Scalars and vectors
Physics divides every measurable quantity into two camps. A scalar is specified by a magnitude alone — a single number with its unit. The mass of a body, the temperature of a room, the time between two events, the distance covered along a winding road: all are scalars. Scalars combine by the ordinary rules of arithmetic, with the caveat that addition and subtraction make sense only for scalars sharing the same dimension.
A vector carries both magnitude and direction, and it obeys the triangle law — equivalently the parallelogram law — of addition. Displacement, velocity, acceleration, force, and momentum are all vectors. In printed text a vector is set in boldface (A); in handwritten work an arrow is drawn over the symbol. The magnitude of a vector is its absolute value, written |A| or simply A. A vector is unchanged when shifted parallel to itself, so vectors of the kind we use in kinematics are free vectors.
Two special objects deserve their names. Two vectors are equal only if they have the same magnitude and the same direction — equal length alone is not enough. The null vector 0 has zero magnitude, no defined direction, and arises naturally when a particle returns to its starting point or when a vector is multiplied by zero. Two equal-and-opposite vectors sum to a null vector: A + (−A) = 0.
Multiplication of a vector by a scalar
When a vector A is multiplied by a positive real number λ, the result λA is a vector of magnitude λ|A| pointing in the same direction as A. Multiplication by a negative scalar reverses the direction; for example, −1.5 A has magnitude 1.5 |A| and points opposite to A. The scaling factor need not be dimensionless — multiplying a velocity vector by a time interval (a scalar with units of seconds) produces a displacement vector with units of metres. This is the operation that underpins every kinematic equation that follows in this chapter.
Vector addition — the graphical method
Two vectors are added geometrically by the triangle (head-to-tail) law. Place the tail of B at the head of A; the resultant R = A + B is then drawn from the tail of A to the head of B. The same operation can be drawn as a parallelogram: bring the tails of A and B to a common origin O, complete the parallelogram by drawing lines parallel to each vector through the head of the other, and the diagonal from O represents R. The two constructions are mathematically equivalent.
Vector addition obeys two algebraic laws. It is commutative: A + B = B + A. It is associative: (A + B) + C = A + (B + C). Vector subtraction is defined through addition — A − B is shorthand for A + (−B), where −B is the same length as B but reversed in direction.
Triangle law
Head-to-tail
two vectors, one diagonal
Tail of B placed at head of A; R closes the triangle from start to finish.
Parallelogram law
Common origin
diagonal = resultant
Tails of A and B brought to O; the diagonal of the completed parallelogram is R.
Polygon law
n vectors
closure to origin
If n vectors form a closed polygon when laid head-to-tail, their sum is 0. Used in equilibrium problems.
Subtraction
A − B
add the reverse
Reverse B to get −B, then apply the triangle law: A − B = A + (−B).
Resolution of vectors
A vector can be split into two — or three — component vectors along any chosen pair of non-parallel directions in its plane. In rectangular coordinates the chosen directions are the x- and y-axes, marked by the unit vectors î and ĵ. A unit vector has magnitude 1, is dimensionless, and serves only to specify a direction; the three axial unit vectors satisfy |î| = |ĵ| = |k̂| = 1 and are mutually perpendicular.
If A makes an angle θ with the x-axis, its rectangular components are Ax = A cos θ and Ay = A sin θ. Conversely, knowing Ax and Ay recovers the magnitude as A = √(Ax² + Ay²) and the direction as tan θ = Ay/Ax. The same procedure extends to three dimensions, where the vector reads A = Axî + Ayĵ + Azk̂ and the magnitude is the square root of the sum of squares.
Ax = A cos θ, Ay = A sin θ, A = √(Ax² + Ay²)
The resolution identities — every kinematics problem starts here
Vector addition — the analytical method
The graphical method is intuitive but tedious and inexact. The analytical method uses components. For two vectors A = Axî + Ayĵ and B = Bxî + Byĵ, the resultant is R = A + B with Rx = Ax + Bx and Ry = Ay + By. Each component of the resultant is simply the sum of the corresponding components of the summands. The same idea extends to any number of vectors in any number of dimensions.
When two vectors are added at an angle θ between them, the magnitude of the resultant follows the law of cosines, R² = A² + B² + 2AB cos θ, and its direction is given by the law of sines. These two laws — derived from the parallelogram construction — are the workhorses of every "find the resultant" problem on the NEET paper.
Dot and cross products — an introduction
Two operations multiply one vector by another. The scalar (dot) product returns a scalar: A · B = AB cos θ, where θ is the angle between the two vectors. It is commutative — A · B = B · A — and yields zero whenever the vectors are perpendicular. In component form, A · B = AxBx + AyBy + AzBz. The work done by a force is the dot product of force and displacement.
The vector (cross) product returns a vector: A × B has magnitude AB sin θ and a direction perpendicular to the plane containing A and B, fixed by the right-hand rule. It is anti-commutative — A × B = −B × A — and vanishes whenever the two vectors are parallel. Torque, angular momentum, and the magnetic Lorentz force are all defined through cross products. The full machinery of products is developed in later chapters; for now it is enough to recognise that the two product types exist and what they return.
Motion in a plane — position, velocity, acceleration
Once vectors are in hand, motion in two dimensions becomes a near-mechanical exercise. The position vector of a particle in the x-y plane is r = xî + yĵ. As the particle moves, its displacement between two instants is Δr = r′ − r = Δx î + Δy ĵ. The average velocity over an interval is Δr/Δt; the instantaneous velocity is the limit as Δt → 0, written as dr/dt. At every point on the path, the instantaneous velocity is tangent to the path and pointing in the direction of motion.
In component form, vx = dx/dt and vy = dy/dt; the speed is √(vx² + vy²) and the angle with the x-axis is arctan(vy/vx). Average and instantaneous acceleration are defined analogously: a = Δv/Δt and dv/dt respectively, with components ax = dvx/dt and ay = dvy/dt. Unlike straight-line motion, velocity and acceleration in two dimensions can make any angle between 0° and 180° with each other.
Motion in a plane with constant acceleration
When the acceleration vector is constant, plane motion decomposes cleanly into two independent one-dimensional motions along perpendicular axes. The vector equations of kinematics take the familiar shape:
v = v0 + at, r = r0 + v0t + ½at²
Vector form of the kinematic equations — independent of axis choice
Resolving along the axes yields vx = v0x + axt, vy = v0y + ayt, x = x0 + v0xt + ½axt², and y = y0 + v0yt + ½ayt². The key insight, which Galileo first identified, is that the motions along the two perpendicular directions can be treated as completely independent. Whatever the x-component of velocity is doing has no bearing on what the y-component is doing — they merely happen in the same particle at the same time.
Projectile motion
The single most exam-relevant application of plane motion with constant acceleration is the projectile — any object launched into the air and acted upon only by gravity. Air resistance is neglected throughout the NCERT treatment. The motion splits into a horizontal component at constant velocity (ax = 0) and a vertical component with constant downward acceleration (ay = −g). A projectile launched with speed u at angle θ to the horizontal has initial components ux = u cos θ and uy = u sin θ.
The three signature results of projectile motion are the time of flight, the maximum height, and the horizontal range. Each is expressed in terms of the launch speed u, the launch angle θ, and the gravitational acceleration g.
T = (2u sin θ) / g, hm = (u sin θ)² / (2g), R = u² sin 2θ / g
Time of flight, maximum height, horizontal range
The range expression hides a beautiful result. Since sin 2θ reaches its largest value (unity) at 2θ = 90°, the horizontal range is maximised when the projectile is launched at θ = 45°. At this angle, Rmax = u²/g. Galileo proved a companion result: two angles that exceed and fall short of 45° by the same amount give equal ranges, because sin(90° + 2α) = sin(90° − 2α). A projectile launched at 30° therefore covers the same horizontal distance as one launched at 60°.
Uniform circular motion
The third great application of plane motion is the uniform circular motion — an object moving along a circular path at constant speed. The word "uniform" refers to the speed, not the velocity: the direction of motion is changing continuously, so the velocity vector is changing, and the object is therefore accelerating. The acceleration has a special name — centripetal acceleration — and a special direction: always along the radius, pointing inward toward the centre of the circle.
For an object moving with speed v in a circle of radius R, geometric similarity between the position-vector triangle and the velocity-vector triangle delivers the central result of this section:
ac = v² / R = ω² R = 4π² ν² R
Centripetal acceleration — three equivalent forms
Here ω = 2π/T is the angular speed (rate of change of angular position), ν = 1/T is the frequency, and T is the time period of revolution. The linear speed v is related to the angular speed by v = ωR. The magnitude of the centripetal acceleration is constant — but because the direction changes continuously, the centripetal acceleration is not a constant vector. The word centripetal was coined by Newton from Greek roots meaning "centre-seeking"; a thorough analysis was first published in 1673 by Christiaan Huygens.
Relative velocity in two dimensions
An object's velocity depends on the reference frame from which it is observed. The velocity of A relative to B is defined by the vector difference
vAB = vA − vB
Relative velocity in any reference frame
where vA and vB are measured in a common frame, typically the ground. The classic NCERT illustration is the rain-and-umbrella problem: rain falling vertically at 35 m/s combines with a horizontal wind at 12 m/s to give a resultant velocity that makes an angle of arctan(12/35) ≈ 19° with the vertical. To stay dry, the boy must tilt his umbrella by that angle in the direction the wind is blowing toward. NEET 2021 sat the same logic into a moving-car-and-falling-ball problem: a ball dropped from a car already moving at 20 m/s acquires a horizontal velocity of 20 m/s relative to the ground while gravity adds a vertical velocity — the resultant is the vector sum.
NEET PYQ Snapshot
Real NEET previous-year questions — solve before moving on.
A bullet is fired from a gun at the speed of 280 m/s in the direction 30° above the horizontal. The maximum height attained by the bullet is (g = 9.8 m/s², sin 30° = 0.5):
Answer: (4) 1000 mWhy: hm = (u sin θ)² / 2g = (280 × 0.5)² / (2 × 9.8) = (140)² / 19.6 = 19600 / 19.6 = 1000 m.
A ball is projected with a velocity 10 m/s at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be:
Answer: (1) 5√3 m/sWhy: An angle of 60° with the vertical equals 30° with the horizontal. At the highest point, vy = 0 and the speed equals the horizontal component, which stays constant: vx = u cos 30° = 10 × (√3/2) = 5√3 m/s.
A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s²)
Answer: (1) 20√2 m/s, 10 m/s²Why: At t = 4 s the car's velocity is v = 0 + 5 × 4 = 20 m/s, which the ball inherits horizontally on release. Two seconds later, vx = 20 m/s (no horizontal force) and vy = g × 2 = 20 m/s. Resultant speed = √(20² + 20²) = 20√2 m/s; acceleration is just g = 10 m/s² (free fall).
The x and y coordinates of a particle at any time are x = 5t − 2t² and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is:
Answer: (4) −4 m/s²Why: vx = dx/dt = 5 − 4t, vy = dy/dt = 10. So ax = dvx/dt = −4 m/s² and ay = 0. The acceleration vector is purely along −x with magnitude 4 m/s². Independent of t, so the answer at t = 2 s is −4 m/s².
If the magnitude of sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is:
Answer: (1) 90°Why: |A + B|² = A² + B² + 2AB cos θ and |A − B|² = A² + B² − 2AB cos θ. Setting them equal forces 4AB cos θ = 0, hence cos θ = 0 and θ = 90°. Perpendicular vectors are the only pair whose sum and difference have equal magnitude.
Expert FAQs
Questions NEET has asked from this chapter, answered straight.
At what angle of projection is the horizontal range of a projectile maximum?
Why is the path of a projectile a parabola?
What is centripetal acceleration and in which direction does it point?
Is the horizontal velocity of a projectile constant?
What is the difference between scalar and vector quantities?
What does sin 2θ in the range formula imply about complementary angles?
What is relative velocity in two dimensions?
Why is uniform circular motion an example of accelerated motion despite constant speed?
Go Deeper
Drill into the subtopics that NEET asks most often.