The calculus toolkit: differentiate to go down the ladder, integrate to come back up, and use a = v dv/dx whenever time is missing from the question.
NCERT 2.2 & 2.3Priority 120 worked questions4 PYQs3 animated figures
1 · The concept in plain language
The one idea
Average velocity is a chord. Instantaneous velocity is a tangent. Everything in this topic
is that one sentence, dressed up in calculus notation.
Suppose a car’s position between 3 s and 5 s changed by 7.84 m. Divide, and you get
3.92 m s⁻¹. But that number is a report on a whole two-second window — it tells you nothing about
whether the car was crawling at 3 s and racing at 5 s. To find out how fast it was moving at the
instant t = 4 s, shrink the window: take 3.5–4.5 s, then 3.9–4.1 s, then 3.995–4.005 s.
The answers march towards 3.84 m s⁻¹ and settle there. That limiting value is the
instantaneous velocity, written v = dx/dt.
animated The limiting process from NCERT Table 2.1, drawn to scale on x = 0.08 t³. The chord P₁P₂ spans 3–5 s (average 3.92 m s⁻¹), then Q₁Q₂ spans 3.5–4.5 s (3.86 m s⁻¹), and in the limit the chord becomes the tangent at P, whose slope is 3.84 m s⁻¹.
Speed vs velocity — where the asymmetry lives
Velocity carries a sign; speed does not. A velocity of −24 m s⁻¹ and one of +24 m s⁻¹
both correspond to a speed of 24 m s⁻¹. Now here is the subtlety NCERT plants and NEET harvests:
Over a finite interval: average speed ≥ |average velocity|. Walk 5 m east then 5 m west in
10 s and your average velocity is zero while your average speed is 1 m s⁻¹.
At an instant: instantaneous speed = |instantaneous velocity|, always, with no inequality.
Why? Because over a vanishingly small interval the path travelled and the straight-line displacement
become the same thing — there is no time to double back.
Acceleration is the same trick, one floor up
Velocity tells you how fast position changes. Acceleration tells you how fast velocity changes:
a = dv/dt = d²x/dt². Because velocity has both magnitude and direction, a body
accelerates if it speeds up, if it slows down, or if it merely turns while keeping the
same speed. In a straight line only the first two apply.
animated Differentiate to go right, integrate to go left. Going left is the harder direction because integration always leaves a constant behind — you must supply the initial velocity or initial position from the problem statement.
The third face of acceleration
Some questions never mention time at all: they hand you velocity as a function of position,
like v = 2√x or v² = 108 − 9x. Writing
a = dv/dt gets you nowhere. Use the chain rule instead:
a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx)
This is not an approximation and it does not need constant acceleration. It is simply
dv/dt re-routed through x. Recognising which of the three forms a question wants
is most of the skill in this topic.
animated Match the form of acceleration to the form of the data. Choosing the wrong one is not wrong physics — it is just a dead end that eats two minutes of exam time.
Why this topic is Priority 1
These questions are pure mechanism. No diagram to interpret, no sign convention to set up, no
physical situation to picture. Differentiate, substitute, done — usually in under forty seconds.
That makes them the cheapest marks in the whole chapter and the best possible use of the first two
minutes of the physics section.
2 · Formula sheet
Instantaneous velocity
v = limΔt→0 Δx/Δt = dx/dt
Slope of the tangent to the x–t graph. Exact — no condition on a.
Instantaneous speed
|v|
Magnitude only. Never negative. Equals |instantaneous velocity| exactly.
Average velocity
v̄ = Δx/Δt
Slope of the chord. Uses end points only.
Average speed
total path length / Δt
Always ≥ |average velocity|; equal only if the motion never reverses.
Average acceleration
ā = (v₂ − v₁)/(t₂ − t₁) = Δv/Δt
Instantaneous acceleration
a = limΔt→0 Δv/Δt = dv/dt = d²x/dt²
Slope of the tangent to the v–t graph.
Chain-rule form
a = v (dv/dx)
Use when v is given as a function of x. Comes from dv/dt = (dv/dx)(dx/dt).
Going backwards
v = v₀ + ∫ a dt ; x = x₀ + ∫ v dt
The integral gives the change. You must add the initial value yourself.
Power rule
d/dt (ktn) = nktn−1 ; ∫ ktn dt = ktn+1/(n+1)
n ≠ −1 for the integral.
Handy pattern
v = k√x ⇒ a = k²/2 (constant)
Because v² = k²x matches v² = 2ax.
Handy pattern
x ∝ (t + c)n ⇒ a ∝ v(n−2)/(n−1)
Covers the whole family of AIPMT 2010-type questions.
3 · Exceptions and traps
Trap 1 — using v = u + at when a is not constant
If the question gives a = 3t² or a = −kv, the three
kinematic equations are illegal. They were derived by assuming a is a fixed number. The moment a
depends on t, v or x, you must integrate. NCERT says this explicitly in Points to Ponder 6.
Trap 2 — final position mistaken for distance travelled
x = 40 + 12t − t³ gives x = 56 m when the particle stops, but it travelled
only 16 m, because it started at 40 m. Subtract the starting coordinate.
Trap 3 — ignoring a reversal inside the interval
If v changes sign between the two times, distance ≠ |displacement|. Find the root of v = 0, split the
interval there, and add the legs as positive numbers.
Trap 4 — forgetting the constant of integration
∫a dt gives Δv, not v. Add v₀. Likewise
∫v dt gives Δx, not x.
Trap 5 — retardation reported with a minus sign
‘Retardation’ and ‘deceleration’ are already magnitudes. If a = −12 m s⁻²,
the retardation is 12 m s⁻². Papers regularly place both −12 and 12 in the options.
Exception — zero velocity does not mean zero acceleration
A ball at the top of its flight has v = 0 and a = −g simultaneously. Zero velocity is an instant;
acceleration describes what happens next. NCERT Points to Ponder 4 states this outright, and NEET has
asked it as a true/false item.
4 · Numbers to remember
Fact
Value
NCERT Table 2.1 limiting velocity for x = 0.08t³ at t = 4 s
3.84 m s⁻¹
Same, average over 3–5 s / 3.5–4.5 s
3.92 / 3.86 m s⁻¹
SI unit and dimension of velocity
m s⁻¹ [LT⁻¹]
SI unit and dimension of acceleration
m s⁻² [LT⁻²]
NCERT Example 2.1: x = 8.5 + 2.5t², v̄ from 2 s to 4 s
15 m s⁻¹
Same example, v at t = 2.0 s
10 m s⁻¹
36 km h⁻¹ converted to SI (the conversion you will use most)
10 m s⁻¹
Multiply km h⁻¹ by this to get m s⁻¹
5/18
5 · Scientists NEET names
Galileo Galilei (1564–1642)
Settled the argument that acceleration should be defined as rate of change of velocity with
time, not with distance. He showed by experiments on freely falling bodies and on inclined planes
that the rate of change of velocity with time is constant for all bodies in free fall, whereas the change
of velocity with distance is not — it decreases as the fall gets longer. NCERT credits him as the
first to make quantitative studies of free fall.
Isaac Newton and Gottfried Leibniz
Not named in this chapter, but the notation is theirs: dx/dt is Leibniz’s.
Worth knowing only because ‘differential coefficient’, the phrase NCERT uses in section 2.2,
is old Newtonian vocabulary for the derivative.
6 · Twenty worked questions
Red left-border = previous-year question. Work each one on paper
first, then read only the Baby steps line you got stuck on.
Q1NEET-style
The position of a particle moving along the x-axis is x = 2t³ − 3t² + 4t + 5, with x in metre and t in second. Its velocity at t = 2 s is
(a) 16 m s⁻¹
(b) 12 m s⁻¹
(c) 20 m s⁻¹
(d) 24 m s⁻¹
Given
x = 2t³ − 3t² + 4t + 5
Asked
instantaneous velocity at t = 2 s
Concept to use
Velocity is the first time-derivative of position. The constant 5 m only shifts the origin — it cannot affect velocity.
Formula or rule
v = dx/dt
Baby steps
Differentiate term by term: v = 6t² − 6t + 4.
Put t = 2: 6(4) − 6(2) + 4.
= 24 − 12 + 4 = 16.
Answer
16 m s⁻¹
Shortcut
The constant term dies on the first differentiation and the linear term dies on the second. Read only the powers you actually need.
Trap
Substituting t = 2 into x instead of v. That gives a position (17 m), not a velocity.
Q2PYQ · AIPMT 2010
A particle moves a distance x in time t according to x = (t + 5)⁻¹. The acceleration of the particle is proportional to
(a) (velocity)2/3
(b) (velocity)3/2
(c) (velocity)³
(d) (velocity)²
Given
x = (t + 5)⁻¹
Asked
the power of v that a is proportional to
Concept to use
Differentiate twice, then eliminate (t + 5) between the two results. Never try to guess the power — always eliminate.
Formula or rule
v = dx/dt, a = dv/dt
Baby steps
v = −(t + 5)⁻².
a = +2(t + 5)⁻³.
From the first line, (t + 5)⁻¹ = |v|1/2, so (t + 5)⁻³ = |v|3/2.
Therefore a ∝ v3/2.
Answer
a ∝ (velocity)3/2
Shortcut
For x ∝ (t+c)n: v ∝ (t+c)n−1, a ∝ (t+c)n−2, so a ∝ v(n−2)/(n−1). Here n = −1 gives 3/2 straight away.
Trap
Forgetting that the exponent index in v is −2, not −1. The sign of the exponent flips the ratio and hands you 2/3 instead.
Q3PYQ · NEET 2016 Phase-II
If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1 s and 2 s is
(a) (3/2)A + 3B
(b) 3A + 7B
(c) (A/2) + (B/3)
(d) (3/2)A + (7/3)B
Given
v = At + Bt², interval t = 1 s to t = 2 s
Asked
distance travelled in that interval
Concept to use
Distance is the integral of velocity. v is positive throughout 1–2 s (for positive A, B), so distance and displacement are equal here.
Formula or rule
s = ∫ v dt
Baby steps
s = ∫12 (At + Bt²) dt.
= [At²/2 + Bt³/3]12.
= A(4 − 1)/2 + B(8 − 1)/3.
= 3A/2 + 7B/3.
Answer
(3/2)A + (7/3)B
Shortcut
Powers go up by one and divide by the new power. Then just plug 2 and 1 into t² and t³: differences 3 and 7.
Trap
Multiplying by the interval (1 s) instead of integrating. That is only legal if v were constant over the interval.
Q4PYQ · AIPMT 2011
The motion of a particle along a straight line is described by x = 8 + 12t − t³ (x in metre, t in second). The retardation of the particle when its velocity becomes zero is
(a) 12 m s⁻²
(b) 6 m s⁻²
(c) 24 m s⁻²
(d) zero
Given
x = 8 + 12t − t³
Asked
magnitude of acceleration at the instant v = 0
Concept to use
Two-step: find the time at which v = 0, then evaluate a at that time. Retardation means the magnitude of a negative acceleration.
Formula or rule
v = dx/dt, a = dv/dt
Baby steps
v = 12 − 3t².
Set v = 0: t² = 4, so t = 2 s (reject t = −2 s).
a = −6t.
At t = 2 s: a = −12 m s⁻², i.e. retardation of 12 m s⁻².
Answer
12 m s⁻²
Shortcut
For x = α + βt − γt³, v = 0 at t = √(β/3γ) and the retardation there is 6γ√(β/3γ).
Trap
Reporting −12 m s⁻² when the question says retardation. Retardation is already a magnitude, so the minus sign must be dropped.
Q5NEET-style
A particle moves along the x-axis with x = 40 + 12t − t³ (SI units). How far does it travel before momentarily coming to rest?
(a) 24 m
(b) 40 m
(c) 56 m
(d) 16 m
Given
x = 40 + 12t − t³, particle starts at t = 0
Asked
distance covered up to the instant v = 0
Concept to use
The question asks for distance travelled, not the final coordinate. Distance = |x(final) − x(initial)| here, because the particle does not reverse before stopping.
Formula or rule
v = dx/dt = 0 at the turning instant
Baby steps
v = 12 − 3t² = 0 ⇒ t = 2 s.
x(0) = 40 m.
x(2) = 40 + 24 − 8 = 56 m.
Distance = 56 − 40 = 16 m.
Answer
16 m
Shortcut
The 40 is just where the particle was parked at t = 0. Subtract it once and forget it.
Trap
Answering 56 m — that is the final position, not the distance travelled. This is the single most common slip in this question type.
Q6NEET-style
A body starts from rest and moves with acceleration a = 3t² m s⁻². Its velocity at t = 2 s is
(a) 8 m s⁻¹
(b) 4 m s⁻¹
(c) 12 m s⁻¹
(d) 6 m s⁻¹
Given
a = 3t², v = 0 at t = 0
Asked
v at t = 2 s
Concept to use
Acceleration is not constant, so v = u + at is illegal. Integrate instead.
Formula or rule
v = ∫ a dt + v₀
Baby steps
v = ∫0t 3t² dt = t³.
The constant of integration is v₀ = 0 because the body starts from rest.
At t = 2 s: v = 2³ = 8 m s⁻¹.
Answer
8 m s⁻¹
Shortcut
Whenever a depends on t, the answer is an integral. a = ktn ⇒ v = ktn+1/(n+1) from rest.
Trap
Plugging into v = u + at with a = 3(2)² = 12, giving 24 m s⁻¹. That equation requires constant a.
Q7NEET-style
The velocity of a particle moving along the x-axis satisfies v² = 108 − 9x (all quantities in SI). Its acceleration is
(a) −4.5 m s⁻², constant
(b) −9 m s⁻², constant
(c) −2.25 m s⁻², constant
(d) not constant — it depends on x
Given
v² = 108 − 9x
Asked
the acceleration
Concept to use
When v is given as a function of x, use a = v dv/dx. Even faster: differentiate the whole relation with respect to x.
Formula or rule
a = v (dv/dx)
Baby steps
Differentiate both sides w.r.t. x: 2v (dv/dx) = −9.
So v (dv/dx) = −4.5.
But a = v dv/dx, hence a = −4.5 m s⁻², independent of x.
Answer
−4.5 m s⁻², constant
Shortcut
Compare with v² = u² + 2ax: the coefficient of x is 2a. Here that coefficient is −9, so a = −4.5 at once.
Trap
Using a = dv/dt and getting stuck because there is no t in the problem. Match the form of the data to the form of the derivative.
Q8NEET-style
The velocity of a particle depends on its position as v = 2√x (SI units). Its acceleration is
(a) 4 m s⁻²
(b) 2 m s⁻²
(c) √x m s⁻²
(d) zero
Given
v = 2√x = 2x1/2
Asked
acceleration
Concept to use
Velocity given in terms of position → chain rule form of acceleration.
Formula or rule
a = v dv/dx
Baby steps
dv/dx = 2 × ½ x−1/2 = x−1/2.
a = v (dv/dx) = 2x1/2 × x−1/2.
= 2 m s⁻², a constant.
Answer
2 m s⁻²
Shortcut
v = k√x always means uniform acceleration a = k²/2. Here k = 2, so a = 2. It is exactly v² = 2ax in disguise.
Trap
Differentiating with respect to t and treating √x as a constant. x itself is a function of time.
Q9NEET-style
A particle’s velocity varies with position as v = 10 − 2x (SI units). Its acceleration at x = 2 m is
(a) −20 m s⁻²
(b) −12 m s⁻²
(c) +12 m s⁻²
(d) −4 m s⁻²
Given
v = 10 − 2x
Asked
a at x = 2 m
Concept to use
Again v as a function of x. Note the acceleration here is not constant — it varies with x.
Formula or rule
a = v dv/dx
Baby steps
dv/dx = −2.
At x = 2: v = 10 − 4 = 6 m s⁻¹.
a = (6)(−2) = −12 m s⁻².
Answer
−12 m s⁻²
Shortcut
For v = A − Bx, a = −B(A − Bx) = −Bv. Acceleration is proportional to velocity, so both die out together.
Trap
Reading dv/dx = −2 as the acceleration. You must still multiply by v.
Q10NEET-style
The retardation of a particle moving in a straight line is proportional to the square root of its speed: dv/dt = −2.5√v (SI). If the initial speed is 6.25 m s⁻¹, the time taken for the particle to stop is
(a) 1 s
(b) 4 s
(c) 2 s
(d) 6.25 s
Given
dv/dt = −2.5√v, v₀ = 6.25 m s⁻¹
Asked
time until v = 0
Concept to use
Separable differential equation: put all the v terms on one side, all the t terms on the other, then integrate between the stated limits.
Formula or rule
∫ dv/√v = −2.5 ∫ dt
Baby steps
Separate: v−1/2 dv = −2.5 dt.
Integrate from v = 6.25 to 0 and t = 0 to T: [2√v]6.250 = −2.5T.
0 − 2(2.5) = −2.5T.
T = 5/2.5 = 2 s.
Answer
2 s
Shortcut
For dv/dt = −k√v, stopping time is T = 2√v₀/k. Here 2(2.5)/2.5 = 2 s.
Trap
Treating −2.5√v as a constant retardation and using t = v₀/a. The retardation shrinks as the body slows.
Q11NEET-style
For a particle moving with x = t³ − 6t² + 9t (SI units), the velocity at the instant when the acceleration is zero is
(a) 0
(b) +3 m s⁻¹
(c) +9 m s⁻¹
(d) −3 m s⁻¹
Given
x = t³ − 6t² + 9t
Asked
v at the instant a = 0
Concept to use
Differentiate twice, set the second derivative to zero to get the time, then feed that time back into v. a = 0 is the point of inflection of the x–t curve, i.e. where v is extreme.
Formula or rule
v = dx/dt, a = d²x/dt²
Baby steps
v = 3t² − 12t + 9.
a = 6t − 12 = 0 ⇒ t = 2 s.
v(2) = 12 − 24 + 9 = −3 m s⁻¹.
Answer
−3 m s⁻¹
Shortcut
a = 0 sits at the vertex of the v–t parabola, so the answer is the minimum velocity: vmin = 9 − 12²/12 = −3.
Trap
Assuming a = 0 means v = 0. Zero acceleration means constant velocity at that instant, not zero velocity.
Q12NEET-style
For the car whose motion NCERT plots in Fig. 2.1, x = 0.08 t³ (SI). The instantaneous velocity at t = 4.0 s is
(a) 5.12 m s⁻¹
(b) 3.84 m s⁻¹
(c) 1.28 m s⁻¹
(d) 7.68 m s⁻¹
Given
x = 0.08 t³, t = 4.0 s
Asked
instantaneous velocity
Concept to use
This is the worked table in the textbook: as Δt shrinks from 2.0 s to 0.01 s the average velocity closes in on 3.84 m s⁻¹, which is exactly what the derivative gives instantly.
Formula or rule
v = dx/dt = 0.24 t²
Baby steps
v = 0.24 t².
At t = 4: 0.24 × 16 = 3.84 m s⁻¹.
Cross-check with Table 2.1: Δt = 0.01 s gives 3.8400 m s⁻¹.
Answer
3.84 m s⁻¹
Shortcut
Calculus does in one line what the table does in five rows. Never build the table in an exam.
Trap
Computing x(4) = 5.12 m and offering it as the velocity. Option (a) is planted exactly for that error.
Q13NEET-style
A particle has acceleration a = (4t − 2) m s⁻² and its velocity at t = 0 is 3 m s⁻¹. Its velocity at t = 3 s is
(a) 12 m s⁻¹
(b) 18 m s⁻¹
(c) 15 m s⁻¹
(d) 21 m s⁻¹
Given
a = 4t − 2, v₀ = 3 m s⁻¹
Asked
v at t = 3 s
Concept to use
Non-constant acceleration → integrate, and remember to add the initial velocity as the constant of integration.
Formula or rule
v = v₀ + ∫0t a dt
Baby steps
∫03 (4t − 2) dt = [2t² − 2t]03.
= 18 − 6 = 12 m s⁻¹ (this is the change in velocity).
v = 3 + 12 = 15 m s⁻¹.
Answer
15 m s⁻¹
Shortcut
The integral of a over an interval is Δv, never v itself. Add v₀ at the end, every time.
Trap
Dropping v₀ and answering 12 m s⁻¹. Option (a) is that mistake.
Q14NEET-style
For a particle with x = 3t² + 5t (SI), the difference between its instantaneous velocity at t = 2 s and its average velocity over 0 to 2 s is
(a) 11 m s⁻¹
(b) 17 m s⁻¹
(c) 6 m s⁻¹
(d) 28 m s⁻¹
Given
x = 3t² + 5t; interval 0 to 2 s
Asked
v(2) − v̄
Concept to use
Average velocity uses end points only; instantaneous velocity uses the derivative. For uniformly accelerated motion the average over an interval equals the instantaneous value at the mid-time, not the end.
Formula or rule
v̄ = Δx/Δt ; v = dx/dt
Baby steps
x(0) = 0, x(2) = 12 + 10 = 22 m.
v̄ = 22/2 = 11 m s⁻¹.
v = 6t + 5, so v(2) = 17 m s⁻¹.
Difference = 17 − 11 = 6 m s⁻¹.
Answer
6 m s⁻¹
Shortcut
Here a = 6 m s⁻² is constant, so v̄ = v at the mid-time t = 1 s = 11 m s⁻¹. The gap to t = 2 s is a×1 s = 6 m s⁻¹.
Trap
Confusing the two velocities, then reporting 11 or 17. Read whether the question wants a slope of a chord or of a tangent.
Q15NEET-style
A particle has a velocity of −24 m s⁻¹ at one instant and +24 m s⁻¹ at another. Which statement is correct?
(a) Its speed is −24 m s⁻¹ at the first instant
(b) Its average speed over the interval must be zero
(c) Its acceleration must be zero
(d) Its instantaneous speed is 24 m s⁻¹ at both instants
Given
v = −24 m s⁻¹ and v = +24 m s⁻¹ at two instants
Asked
the correct statement about speed
Concept to use
Instantaneous speed is the magnitude of instantaneous velocity, so it is never negative. Average speed and average velocity are different animals, though.
Formula or rule
speed = |v| at every instant
Baby steps
Speed cannot carry a sign, so (a) is out.
|−24| = |+24| = 24 m s⁻¹, so (d) is right.
Average velocity could be zero if the particle returned to its start, but average speed = path length/time > 0, so (b) is out.
The velocity reversed sign, so acceleration was certainly non-zero: (c) is out.
Answer
Its instantaneous speed is 24 m s⁻¹ at both instants
Shortcut
NCERT states it flatly: average speed ≥ |average velocity|, but instantaneous speed = |instantaneous velocity|. Over a vanishing interval the path and the chord coincide.
Trap
Carrying the ‘average speed ≥ average velocity’ inequality over to instantaneous quantities, where it becomes a strict equality.
Q16NEET-style
A particle moving in a straight line has acceleration a = −kv, where k is a positive constant and v is its speed. If its speed at t = 0 is v₀, then
(a) v = v₀ − kt
(b) v = v₀/(1 + kt)
(c) v = v₀e−kt²
(d) v = v₀e−kt
Given
dv/dt = −kv, v = v₀ at t = 0
Asked
v as a function of t
Concept to use
Separable equation again. When the rate of decrease is proportional to the quantity itself, the result is always an exponential decay.
Formula or rule
dv/v = −k dt
Baby steps
Separate and integrate: ∫v₀v dv/v = −k∫0t dt.
ln(v/v₀) = −kt.
v = v₀e−kt.
Answer
v = v₀e−kt
Shortcut
Rate ∝ the quantity → exponential. The body slows forever and never formally stops — same shape as radioactive decay and RC discharge.
Trap
Using v = u + at with a = −kv₀. That gives option (a), which wrongly predicts the particle stopping at t = 1/k.
Q17NEET-style
The position of an object moving along the x-axis is x = a + bt² with a = 8.5 m and b = 2.5 m s⁻². Its average velocity between t = 2.0 s and t = 4.0 s is
(a) 5 m s⁻¹
(b) 10 m s⁻¹
(c) 15 m s⁻¹
(d) 20 m s⁻¹
Given
x = 8.5 + 2.5t²; interval 2.0 s to 4.0 s
Asked
average velocity
Concept to use
Average velocity is the slope of the chord: total displacement over total time. The constant a = 8.5 m cancels in the subtraction.
Formula or rule
v̄ = [x(t₂) − x(t₁)] / (t₂ − t₁)
Baby steps
x(4) = 8.5 + 2.5(16) = 48.5 m.
x(2) = 8.5 + 2.5(4) = 18.5 m.
v̄ = (48.5 − 18.5)/2 = 30/2 = 15 m s⁻¹.
Answer
15 m s⁻¹
Shortcut
v̄ = b(t₂ + t₁) = 2.5(6) = 15. Equivalently, a = 5 m s⁻² is constant, so v̄ is the instantaneous velocity at the mid-time t = 3 s: 5×3 = 15.
Trap
Using v = 5t at t = 4 (giving 20) or at t = 2 (giving 10). Those are the end-point velocities, not the average. This is NCERT Example 2.1.
Q18NEET-style
A particle’s velocity is v = 3t² − 6t (SI). The particle is momentarily at rest at
(a) t = 0 and t = 2 s
(b) t = 2 s only
(c) t = 0 and t = 6 s
(d) t = 3 s only
Given
v = 3t² − 6t
Asked
all instants at which v = 0
Concept to use
‘Momentarily at rest’ simply means v = 0. Factorise rather than using the quadratic formula, and do not discard t = 0 unless the question rules it out.
Formula or rule
v = 0
Baby steps
3t(t − 2) = 0.
t = 0 or t = 2 s.
Both are physically admissible (t ≥ 0), so both count.
Answer
t = 0 and t = 2 s
Shortcut
Factorising out t immediately exposes the t = 0 root, which the quadratic formula hides in arithmetic.
Trap
Dropping the t = 0 root out of habit. The particle really does start from rest here.
Q19NEET-style
For a particle with x = t² − 4t + 3 (SI), the distance travelled in the first 4 s is
(a) zero
(b) 8 m
(c) 4 m
(d) 12 m
Given
x = t² − 4t + 3, interval 0 to 4 s
Asked
distance travelled (not displacement)
Concept to use
The particle reverses inside the interval, so distance ≠ |displacement|. Break the interval at the turning point and add the two legs as positive lengths.
Formula or rule
v = 0 locates the turn; distance = Σ |Δx| over each leg
Baby steps
v = 2t − 4 = 0 ⇒ t = 2 s (turning point inside 0–4 s).
x(0) = 3 m, x(2) = −1 m, x(4) = 3 m.
Leg 1: |−1 − 3| = 4 m. Leg 2: |3 − (−1)| = 4 m.
Distance = 4 + 4 = 8 m (displacement is zero).
Answer
8 m
Shortcut
Always check whether v changes sign inside the interval before integrating. If it does, split at the root.
Trap
Answering ‘zero’ because x(0) = x(4). That is the displacement. Distance can never be zero unless the body never moved.
Q20NEET-style
The velocity of a particle varies with its position as v = k/x, where k is a positive constant. Its acceleration is
(a) −k/x²
(b) k²/x³
(c) −k²/x³
(d) −k²/x²
Given
v = k/x
Asked
acceleration as a function of x
Concept to use
Third form of acceleration once more — v is handed to you in terms of x, so a = v dv/dx is the only clean route.
Formula or rule
a = v dv/dx
Baby steps
dv/dx = −k/x².
a = (k/x)(−k/x²).
= −k²/x³.
Answer
−k²/x³
Shortcut
Multiply exponents mentally: x⁻¹ × x⁻² = x⁻³, and k appears once in each factor, hence k². The minus sign survives.
Trap
Stopping at dv/dx = −k/x² and calling it the acceleration. Option (a) is exactly that unfinished answer.
Answer key
aQ1
bQ2
dQ3
aQ4
dQ5
aQ6
aQ7
bQ8
bQ9
cQ10
dQ11
bQ12
cQ13
cQ14
dQ15
dQ16
cQ17
aQ18
bQ19
cQ20
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.