Motion in a Straight Line — NEET Notes

Frame of reference and the point-object idealisation

Motion is the change in the position of a body with time. Position itself has meaning only relative to a chosen reference point, called the origin, on a chosen axis. NCERT's first move is to set up a one-dimensional axis along the line of motion: positions to the right of the origin are positive, those to the left are negative. Every quantity that follows — displacement, velocity, acceleration — inherits this sign convention from the axis. Choose the axis first; assign signs second.

NCERT also makes a second, equally important assumption: bodies in motion are treated as point objects. This idealisation is justified whenever the size of the body is much smaller than the distance it traverses in the time being considered. A car covering several kilometres on a highway is fairly treated as a point; a tumbling beaker sliding off a table is not, because rotation and finite size matter on that length scale. The chapter restricts itself further to rectilinear motion — motion along a single straight line — and within rectilinear motion, mostly to the case of constant acceleration, where simple algebraic equations replace the calculus.

Position, path length, and displacement

Three distinct quantities describe where a body has been. Position is the coordinate of the body on the chosen axis at a given instant — a single number with a sign. Path length, often called distance, is the total length actually traced out by the body during a time interval; it is a scalar and can only increase or stay the same. Displacement is the vector from the initial position to the final position — in one dimension, it reduces to the signed difference Δx = x₂ − x₁. The magnitude of displacement is the shortest distance between the two positions and is bounded above by the path length.

Consider NCERT's set-piece example (Exercise 2.10): a man walks 2.5 km to the market in half an hour, finds it shut, and walks back in twenty minutes. Over the round trip (50 min), his path length is 5 km but his displacement is zero. Over the outbound leg alone (30 min), both are equal in magnitude. The two quantities collide only when motion is unidirectional — once the body reverses, path length runs ahead of displacement magnitude permanently.

Two corollaries follow. First, average speed (total path length divided by time) is always greater than or equal to the magnitude of average velocity (displacement divided by time). Equality holds only when the motion is one-directional. Second, when both quantities are taken as instantaneous limits — i.e. over an infinitesimal Δt — the body cannot reverse direction within the interval, so path length and magnitude of displacement coincide. Instantaneous speed therefore always equals the magnitude of instantaneous velocity. NCERT highlights this asymmetry in §2.2 and again in Exercise 2.11.

Average vs instantaneous velocity

Average velocity over a time interval is the displacement divided by the duration: v̄ = Δx / Δt = (x₂ − x₁) / (t₂ − t₁). It is a single number that summarises the whole interval and tells nothing about variations inside it. Instantaneous velocity, by contrast, is the velocity at a particular instant — formally the limit of the average velocity as Δt shrinks to zero. In calculus notation, v = dx/dt. On a position-time graph, the average velocity over an interval is the slope of the chord joining the two endpoints; the instantaneous velocity at any instant is the slope of the tangent at that point.

NCERT Table 2.1 makes the limit concrete for the trajectory x = 0.08 t³. Starting with Δt = 2.0 s centred at t = 4.0 s, the average velocity is calculated; then Δt is reduced to 1.0 s, 0.5 s, 0.1 s and finally 0.01 s. The ratios approach 3.84 m s⁻¹ — the value of dx/dt at t = 4.0 s. The tangent and the secant agree in the limit. Speed, finally, is the magnitude of velocity: a body with velocity −24 m s⁻¹ has speed 24 m s⁻¹. Speed is always non-negative; velocity carries a sign.

Acceleration

Acceleration is the rate of change of velocity with time. The average acceleration over an interval is ā = Δv / Δt = (v₂ − v₁) / (t₂ − t₁); the instantaneous acceleration is its limit, a = dv/dt. The SI unit is m s⁻². On a velocity-time graph, average acceleration is the slope of the chord and instantaneous acceleration is the slope of the tangent. Galileo's contribution, NCERT recalls, was to settle a long-standing dispute: should the change of velocity be tracked against distance or against time? His inclined-plane and free-fall experiments established that the rate of change with respect to time is constant for freely falling bodies, fixing the modern definition.

Acceleration can be positive, negative, or zero, and its sign behaviour mirrors that of velocity. Because velocity has both magnitude and direction, acceleration can arise from a change in speed, a change in direction, or both. The chapter restricts attention to motion along a fixed line, so direction changes only by reversal — meaning acceleration here is just the rate of change of the signed scalar velocity. Two subtle facts deserve emphasis. A particle with zero instantaneous velocity can have non-zero acceleration: at the top of its flight, a ball thrown upward is momentarily at rest yet continues to accelerate at g downward. And acceleration is a kinematic quantity defined at every smooth point in time — it does not require knowledge of force, though Newton's second law (next chapter) connects them.

Kinematic equations for uniform acceleration

For motion with constant acceleration along a straight line, three algebraic equations relate the five core kinematic variables: initial velocity u (NCERT writes v₀), final velocity v, acceleration a, time t, and displacement x (or s) measured from the starting point. They can be derived from the v-t graph (a straight line, with displacement equal to the area beneath it) or by integrating dv/dt = a once and then twice. Both routes give the same result. These equations are the workhorses of NEET kinematics: roughly half the PYQs from this chapter reduce to a single substitution.

A vital caveat: these equations are valid only when acceleration is constant in both magnitude and direction during the interval. If a changes mid-motion — for instance, a vehicle that first accelerates and then decelerates — the equations must be applied piecewise to each sub-interval. The general definitions v = dx/dt and a = dv/dt are exact for every motion; the algebraic kinematic equations are not.

Reading the graphs

Position-time and velocity-time graphs encode every kinematic quantity geometrically. On the x-t graph: the slope of the curve at any instant is the velocity, a straight horizontal line implies rest, a straight inclined line implies uniform velocity, and a parabola implies uniform acceleration. On the v-t graph: the slope is the acceleration and the area between the curve and the time axis is the displacement. A v-t graph parallel to the time axis means uniform velocity; a v-t graph that is a straight inclined line means uniform acceleration. NCERT's Fig. 2.3 organises four cases: positive-positive, positive-negative, negative-negative, and a case where the body reverses direction at some t₁. Reading these graphs is a small but recurring chunk of NEET asks — NEET 2022's question on displacement-time slopes (angles 30° and 45°) reduces to one line: slope = tan θ = v.

Relative velocity in one dimension

All velocities are relative. To specify "the velocity of B" without saying "with respect to what" is to leave a quantity half-defined. For two bodies A and B moving along the same straight line with velocities v_A and v_B (each measured with respect to the ground), the velocity of B relative to A is

v_BA  =  v_B − v_A

Relative velocity in one dimension

Two facts follow. When A and B move in the same direction, |v_BA| is small — the slow-overtake situation. When they move in opposite directions, one velocity is negative, so the magnitudes add and |v_BA| is large — this is why a passing train in the opposite direction appears to streak past while a train alongside seems to crawl. NIOS illustrates this with two trains moving north-to-south at 60 km h⁻¹ and south-to-north at 70 km h⁻¹: relative velocity = 70 − (−60) = 130 km h⁻¹. Same logic, with one well-placed minus sign.

Relative velocity simplifies two-body problems into one-body problems: switch into the rest frame of A, and B's motion is governed by v_BA alone. NCERT Exercise 2.14 — police-van at 30 km h⁻¹ firing a bullet (muzzle 150 m s⁻¹) at a thief's car at 192 km h⁻¹ — is solved cleanly with this trick: convert both to the same units, add the bullet's muzzle speed to the van's speed (the bullet's ground velocity) and subtract the thief's velocity to get the bullet's velocity relative to the thief's car. The relevant damage-causing quantity is the relative speed at impact, not the bullet's ground speed.

Motion under gravity — free fall

A body released near the Earth's surface and subject only to gravity is said to be in free fall. Neglecting air resistance, the magnitude of acceleration is the same for every body irrespective of mass — Galileo's foundational result. The magnitude is denoted g and equals approximately 9.8 m s⁻² at the Earth's surface, treated as constant whenever the height of fall is small compared to the Earth's radius. Free fall is therefore a textbook case of uniform-acceleration motion, and the kinematic equations apply directly with a = ±g, the sign chosen by the convention for the axis.

NCERT adopts upward-positive throughout. Then a = −g. For a body released from rest (v₀ = 0) at y = 0, the equations of motion become v = −gt, y = −½gt², and v² = −2gy. The downward motion produces negative values of v and y, consistent with the chosen sign. A body thrown vertically upward decelerates at −g during ascent, reaches v = 0 at the apex, and then accelerates downward at the same −g — the acceleration is unchanged throughout. The total time of flight for a body thrown upward with initial speed u₀ and returning to the launch level is t = 2u₀/g, and the maximum height attained is H = u₀²/(2g) — both immediate consequences of the kinematic equations.

Galileo's law of odd numbers deserves a separate look because it appears verbatim on NEET. Starting from y = ½gt² for a body falling from rest, the distance covered between t = (n−1)τ and t = nτ is s_n = ½g(2n − 1)τ². For τ fixed, the distances form the ratio 1 : 3 : 5 : 7 : 9 … — the consecutive odd natural numbers. NEET 2022's Q.5 asks exactly this ratio; the answer is option (2). NEET 2021 generalises it to s_n / s_{n+1} = (2n − 1) / (2n + 1), tested on a block sliding down a frictionless incline (a = g sin θ), which behaves kinematically just like free fall with g replaced by g sin θ.