1🗓️ How to Prepare
NEET weightage: 1–2 questions, and the chapter is the foundation for Laws of Motion, Work-Energy and every graph question in the paper. Most questions are graph-reading or a direct kinematic-equation substitution — 45-second marks once the sign convention is automatic.
Day 1 — Language
Distance vs displacement, average vs instantaneous, sign convention. Watch Animation 1 until x-t / v-t feel obvious.
Day 2 — Equations
Three kinematic equations + nth-second formula. 12 substitution problems with signs.
Day 3 — Free Fall
Ball up/down (Animation 2), Galileo's odd numbers, drops from height. 10 problems.
Day 4 — Graphs + Relative
All graph shapes below, area/slope rules, catch-up problems (Animation 3).
Day 5 — Mock 🧪
Timed 20-question chapter test. Wrong answers → error notes.
2🧠 Concepts
1 · Position, distance, displacement foundation
Distance = actual path length (scalar, never negative, never decreases). Displacement = change in position, x₂ − x₁ (vector; in 1-D just a signed number). Displacement can be zero or negative; distance ≥ |displacement| always.
2 · Average vs instantaneous
Average velocity = displacement/time = (x₂−x₁)/(t₂−t₁). Average speed = distance/time. Instantaneous velocity = dx/dt — the slope of the x-t curve at that instant. Instantaneous speed = |instantaneous velocity| (they are always equal in magnitude at an instant — the average versions are not).
3 · Acceleration
a = dv/dt = slope of the v-t graph. Negative a does not automatically mean slowing down — the body slows only when a and v have opposite signs. Speeding up in the −x direction also has negative a.
4 · The three kinematic equations constant a only
v = u + at · s = ut + ½at² · v² = u² + 2as. Valid only for constant acceleration. Each hides one variable: first hides s, second hides v, third hides t — pick the one whose missing variable you don't need.
5 · Free fall
Constant a = g downward (≈ 9.8 m s⁻²), independent of mass. Take up as +: a = −g throughout the flight — including at the top, where v = 0 but a is still g. Time up = time down; speed of return = speed of launch (no air resistance).
6 · Graph dictionary
- x-t: slope = velocity. Straight line → uniform velocity; parabola → uniform acceleration; horizontal → rest.
- v-t: slope = acceleration; area under curve = displacement (area below the t-axis counts negative).
- a-t: area = change in velocity.
7 · Relative velocity (1-D)
vAB = vA − vB. Same direction → magnitudes subtract; opposite directions → magnitudes add. Catch-up time = initial gap ÷ relative speed.
8 · When equations fail → calculus
If a depends on time (a = f(t)), integrate: v = ∫a dt, x = ∫v dt. The three SUVAT equations are the special case where the integrals are trivial.
3🎬 Animations — watch the graphs move
Animation 1 · One motion, three graphs
Set u and a, press play. The dot moves on the track while x-t and v-t are drawn in real time — the x-t curve is the slope-story of the v-t line.
Animation 2 · Ball thrown straight up
The v-t graph of a vertical throw is one straight line of slope −g. At the top v = 0 — but the line's slope never changes: gravity does not pause.
Animation 3 · Catch-up — relative velocity
A police van chases a bike with a 60 m head start. The gap closes at the relative speed — the ground speeds don't matter separately.
4🧮 Formula Bank
| Situation | Formula | Note |
|---|---|---|
| Velocity after time t | v = u + at | no s |
| Displacement | s = ut + ½at² | no v |
| Velocity–displacement | v² = u² + 2as | no t |
| Displacement in nᵗʰ second | sₙ = u + a(2n−1)/2 | units are m (per that second) |
| Average velocity (const a) | v̄ = (u+v)/2 = u + at/2 | only for constant a |
| Equal distances at v₁ then v₂ | v̄ = 2v₁v₂/(v₁+v₂) | harmonic mean — never (v₁+v₂)/2 |
| Equal times at v₁ then v₂ | v̄ = (v₁+v₂)/2 | arithmetic mean |
| Free fall from rest | v = gt · h = ½gt² · v² = 2gh | downward + |
| Time of flight (throw up, return to start) | T = 2u/g · Hmax = u²/2g | tup = u/g |
| Galileo's odd-number rule | distances in successive seconds → 1 : 3 : 5 : 7… | from rest, constant a |
| Relative velocity | vAB = vA − vB | 1-D: signed numbers |
| Stopping distance | s = u²/2a | double u → 4× distance |
| Variable acceleration | v = ∫a dt · x = ∫v dt · a = v dv/dx | use when a is not constant |
5📋 Formula Sheet — one glance before the exam
6✍️ Derivations
D1 · v = u + at (from the definition)
a = dv/dt is constant → ∫uvdv = a∫0tdt → v − u = at. Graphically: v-t is a straight line starting at u with slope a.
D2 · s = ut + ½at² (area under v-t)
Displacement = area under the v-t line = rectangle (u·t) + triangle (½·t·at) = ut + ½at². This "area" picture is worth more in NEET than the calculus.
D3 · v² = u² + 2as (eliminate t)
From D1, t = (v−u)/a. Substitute into s = [(u+v)/2]·t (average velocity × time): s = (v+u)(v−u)/2a → v² = u² + 2as.
D4 · Distance in the nᵗʰ second
sₙ = s(n) − s(n−1) = [un + ½an²] − [u(n−1) + ½a(n−1)²] = u + a(2n−1)/2. From rest (u=0): s₁:s₂:s₃ = 1:3:5 — Galileo's odd numbers.
D5 · a = v dv/dx (the chain-rule form)
a = dv/dt = (dv/dx)(dx/dt) = v·dv/dx. Use when acceleration is given as a function of position.
7🧩 Problem Types — the 10 ways NEET asks this chapter
T1 · Direct SUVAT substitution
Given three of u, v, a, s, t → pick the equation missing the fourth. Signs first, numbers second.
T2 · Ball thrown up / dropped
Set up as + once, then a = −g everywhere. Height, time of flight, speed at a level, crossing times.
T3 · nᵗʰ-second distance
"Distance in the 5ᵗʰ second" → sₙ formula, not s(5). From rest the answers sit in ratio 1:3:5:7.
T4 · Average speed over two legs
Equal distances → harmonic mean; equal times → arithmetic mean. NEET alternates these deliberately.
T5 · Graph reading
Given x-t or v-t: find velocity/acceleration/displacement/distance. Slope for rates, area for totals; area below axis is negative displacement but positive distance.
T6 · Graph shape matching
Match x-t ↔ v-t ↔ a-t for the same motion. Differentiate going down the ladder (x→v→a).
T7 · Catch-up / overtake
Relative speed closes the gap: t = gap / (v₂ − v₁). Trains passing: use sum of lengths as the "gap".
T8 · Stopping distance & reaction time
Total = reaction distance (u·tr, no braking) + braking distance (u²/2a).
T9 · Variable acceleration (calculus)
a(t) given → integrate; v(x) given → a = v dv/dx. Check limits match the initial conditions.
T10 · Distance vs displacement bookkeeping
Motion that reverses direction: split at the turning point (v = 0), add magnitudes for distance, add signed values for displacement.
8📈 Graphs
G1 · The six x-t shapes to recognise on sight
Slope = velocity. A vertical segment would mean two positions at one instant — no x-t graph may ever be vertical (or double back in time).
G2 · v-t: area = displacement, below the axis counts negative
Where the v-t line crosses the axis, the body reverses. Displacement subtracts the red region; distance adds it.
G3 · Free fall v-t — one unbroken line of slope −g
The most-tested idea in the chapter: nothing special happens to the graph at the top. Slope (−g) is constant from launch to catch.
9📏 Units & Dimensions
| Quantity | SI unit | Dimensional formula |
|---|---|---|
| Position, displacement, distance | m | [M⁰LT⁰] |
| Velocity, speed | m s⁻¹ | [M⁰LT⁻¹] |
| Acceleration | m s⁻² | [M⁰LT⁻²] |
| Slope of x-t graph | m s⁻¹ | velocity |
| Slope of v-t graph | m s⁻² | acceleration |
| Area under v-t graph | m | displacement |
| Area under a-t graph | m s⁻¹ | change in velocity |
| Jerk (da/dt) | m s⁻³ | [M⁰LT⁻³] |
10🔢 Standard Values
| Value | Remember |
|---|---|
| g (use in problems) | 9.8 m s⁻² — take 10 unless told otherwise |
| 1 km h⁻¹ → m s⁻¹ | × 5/18 (36 km/h = 10 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s) |
| Free-fall distances from rest (g = 10) | 5, 20, 45, 80 m after 1, 2, 3, 4 s |
| Successive-second distances from rest | 5, 15, 25, 35 m — the 1:3:5:7 ladder |
| Typical reaction time | ≈ 0.2 s (used in stopping-distance questions) |
11⚡ Shortcuts
List what you have and what you want; use the SUVAT equation that omits the variable you neither have nor want. Zero algebra wasted.
u = 20 → top in 2 s, H = 20 m, flight 4 s. u = 30 → 3 s, 45 m, 6 s. Pattern: t = u/10, H = u²/20.
Half distance at v₁, half at v₂ → v̄ = 2v₁v₂/(v₁+v₂). 40 & 60 → 48, never 50. If the options contain both, the arithmetic mean is the trap.
Ratio of 1ˢᵗ, 2ⁿᵈ, 3ʳᵈ-second distances = 1:3:5. "Distance in 3ʳᵈ s ÷ distance in 1ˢᵗ s" = 5, done in 5 seconds.
Double the speed → 4× the braking distance; triple → 9×. Same brakes assumed.
Fraction of height fallen in the last second of an n-second drop = (2n−1)/n². For n = 3: 5/9 of the height.
12⚠️ Traps
Unless the times are equal. For equal distances it's the harmonic mean — always less than the arithmetic mean.
Top of a vertical throw: v = 0, a = g. A body can be momentarily at rest while accelerating.
Slowing needs a and v of opposite signs. A car speeding up while moving in −x has negative acceleration.
The units look like m/s but the formula is the displacement in that one second — dimensionally consistent because "1 second" is multiplied in silently.
Displacement = signed area; distance = total unsigned area. Questions that give a v-t crossing the axis are testing exactly this.
The parabola in an x-t graph is a shape in the position-time plane, not a trajectory in space. Motion is still along one straight line.
"Dropped" → u = 0 relative to the ground only if the support is at rest. Dropped from a rising balloon → u = balloon's velocity, upward.
v² = u² + 2as already assumes constant a. For variable a, none of SUVAT applies — integrate or use a = v dv/dx.
Catch-up: set x₁(t) = x₂(t). Equal speeds is the moment the gap stops changing — often the moment of maximum gap.
Ball up at u comes back past the launch point at u (downward). Energy symmetry; air resistance breaks it (then return is slower).
13🧵 Mnemonics
S-displacement, U-initial, V-final, A-acceleration, T-time. Each equation drops exactly one letter: "no S", "no V", "no T", pick your poison.
x → v → a: going down take slopes (differentiate). a → v → x: coming up take areas (integrate).
From rest, each successive second covers the next odd multiple of the first second's distance.
Half-distance → Harmonic mean. Equal Time → ariThmetic mean.
First sentence of every solution: "taking up/right as positive". Half the chapter's wrong answers die right there.
14🛠️ Directions & Graphs Repair — fixing the classic misreads
Before answering any graph question, write the axis labels in the margin. A straight rising x-t line means constant velocity, zero acceleration — not "accelerating because it goes up".
Choose + direction at the start, apply it to u, a, s and the answer. Switching midway (up + for the rise, down + for the fall) is the classic self-inflicted error.
Word problems give magnitudes; your equations need signs. Deceleration opposes v — insert the minus yourself.
Trace the v-t curve, mark axis crossings, treat each lobe separately, then combine per what is asked.
In catch-up problems drawn on one line, faster-behind catches slower-ahead only if the relative velocity points from chaser to target. If not, the gap grows — "never meets" is a legitimate NEET option.
When a question asks for acceleration "at the highest point" of a throw, the answer is g downward — the velocity is what vanishes, not gravity.
15🚨 Exceptions — the odd ones out NEET loves
Equality holds only for one-way straight-line motion. Any reversal breaks it.
A round trip has average velocity zero but nonzero average speed: the extreme case.
The average pair can differ; the instantaneous pair cannot — a favourite true/false statement.
Zero acceleration is not rest. Conversely a body at rest can have a ≠ 0 (ball at the top).
Air resistance, position-dependent forces, a(t) given — integrate instead. The equations are a special case, not laws.
Crossing means they meet at that instant; equal slopes would mean equal velocities — different questions entirely.
a shrinking but same sign as v → still speeding up, just more gently. "Decreasing acceleration = slowing" is false.