^2 + 2AB

= B A + B

Here is the angle between A and B , and is the angle the resultant makes with A .

• The parallelogram law and triangle law give the same result — they're equivalent geometrically
• At = 90° : R = A^2 + B^2 — this is just Pythagoras' theorem
• If | A | = | B | , the resultant bisects the angle between them
• If R^2 = A^2 + B^2 , then the vectors are perpendicular ( = 90° )

Quick recall

Two forces of 3 N and 4 N act at an angle of 90°. Find the resultant.

R = 3^2 + 4^2 = 9 + 16 = 25 = 5 N. Direction: = 4/3 → = 53° from the 3 N force.

Polygon Law

An extension of the triangle law to any number of vectors. Place them all head-to-tail; the resultant closes the polygon.

To add A + B + C + ... : place all vectors head-to-tail sequentially. The resultant goes from the tail of the first vector to the head of the last vector.

• Polygon law is just repeated application of the triangle law
• Order doesn't matter — you can arrange the vectors in any sequence (commutativity)
• A closed polygon of vectors means the vector sum is zero → equilibrium condition
• Useful in problems with 3+ forces acting on a body

Vector Subtraction

Subtracting a vector means adding its negative (same magnitude, opposite direction). This converts subtraction into addition, so the same parallelogram/triangle laws apply.

A - B = A + (- B ) . Just reverse B (flip its direction) and then add. The magnitude formula becomes:

| A - B | = A^2 + B^2 - 2AB

Note the minus sign before 2AB — because the angle between A and - B is (180° - ) , and (180° - ) = - .

• Vector subtraction is NOT commutative: A - B B - A (they are negatives of each other)
• If | A + B | = | A - B | , then = 0 → = 90° . Classic NEET question!
• Change in velocity = v _f - v _i — this is vector subtraction, not scalar

Quick recall

If | A | = | B | and the angle between them is 120°, find | A - B | .

| A - B | = A^2 + A^2 - 2A^2 120° = 2A^2 - 2A^2(- 1 2 ) = 3A^2 = A 3 .

Resolution of Vectors & Vector Products

Any vector can be broken into components along chosen axes — this is resolution, and it converts 2D vector problems into simple 1D arithmetic. Combined with dot product and cross product, these tools let you handle every vector calculation NEET throws at you.

A_x = A , A_y = A | A B = AB | | A B | = AB

Resolution into Components

Any vector in a plane can be split into two perpendicular components along the x and y axes. This is the single most useful skill in 2D physics.

A vector A making angle with the x-axis can be written as: A = A_x i + A_y j , where A_x = A and A_y = A .

A_x = A , A_y = A

A = A_x^2 + A_y^2 , = ^ -1 ( A_y A_x )

• is always measured from the axis you take cosine with — usually the x-axis
• If is from the y-axis instead, then A_x = A and A_y = A — the trig functions swap
• Components can be negative — a vector pointing left has A_x < 0
• Adding vectors by components: R = (A_x + B_x) i + (A_y + B_y) j — just add x's and y's separately

Unit Vectors i , j , k

The standard unit vectors along x, y, z axes form the coordinate system's backbone. Every vector in physics is expressed in terms of these three.

i points along the +x axis, j along +y, k along +z. Each has magnitude 1 and is dimensionless. Any vector in 3D space: A = A_x i + A_y j + A_z k .

• | i | = | j | = | k | = 1 — they are dimensionless
• i i = 1 , i j = 0 — dot product of same = 1, different = 0
• i j = k , j k = i , k i = j (cyclic order)
• For NEET: most problems are 2D, so you mainly use i and j . k appears in torque and magnetic force

Scalar (Dot) Product

The dot product gives a scalar that measures how much two vectors 'align.' It equals the projection of one vector onto the other, times the magnitude of the other.

The dot product (scalar product) of A and B is defined as:

A B = AB = A_xB_x + A_yB_y + A_zB_z

The result is a scalar (just a number — no direction). The angle is the angle between the vectors when placed tail-to-tail.

• Dot product is commutative: A B = B A
• If A B (perpendicular), then A B = 0 (since 90° = 0 )
• If A B (parallel), then A B = AB (max value)
• Work = F s = Fs — the most important physical application of dot product
• Power = F v — another NEET application

Quick recall

Find A B if A = 2 i + 3 j and B = 4 i - j .

A B = (2)(4) + (3)(-1) = 8 - 3 = 5. It's a scalar — no i or j in the answer.

Vector (Cross) Product

The cross product gives a new vector perpendicular to both input vectors. Its magnitude equals the area of the parallelogram formed by the two vectors. Torque and magnetic force both use cross products.

The cross product (vector product) of A and B is a vector perpendicular to both, with magnitude:

| A B | = AB

Direction: given by the right-hand rule. Curl the fingers of your right hand from A toward B — your thumb points along A B .

• Cross product is NOT commutative: A B = -( B A ) (anti-commutative)
• If A B , then A B = 0 (since 0° = 0 )
• If A B , then | A B | = AB (maximum)
• Physical applications: Torque = r F , magnetic force F = q( v B ) , angular momentum L = r p

Motion in a Plane — Position, Velocity & Acceleration Vectors

In 2D, every kinematic quantity (position, displacement, velocity, acceleration) becomes a vector with x and y components. The key insight: x and y motions are independent — you solve two separate 1D problems and combine the results.

v = d r dt = v_x i + v_y j | a = d v dt = a_x i + a_y j

Position Vector & Displacement in 2D

The position vector points from the origin to the object's location. Displacement is the change in position vector — a vector from the initial to the final point.

The position vector r of a point P at coordinates (x, y) is: r = x i + y j . It points from the origin O to the point P.

r = x i + y j | r | = x^2 + y^2

Displacement in 2D: r = r _2 - r _1 = (x_2 - x_1) i + (y_2 - y_1) j . It's the straight-line vector from the initial to the final position — independent of the path.

• Position vector depends on choice of origin; displacement does NOT
• Displacement is a vector — has both magnitude and direction
• | r | total path length (distance), with equality only for straight-line motion
• In 2D, we need TWO coordinates (x, y) to specify position — unlike 1D which needed just one

Velocity in 2D

In 2D, velocity has two components. The velocity vector is tangent to the path at every point. Its magnitude is the speed; its direction tells you where the object is headed at that instant.

Average velocity: v _ avg = r t . Instantaneous velocity: v = d r dt = dx dt i + dy dt j = v_x i + v_y j .

v_x = dx dt , v_y = dy dt , | v | = v_x^2 + v_y^2

• The velocity vector is always tangent to the path of motion
• Speed = | v | = v_x^2 + v_y^2 — it's the magnitude of velocity
• Direction of velocity: = v_y / v_x where is measured from x-axis
• In 2D, speed can be constant while velocity changes (direction changes) — like in circular motion

Acceleration in 2D

Acceleration in 2D is the rate of change of the velocity vector. An object accelerates whenever its velocity changes — in magnitude, direction, or both.

Average acceleration: a _ avg = v t . Instantaneous acceleration: a = d v dt = a_x i + a_y j .

a_x = dv_x dt , a_y = dv_y dt , | a | = a_x^2 + a_y^2

• Acceleration need NOT be along the direction of velocity — it can be at any angle to it
• If a v : speed changes, direction doesn't (straight-line motion)
• If a v : direction changes, speed doesn't (uniform circular motion!)
• If a is at an angle to v : both speed and direction change (general curved motion)

Projectile Motion

A projectile is any object launched into the air with only gravity acting on it (air resistance ignored). The magic trick: horizontal and vertical motions are completely independent. The horizontal component doesn't 'know' about the vertical one. Master the three key formulas (R, H, T) and you'll handle 90% of NEET projectile problems.

R = u^2 2 g | H = u^2 ^2 2g | T = 2u g

What is a Projectile?

A projectile is any body given an initial velocity and then left to move under gravity alone. No engine, no air resistance, no thrust — just gravity pulling it down while it coasts forward.

A projectile is an object that, once launched, moves under the influence of gravity alone. Assumptions: (1) air resistance is negligible, (2) g is constant in magnitude and direction (downward), (3) Earth's rotation is ignored.

• Horizontal acceleration a_x = 0 — no force acts horizontally (air resistance ignored)
• Vertical acceleration a_y = -g = -9.8 m/s ^2 — gravity pulls downward
• The path (trajectory) of a projectile is a parabola — this follows directly from the equations
• The two motions are independent: horizontal motion is uniform ( v_x = constant), vertical motion is uniformly accelerated ( a_y = -g )

Equations of Projectile Motion

Split the initial velocity into horizontal and vertical components. Apply 1D kinematics to each independently. Then combine to get the trajectory equation, range, height, and time of flight.

For a projectile launched with speed u at angle from the horizontal: u_x = u (constant) and u_y = u (changes due to gravity).

Eliminating t between the x and y equations gives the trajectory equation — the shape of the path:

y = x - g x^2 2u^2 ^2

• Trajectory is a parabola — this is the hallmark of projectile motion
• At the highest point: v_y = 0 (vertical velocity is momentarily zero), but v_x 0
• Speed at highest point = v_x = u — NOT zero! Only the vertical component is zero
• The velocity vector rotates during flight but its horizontal component stays fixed

Range, Maximum Height & Time of Flight

These three results are derived from the basic equations and used directly in most NEET problems. They assume the projectile lands at the same height it was launched from (level ground).

For a projectile launched from the ground with speed u at angle , landing on level ground:

T = 2u g (Time of Flight)

H = u^2 ^2 2g (Maximum Height)

R = u^2 2 g (Range)

• Time to reach max height = T/2 = u g (symmetry of the parabola)
• At max height: v_y = 0 , speed = u , KE is minimum (not zero!)
• R = u^2 2 g is maximized when 2 = 1 → = 45°
• Maximum range: R_ max = u^2 g (at = 45° )

Quick recall

A ball is thrown at 20 m/s at 30° to the horizontal. Find T, H, and R. (Take g = 10 m/s ^2 )

T = 2 20 30° 10 = 20 10 = 2 s. H = 400 ^2 30° 20 = 400 0.25 20 = 5 m. R = 400 60° 10 = 400 0.866 10 34.6 m.

Special Cases of Projectile Motion

Beyond the basic formulas, NEET tests tricky variations: complementary angles giving the same range, horizontal projection from a height, and the relationship between R and H. These special cases appear as direct MCQs almost every year.

R( ) = R(90° - ) | R = 4H at = 45°

Complementary Angles

For a given launch speed on level ground, angles theta and (90 - theta) give the same range. This is because sin(2*theta) = sin(2*(90 - theta)). NEET tests this almost every year.

Since R = u^2 2 g and 2 = (180° - 2 ) = 2(90° - ) , two launch angles and (90° - ) give the same range.

• Complementary angle pairs: (15°, 75°), (30°, 60°), (40°, 50°), etc.
• Same range but different heights and times: the steeper angle gives greater height and longer time of flight
• At = 45° , the two angles merge → maximum range = u^2/g
• If R_1 = R_2 for two angles _1 and _2 , then _1 + _2 = 90°

Quick recall

A projectile has range R at angle 30°. What is its range at 60° with the same speed?

Same R — because 30° and 60° are complementary ( 30° + 60° = 90° ). 60° = 120° = (180° - 60°) → same 2 .

Horizontal Projection

A special case where theta = 0: the projectile is launched horizontally from a height. There's no initial vertical velocity. Think of a ball rolling off a table edge.

When a projectile is launched horizontally from height h with speed u : u_x = u , u_y = 0 (no initial vertical velocity). This is like = 0° in the general case.

x = ut, y = 1 2 gt^2, T = 2h g , R = u 2h g

• Time to hit the ground depends ONLY on height: T = 2h/g — horizontal speed doesn't matter
• Horizontal range = R = uT = u 2h/g
• Trajectory equation: y = g 2u^2 x^2 — a parabola opening downward (taking downward as positive y )
• Speed at ground: v = u^2 + 2gh (from energy conservation or adding components)

Quick recall

A ball is thrown horizontally at 10 m/s from a 20 m high building. Find the time to hit the ground and the horizontal distance. ( g = 10 m/s ^2 )

T = 2 20/10 = 4 = 2 s. R = 10 2 = 20 m. Speed at ground: v = 100 + 400 = 500 22.4 m/s.

Useful Relations Between R, H, and T

Several elegant relationships connect R, H, and T that let you solve problems without going back to first principles. These are NEET gold.

From the three main formulas, we can derive relationships that eliminate or u . These save time in MCQs.

H = gT^2 8 (relates H and T directly — no u or needed!)

• R_ max = u^2/g at = 45° — the single most tested projectile fact
• At = 45° : R = 4H → maximum range is 4 times the max height at that angle
• = 4H R — gives the launch angle from height and range
• For horizontal projection: R = u 2h/g and there is no 'time of flight formula' involving

Uniform Circular Motion

An object moving at constant speed in a circle is constantly accelerating — toward the center. This centripetal acceleration keeps bending the velocity vector without changing its magnitude. Understanding this resolves the paradox of 'constant speed but non-zero acceleration' that confuses most students.

a_c = v^2 r = ^2 r | v = r | T = 2 r v = 2

What is Uniform Circular Motion?

Uniform circular motion means moving along a circle at constant speed. The velocity changes because its direction changes continuously, even though the magnitude (speed) stays fixed.

Uniform circular motion (UCM) = motion along a circular path with constant speed. The velocity vector has constant magnitude but continuously changing direction — it's always tangent to the circle.

• Speed = constant, but velocity ≠ constant (direction keeps changing)
• Acceleration ≠ 0 — the velocity vector changes direction, so there IS acceleration
• The acceleration points toward the center of the circle — called centripetal acceleration
• Examples: satellite in orbit, electron in a magnetic field, car turning a curve, tip of a clock hand

Angular Velocity, Period & Frequency

Angular velocity, period, and frequency describe 'how fast' something goes around. They're connected by simple relations that NEET tests frequently.

Angular velocity = rate of change of angle. Period T = time for one complete revolution. Frequency (or f ) = number of revolutions per second.

= 2 T = 2 (rad/s)

v = r = 2 r T (linear speed from angular velocity)

• is measured in rad/s — not degrees/s, not rpm
• To convert rpm to rad/s: = 2 rpm 60
• v = r — the farther from the center, the faster you move (same , larger r )
• All points on a rigid rotating body have the same but different v (depending on r )

Quick recall

The second hand of a clock has length 2 cm. Find and the speed of its tip.

T = 60 s. = 2 /60 = /30 0.105 rad/s. v = r = ( /30)(0.02) 2.09 mm/s.

Centripetal Acceleration

In UCM, the acceleration always points toward the center of the circle with magnitude v^2/r. It changes the direction of velocity without changing its magnitude.

In uniform circular motion, the acceleration is directed radially inward (toward the center) and has magnitude:

a_c = v^2 r = ^2 r = v = 4 ^2 r T^2

This is called centripetal acceleration (from Latin 'centrum' = center, 'petere' = to seek). It's not a new type of force — it's just the net inward acceleration provided by whatever force keeps the object on the circle.

• Direction: always toward the center — perpendicular to velocity at every instant
• Magnitude: constant (= v^2/r ) in UCM, since v and r are both constant
• The acceleration vector rotates along with the object — it always points inward
• Four equivalent forms: v^2/r , ^2 r , v , 4 ^2 r/T^2 — use whichever fits the given data

Quick recall

A car moves at 36 km/h around a curve of radius 25 m. Find centripetal acceleration.

v = 36 km/h = 10 m/s . a_c = v^2/r = 100/25 = 4 m/s ^2 (directed toward center of curve).

Relative Velocity in 2D

Relative velocity is vector subtraction: the velocity of A as seen by B is v_A minus v_B. In 2D, you can't just subtract magnitudes — you must subtract vector components. The classic NEET problems are rain-man and river-boat, and both reduce to drawing a velocity triangle.

v _ AB = v _A - v _B | v_ resultant = v_1^2 + v_2^2 + 2v_1 v_2

Relative Velocity as Vector Subtraction

The velocity of object A relative to object B is the vector difference of their velocities. This transforms problems into a simpler frame where one object is stationary.

The velocity of A relative to B is: v _ AB = v _A - v _B . This is what A's velocity looks like from B's perspective — as if B were stationary.

v _ AB = v _A - v _B and v _ BA = v _B - v _A = - v _ AB

• v _ AB = - v _ BA — relative velocity reverses when you swap the observer
• In 2D, use components: v_ AB,x = v_ A,x - v_ B,x and v_ AB,y = v_ A,y - v_ B,y
• Magnitude: | v _ AB | = (v_ A,x -v_ B,x )^2 + (v_ A,y -v_ B,y )^2
• Same direction → relative speed = |v_A - v_B| . Opposite direction → relative speed = v_A + v_B

Rain-Man Problem

Rain falls vertically, you walk horizontally. From your frame, rain appears to come at an angle. This is 2D relative velocity in its most intuitive form.

Rain falls vertically with velocity v _r = -v_r j (downward). A man walks with velocity v _m = v_m i (horizontal). The velocity of rain relative to the man is:

v _ rm = v _r - v _m = -v_m i - v_r j

| v _ rm | = v_m^2 + v_r^2 , = v_m v_r

The man should tilt his umbrella at angle = ^ -1 (v_m/v_r) from the vertical, in the direction of his motion.

• Umbrella tilt angle from vertical: = ^ -1 (v_m/v_r)
• If the man runs faster, he must tilt more (the rain appears more horizontal)
• If rain already falls at an angle, add that component vectorially before subtracting v _m
• This is a pure vector subtraction problem — draw the triangle to visualize

Quick recall

Rain falls vertically at 10 m/s. A man runs at 10 m/s. At what angle from the vertical does the rain appear to hit him?

= v_m/v_r = 10/10 = 1 → = 45° from vertical. He must tilt his umbrella at 45° in his direction of motion.

River-Boat Problem

A boat crosses a river while the river flows perpendicular to the desired direction. Two classic questions: (1) How to cross in minimum time? (2) How to cross with zero drift?

A river of width d flows with velocity v _r (along x). A boat can move at speed v_b relative to water. The boat wants to cross (along y). The actual velocity of the boat is v _ boat = v _ b,water + v _r .

• For zero drift, v_b > v_r is required — the boat must be faster than the river
• If v_b < v_r : zero drift is impossible — the boat will always drift downstream
• If v_b = v_r : the boat must point directly upstream at 90° to the bank, but then it has zero crossing speed — it never reaches the other side (limiting case)
• For minimum time, the boat always points perpendicular to the banks regardless of v_r

Zero drift: = v_r v_b , t = d v_b = d v_b^2 - v_r^2

Quick recall

A river is 100 m wide, flowing at 3 m/s. A boat can travel at 5 m/s in still water. Find (a) minimum time to cross, (b) time to cross with zero drift.

(a) t_ min = 100/5 = 20 s (drift = 3 20 = 60 m). (b) = 3/5 , = 4/5 . Crossing speed = 5 4/5 = 4 m/s. t = 100/4 = 25 s (zero drift).