NEET 2027 · Priority 2 · Non-uniform acceleration

1 · The concept in plain language

The one idea

The three kinematic equations are not laws. They are what you get when you integrate a constant acceleration. When a is not constant, throw them away and integrate for yourself.

NCERT says this plainly in Points to Ponder 6: the definitions v = dx/dt and a = dv/dt are exact for every motion, while the kinematic equations hold only for uniformly accelerated motion. Almost every hard kinematics question in NEET is built on candidates forgetting that second half of the sentence.

Three routes, chosen by one question

When a varies, the only decision you have to make is: a is given as a function of what? The answer picks your route, and each route is a single separable integral.

animated Match the variable to the route. Time → integrate dv = a dt. Position → integrate v dv = a dx. Velocity → separate as dv/a = dt for time, or v dv/a = dx for position.

The identity that makes route 2 possible

Everything on the right-hand branches rests on one application of the chain rule:

a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx)

It is worth pausing on how much this buys you. It converts an acceleration expressed in position into something you can integrate without ever knowing x(t). And run backwards, it says that d(v²)/dx = 2a — so if a question hands you v² as a function of x, the acceleration is just half the derivative.

A worked case: the square-root curve

Take v = √(40x). It looks like non-uniform motion — the velocity curve bends. But apply the identity and the bending cancels exactly:

animated As the particle advances, v rises and dv/dx falls, in exact proportion. Their product — the acceleration — stays pinned at 20 m/s². Any relation of the form v² = αx + β is uniformly accelerated motion wearing a disguise.

A worked case: resistive retardation

The other family you must recognise on sight is a = −kv — the drag law for slow motion through a fluid. Because the rate of change of v is proportional to v itself, the answer is an exponential, and it has a property that catches people out:

animated v = 20e^(−0.5t). The velocity halves every 1.39 s and never quite reaches zero — so strictly the body never stops. Yet the area under this curve is finite: it travels only 40 m in total. ‘Never stops’ and ‘stops after 40 m’ are both correct statements about the same motion.

Why this is Priority 2 and not Priority 3

The calculations here are rarer than kinematic-equation questions, but they are almost always the hardest item on the paper when they do appear — and the conceptual version (‘which of these relations is always valid?’) shows up far more often than the numerical one. The conceptual mark is cheap; take it.

2 · Formula sheet

The two definitions

v = dx/dt ; a = dv/dt

Exact for every motion, uniform or not. Never fail.

The chain-rule identity

a = dv/dt = v (dv/dx)

The single most useful line in this topic. Converts between time and position pictures.

Route 1 — a = f(t)

v = u + ∫a dt ; x = x₀ + ∫v dt

Integrate once for v, twice for x.

Route 2 — a = f(x)

v² = u² + 2∫a dx

The general form of the third kinematic equation.

Route 3 — a = f(v), for t

t = ∫_u^v dv/a

Gives t(v); invert if v(t) is wanted.

Route 3 — a = f(v), for x

x = ∫_u^v v dv/a

Gives x(v) directly.

Differentiating v²

d(v²)/dx = 2a

So for v² = f(x), the acceleration is half the derivative.

a = −kv

v = u e^−kt ; v = u − kx ; x_total = u/k

Never stops in time, but travels only a finite distance.

a = −kv²

1/v = 1/u + kt ; v = u e^−kx

The reciprocal of v grows linearly. Neither time nor distance is finite.

a = −k√v

t_stop = 2√u/k

This one does stop in finite time.

v = k√x

a = k²/2 — constant

Uniform acceleration from rest, in disguise.

v² = αx + β

a = α/2 — constant

Same disguise, general form.

a ∝ −x

simple harmonic motion

Preview of Chapter 13.

Areas — watch the axis

∫a dt = Δv ; ∫a dx = Δ(v²/2)

Same-looking graph, different meanings.

Maxima

v is maximum where a = 0 ; x is maximum where v = 0

One rung up the ladder each time.

3 · Exceptions and traps

Trap — substituting a variable a into a constant-a equation

Writing v² = u² + 2ax with a = 4x is the single most common error in this topic. That equation was derived by assuming a is constant; substituting a function of x into it is not a shortcut, it is a contradiction.

Trap — a = 0 does not mean at rest

Zero acceleration means the velocity has stopped changing, not that it is zero. At the instant a = 0 the speed is usually at its maximum or minimum — the one moment it is definitely not zero.

Trap — which area, which axis

Area under a–t gives the change in velocity. Area under a–x gives the change in v²/2. The two graphs can look identical. Read the horizontal axis label first.

Trap — t or x?

For a = f(v) there are two different integrals: ∫dv/a gives time, ∫v dv/a gives distance. Questions are set in pairs where the two answers look similar, such as ue^−kt and ue^−kx.

Exception — ‘never stops’ vs ‘travels a finite distance’

Under a = −kv the body never reaches exactly zero speed, yet it covers only u/k in total. Both statements are true. Under a = −k√v it does stop, in time 2√u/k. The stopping behaviour depends on the power of v.

Exception — return time is not double the turning time

The symmetry ‘time up = time down’ is a property of constant acceleration alone. With a time-dependent acceleration the two legs take different times.

4 · Numbers to remember

Given	Result
a = f(t)	integrate dv = a dt
a = f(x)	integrate v dv = a dx
a = f(v)	dv/a = dt or v dv/a = dx
v² = αx + β	a = α/2, constant
v = k√x	a = k²/2, constant
a = −kv	v = ue^−kt, total distance u/k
a = −kv²	1/v = 1/u + kt
a = −k√v	stops at t = 2√u/k
a = −ω²x	simple harmonic motion
x = t³ − 6t² + 3t + 4	a = 0 at t = 2 s, where v = −9 m s⁻¹

5 · Scientists NEET names

Isaac Newton (1642–1727) and Gottfried Wilhelm Leibniz (1646–1716)

Calculus was developed independently by both. The notation you use here is Leibniz’s — dx/dt, the integral sign, the whole apparatus of separable differentials. Newton’s dot notation survives mainly in mechanics. NCERT introduces the derivative in this chapter precisely because non-uniform motion cannot be handled without it.

George Gabriel Stokes (1819–1903)

The retardation law a ∝ −v that generates the exponential decay above is Stokes’ drag, valid for slow motion through a viscous fluid. At higher speeds drag goes as v² instead — which is why both forms appear in question papers. You will meet Stokes again in Mechanical Properties of Fluids.

6 · Fifteen worked questions

Built on the method of NCERT Example 2.2 and Points to Ponder 6. No previous-year attributions are claimed here.

Q1NEET-style

A particle starts from rest and moves with an acceleration a = 3t² m s⁻². Its velocity at t = 2 s is

(a) 4 m s⁻¹
(b) 12 m s⁻¹
(c) 24 m s⁻¹
(d) 8 m s⁻¹

Given

a = 3t², u = 0, t = 2 s

Asked

velocity at t = 2 s

Concept to use

The acceleration is a function of time, so integrate dv = a dt. The kinematic equations are unavailable — they assume a is constant.

Formula or rule

v = u + ∫₀^t a dt

Baby steps

v = 0 + ∫₀² 3t² dt.
= [t³]₀².
= 8 − 0 = 8 m s⁻¹.

Answer

8 m s⁻¹

Shortcut

Integrating a power of t just raises the index and divides: 3t² → t³. Read the answer straight off.

Trap

Substituting t = 2 into v = u + at to get 0 + 3(4)(2) = 24. That equation needs constant a; here a itself grows from 0 to 12 over the interval. The option 24 is there for exactly this reason.

Q2NEET-style

The acceleration of a particle is a = 6t + 4 m s⁻² and its initial velocity is 2 m s⁻¹. The velocity at t = 2 s is

(a) 14 m s⁻¹
(b) 18 m s⁻¹
(c) 22 m s⁻¹
(d) 32 m s⁻¹

Given

a = 6t + 4, u = 2 m s⁻¹, t = 2 s

Asked

velocity at t = 2 s

Concept to use

Same route as before — a is a function of t — but now the constant of integration is the initial velocity, which is not zero.

Formula or rule

v = u + ∫ a dt

Baby steps

∫(6t + 4) dt = 3t² + 4t + C.
At t = 0, v = 2, so C = 2.
v = 3t² + 4t + 2.
At t = 2: 12 + 8 + 2 = 22 m s⁻¹.

Answer

22 m s⁻¹

Shortcut

Use definite limits and the constant handles itself: v − 2 = [3t² + 4t]₀² = 20.

Trap

Dropping the initial velocity, giving 20. Whenever the body does not start from rest, the constant of integration is a real number you must carry.

Q3NEET-style

The position of a particle is x = t³ − 6t² + 3t + 4 metres. Its velocity at the instant when its acceleration is zero is

(a) −9 m s⁻¹
(b) −12 m s⁻¹
(c) 3 m s⁻¹
(d) 9 m s⁻¹

Given

x = t³ − 6t² + 3t + 4

Asked

velocity at the instant a = 0

Concept to use

Differentiate twice. Set the second derivative to zero to find the instant, then substitute that instant into the first derivative.

Formula or rule

v = dx/dt ; a = d²x/dt²

Baby steps

v = 3t² − 12t + 3.
a = 6t − 12.
a = 0 ⇒ t = 2 s.
v(2) = 3(4) − 12(2) + 3 = 12 − 24 + 3 = −9 m s⁻¹.

Answer

−9 m s⁻¹

Shortcut

For a cubic, a = 0 always lands at the vertex of the velocity parabola — so the answer is the minimum velocity of the motion. Here that minimum is negative, meaning the particle is moving backwards fastest at that instant.

Trap

Setting v = 0 instead of a = 0, or stopping at t = 2 and forgetting to substitute. Note that zero acceleration does not mean the particle is at rest — it is moving at 9 m s⁻¹ here.

Q4NEET-style

A particle moves so that its velocity varies with position as v = k√x, where k is a constant. The acceleration of the particle is

(a) zero
(b) proportional to x
(c) proportional to 1/√x
(d) k²/2 — a constant

Given

v = k√x

Asked

the acceleration

Concept to use

a is wanted from a velocity–position relation, so use the chain-rule form a = v·dv/dx. Differentiating with respect to t directly would need x(t), which you do not have.

Formula or rule

a = v (dv/dx)

Baby steps

dv/dx = k/(2√x).
a = v (dv/dx) = k√x × k/(2√x).
The √x cancels: a = k²/2, a constant.

Answer

k²/2 — a constant

Shortcut

v = k√x means v² = k²x. Compare with v² = u² + 2ax: reading off, 2a = k². Any relation of the form v² = αx + β is uniform acceleration in disguise.

Trap

Assuming that because v depends on x, the motion must be non-uniform. It is uniformly accelerated motion starting from rest — the square-root shape is what constant acceleration looks like on a v–x plot.

Q5NEET-style

A body starting from rest moves along a straight line with an acceleration that depends on position as a = 4x (SI units). Its speed when it has travelled 3 m is

(a) 3√2 m s⁻¹
(b) 12 m s⁻¹
(c) 6 m s⁻¹
(d) 18 m s⁻¹

Given

a = 4x, u = 0, x = 3 m

Asked

speed at x = 3 m

Concept to use

a is a function of position, so use v dv = a dx and integrate both sides between matching limits.

Formula or rule

∫v dv = ∫a dx ⇒ v²/2 = ∫a dx

Baby steps

∫₀^v v dv = ∫₀³ 4x dx.
v²/2 = [2x²]₀³ = 18.
v² = 36 ⇒ v = 6 m s⁻¹.

Answer

6 m s⁻¹

Shortcut

v² = u² + 2∫a dx is the general version of the third kinematic equation. When a is constant the integral is ax and you recover v² = u² + 2ax exactly.

Trap

Using v² = u² + 2ax with a = 4x, which gives v² = 2(4×3)(3) = 72. That formula already assumes a is constant — you cannot substitute a varying a into it. The area under the a–x graph is what matters.

Q6NEET-style

A particle moving in a straight line experiences a retardation proportional to its velocity: a = −kv. If its initial speed is u, the total distance it travels before coming to rest is

(a) u²/2k
(b) u/k
(c) 2u/k
(d) infinite — it never stops

Given

a = −kv, initial speed u

Asked

total distance travelled

Concept to use

For distance from a velocity-dependent acceleration, use the position form v dv/dx = a. The v cancels, leaving a beautifully simple integral.

Formula or rule

v (dv/dx) = −kv ⇒ dv = −k dx

Baby steps

Cancel v on both sides: dv = −k dx.
Integrate from v = u to v = 0, and x = 0 to x = s: (0 − u) = −ks.
s = u/k.

Answer

u/k

Shortcut

Because dv/dx = −k is constant, the v–x graph is a straight line from (0, u) to (u/k, 0). Read the intercept off the line rather than integrating.

Trap

Confusing this with the time behaviour. In time, v = ue^−kt never quite reaches zero, so the body technically never stops — but the distance converges to the finite value u/k. Both statements are true and NEET plays them against each other.

Q7NEET-style

For a particle with a = −kv and initial speed u, the velocity at time t is

(a) u − kt
(b) u e^−kt
(c) u/(1 + ukt)
(d) u e^−kx

Given

a = −kv, v = u at t = 0

Asked

v as a function of t

Concept to use

a is a function of v, so separate the variables as dv/a = dt and integrate. A 1/v integrand always produces a logarithm, which exponentiates.

Formula or rule

dv/dt = −kv ⇒ ∫dv/v = −k∫dt

Baby steps

∫_u^v dv/v = −k∫₀^t dt.
ln(v/u) = −kt.
v = u e^−kt.

Answer

u e^−kt

Shortcut

Whenever the rate of change of a quantity is proportional to the quantity itself, the answer is an exponential. The same equation governs radioactive decay and RC discharge — recognise the shape and skip the integration.

Trap

Picking u e^−kx, which is the correct expression for the same motion but as a function of position, not time. Both appear in the options because both are genuine results for this system — read which variable is asked for.

Q8NEET-style

A particle is retarded according to a = −kv², starting with speed u. Its speed after time t is

(a) u − kt
(b) u e^−kt
(c) u/(1 + ukt)
(d) u/(1 + kt)

Given

a = −kv², v = u at t = 0

Asked

v as a function of t

Concept to use

Again a is a function of v, so separate and integrate — but the integrand is now 1/v², which gives a reciprocal rather than a logarithm.

Formula or rule

dv/v² = −k dt

Baby steps

∫_u^v v⁻² dv = −k∫₀^t dt.
−1/v + 1/u = −kt.
1/v = 1/u + kt = (1 + ukt)/u.
v = u/(1 + ukt).

Answer

u/(1 + ukt)

Shortcut

The reciprocal of the velocity grows linearly: 1/v = 1/u + kt. That is the cleanest form to remember, and it is what most questions actually test.

Trap

Choosing u/(1 + kt), which is dimensionally wrong — kt must be dimensionless in that expression, but k here has units of m⁻¹, so it is ukt that is dimensionless. Dimensional analysis eliminates two options instantly.

Q9NEET-style

A body moving with velocity u is subject to a retardation a = −k√v. It comes to rest after a time

(a) u/k
(b) 2√u/k
(c) √u/2k
(d) u²/2k

Given

a = −k√v, initial speed u, final speed 0

Asked

time taken to stop

Concept to use

a is a function of v, so integrate dv/a = dt. Here the integrand is v^−1/2, which integrates to a square root.

Formula or rule

dv/√v = −k dt

Baby steps

∫_u⁰ v^−1/2 dv = −k∫₀^t dt.
[2√v]_u⁰ = −kt.
−2√u = −kt.
t = 2√u/k.

Answer

2√u/k

Shortcut

Unlike the −kv case, this body genuinely stops in finite time, because the retardation does not vanish as fast as the velocity does. Compare the three laws: √v stops in finite time, v and v² do not.

Trap

Setting the limits the wrong way round and losing the sign. Integrate from u down to 0 on the left and 0 up to t on the right — the two minus signs then cancel correctly.

Q10NEET-style

A particle’s velocity is v = 3t² − 2t + 1 m s⁻¹. The displacement between t = 0 and t = 2 s is

(a) 4 m
(b) 6 m
(c) 9 m
(d) 12 m

Given

v = 3t² − 2t + 1, from t = 0 to t = 2 s

Asked

displacement

Concept to use

Displacement is the integral of velocity over time — the area under the v–t curve. Since the discriminant of this quadratic is negative, v is never zero, so displacement and distance agree.

Formula or rule

Δx = ∫_t₁^t₂ v dt

Baby steps

∫(3t² − 2t + 1) dt = t³ − t² + t.
At t = 2: 8 − 4 + 2 = 6.
At t = 0: 0.
Δx = 6 m.

Answer

6 m

Shortcut

Check the discriminant before splitting the integral. Here 4 − 12 < 0, so v never changes sign and one integration is enough — no turning point to hunt for.

Trap

Multiplying v(2) by 2 to get 18, treating the velocity as if it were constant. Also watch for the opposite error: splitting an integral at a ‘turning point’ that does not exist.

Q11NEET-style

The acceleration of a particle is a = 2 − 0.5t m s⁻² and it starts from rest. Its velocity is maximum at

(a) t = 4 s
(b) t = 2 s
(c) t = 8 s
(d) the velocity keeps increasing

Given

a = 2 − 0.5t, u = 0

Asked

instant of maximum velocity

Concept to use

Velocity is maximum where its rate of change — the acceleration — passes through zero and turns negative. No integration is needed to find the instant.

Formula or rule

v is maximum when a = dv/dt = 0

Baby steps

2 − 0.5t = 0.
t = 4 s.
Before that a > 0 so v is rising; after it a < 0 so v falls. It is a genuine maximum.
(For the record, v_max = 2(4) − 0.25(16) = 4 m s⁻¹.)

Answer

t = 4 s

Shortcut

Maximum velocity ⇔ zero acceleration, exactly as maximum displacement ⇔ zero velocity. The pattern repeats one rung up the ladder each time.

Trap

Integrating first and then differentiating — correct, but two unnecessary steps. Also note a = 0 marks a maximum only because a is decreasing through zero; if a were increasing through zero it would mark a minimum.

Q12NEET-style

For a particle in one-dimensional motion, which relation is always valid, whether or not the acceleration is constant?

(a) a = v (dv/dx)
(b) v = u + at
(c) x = ut + ½at²
(d) v² = u² + 2ax

Given

four kinematic relations

Asked

the one that holds for any motion

Concept to use

The three kinematic equations are derived by integrating a constant a. The chain-rule identity is not derived from anything — it is just the chain rule, so it is always true.

Formula or rule

a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx)

Baby steps

dv/dt = (dv/dx)·(dx/dt) by the chain rule.
dx/dt = v by definition.
So a = v (dv/dx), with no assumption about a at all.
The other three all require a to be constant throughout the interval.

Answer

a = v (dv/dx)

Shortcut

NCERT’s Points to Ponder makes exactly this distinction: the definitions of v and a are exact for every motion; the kinematic equations are special cases.

Trap

Assuming the three equations are laws of nature. They are consequences of one assumption — constant acceleration — and they fail the moment it does. This is the single most frequently tested conceptual point in the whole chapter.

Q13NEET-style

The velocity of a particle varies with position as v² = 9 − x² (SI units). Its acceleration at x = 2 m is

(a) −4 m s⁻²
(b) +2 m s⁻²
(c) √5 m s⁻²
(d) −2 m s⁻²

Given

v² = 9 − x², x = 2 m

Asked

acceleration at x = 2 m

Concept to use

Differentiate the whole relation with respect to x. The factor 2v(dv/dx) that appears is exactly 2a by the chain-rule identity — so you never need v itself.

Formula or rule

d(v²)/dx = 2v(dv/dx) = 2a

Baby steps

Differentiate: d(v²)/dx = −2x.
But d(v²)/dx = 2a.
So a = −x.
At x = 2: a = −2 m s⁻².

Answer

−2 m s⁻²

Shortcut

Differentiating v² with respect to x gives 2a directly. That one line handles every question phrased as ‘v² = f(x)’ — a is half the derivative.

Trap

Solving for v = √5 first and then trying to differentiate the square root. It works but is slower and error-prone. Note also that a = −x is simple harmonic motion — a preview of Chapter 13.

Q14NEET-style

A particle’s acceleration–position graph is a straight horizontal line at a = 5 m s⁻² from x = 0 to x = 4 m. If the particle starts from rest at x = 0, its speed at x = 4 m is

(a) 2√10 m s⁻¹
(b) √10 m s⁻¹
(c) 20 m s⁻¹
(d) 40 m s⁻¹

Given

a = 5 m s⁻² over 0 ≤ x ≤ 4 m, u = 0

Asked

speed at x = 4 m

Concept to use

The area under an a–x graph equals v²/2 − u²/2. That is the position-space analogue of ‘area under a–t gives Δv’.

Formula or rule

∫a dx = (v² − u²)/2

Baby steps

Area under the graph = 5 × 4 = 20.
v²/2 − 0 = 20.
v² = 40 ⇒ v = √40 = 2√10 ≈ 6.3 m s⁻¹.

Answer

2√10 m s⁻¹

Shortcut

Two different areas, two different meanings: area under a–t gives Δv; area under a–x gives Δ(v²/2). Check the horizontal axis before you use an area.

Trap

Reading the area as Δv and answering 20 m s⁻¹. The horizontal axis here is position, not time. Both 20 and 40 appear in the options to catch the two halves of this mistake.

Q15NEET-style

A particle starts from the origin with a velocity of 10 m s⁻¹ and moves with an acceleration a = −2t m s⁻². It returns to the origin at

(a) t = √10 s
(b) t = 5 s
(c) t = √30 s
(d) t = √15 s

Given

u = 10 m s⁻¹, a = −2t, starts at x = 0

Asked

time at which x returns to zero

Concept to use

Integrate twice: once for v(t), once for x(t). Then set x back to zero and reject the trivial root.

Formula or rule

v = u + ∫a dt ; x = ∫v dt

Baby steps

v = 10 + ∫(−2t) dt = 10 − t².
x = ∫(10 − t²) dt = 10t − t³/3 (no constant, since x = 0 at t = 0).
x = 0 ⇒ t(10 − t²/3) = 0.
Rejecting t = 0: t² = 30 ⇒ t = √30 ≈ 5.5 s.

Answer

t = √30 s

Shortcut

The particle is momentarily at rest at t = √10 ≈ 3.16 s — that is the turning point, not the return. Returning always takes longer than turning, so any answer smaller than √10 can be struck out at once.

Trap

Answering √10, the instant of maximum displacement. Because the retardation strengthens with time, the return leg is faster than the outward leg, so the return time is not double the turning time — that symmetry belongs only to constant acceleration.

Answer key

dQ1

cQ2

aQ3

dQ4

cQ5

bQ6

bQ7

cQ8

bQ9

bQ10

aQ11

aQ12

dQ13

aQ14

cQ15

NEET 2027 preparation — built for Aamirah Fathima. Source: NCERT Physics Class XI, Chapter 2 (Reprint 2026–27). Single self-contained file — print directly from any browser.