Kinematic equations for uniformly accelerated motion
Three equations, five quantities, one condition. Plus the nᵗʰ-second formula, stopping distance and the mid-point trap.
NCERT 2.4Priority 120 worked questions4 PYQs3 animated figures
1 · The concept in plain language
The one idea
Five quantities describe uniformly accelerated motion: x, v₀, v, a, t. Each equation leaves
one of them out. Look at what the question gives you, spot the missing one, and the equation picks itself.
There is nothing mysterious about the three kinematic equations — they are just the
definitions of velocity and acceleration, rearranged for the special case where a never changes. NCERT
derives them from a picture rather than from algebra, and that picture is worth carrying in your head.
On a velocity–time graph, the area under the curve is the displacement. For uniform
acceleration the graph is a straight line, so the area is a trapezium. Split that trapezium into a
rectangle of height v₀ and a triangle of height (v − v₀) = at, and you
have read the second equation straight off the diagram:
animated NCERT Fig. 2.5 rebuilt. Rectangle OACD contributes v₀t (the distance you would cover with no acceleration at all) and triangle ABC contributes ½at² (the bonus the acceleration buys you). Their sum is the displacement.
Choosing the right equation
The question does not mention…
Use
Because
displacement
v = v₀ + at
x is absent from it
final velocity
x = v₀t + ½at²
v is absent from it
time
v² = v₀² + 2ax
t is absent from it
acceleration (but gives both velocities)
x = [(v+v₀)/2]t
a is absent from it
Distance in the nth second
This phrase confuses more students than any other in the chapter. “Distance in 5 seconds”
means everything from t = 0 to t = 5 s. “Distance in the 5th second” means only the
slice between t = 4 s and t = 5 s. Subtracting the two totals gives:
Sn = v₀ + (a/2)(2n − 1)
Starting from rest, these slices are in the ratio 1 : 3 : 5 : 7 — the odd numbers —
because the totals are in the ratio 1 : 4 : 9 : 16 and consecutive squares differ by consecutive odd
numbers. That is Galileo’s law of odd numbers, and NCERT proves it in Example 2.5.
animated A body released from rest passes the 1 m, 4 m, 9 m and 16 m marks at 1, 2, 3 and 4 seconds. The gaps between marks grow as 1 : 3 : 5 : 7. Watch the dot: it clears the last gap in the same one second it took to clear the first, tiny one.
Stopping distance — the square law
Set v = 0 in the third equation and you get ds = −v₀²/2a.
The initial speed appears squared. Drive at twice the speed and you need four times the road to
stop, with the same brakes. NCERT quotes real data for one make of car: 10 m, 20 m, 34 m and 50 m at
11, 15, 20 and 25 m s⁻¹. This is the physics behind school-zone speed limits.
animated Two identical cars, identical brakes, speeds in the ratio 1:2. The red car needs four times the road. Stopping time, by contrast, only doubles.
Why this topic is Priority 1
It has the highest raw question count in the chapter, and it is load-bearing for everything after it:
Laws of Motion, Work–Energy–Power and Projectile Motion all assume you can apply these three
equations without thinking. Time spent here pays out three more times later in the syllabus.
2 · Formula sheet
First equation (no x)
v = v₀ + at
Links v, v₀, a, t.
Second equation (no v)
x = v₀t + ½at²
Links x, v₀, a, t.
Third equation (no t)
v² = v₀² + 2ax
Links v, v₀, a, x. The workhorse.
Average-velocity form
x = [(v + v₀)/2] t = v̄t
Constant acceleration only. NCERT Eq. 2.7a/2.7b.
Non-zero start position
x = x₀ + v₀t + ½at² ; v² = v₀² + 2a(x − x₀)
Replace x by (x − x₀) everywhere.
Distance in the nth second
Sn = v₀ + (a/2)(2n − 1)
A one-second slice, not a total. From rest this gives the 1:3:5:7 ratios.
Stopping distance
ds = −v₀²/2a
∝ v₀². Double the speed, quadruple the distance.
Stopping time
t = v₀/|a| = 2ds/v₀
Linear in v₀.
Distance in the last second
a/2
Independent of the initial speed. Comes from reversing time.
Velocity at mid-time
(u + v)/2
Arithmetic mean.
Velocity at mid-point
√[(u² + v²)/2]
Root-mean-square. Always the larger.
Equal distances at two speeds
v̄ = 2v₁v₂/(v₁ + v₂)
Harmonic mean.
Equal times at two speeds
v̄ = (v₁ + v₂)/2
Arithmetic mean.
Accelerate at α then brake at β in total time t
vmax = αβt/(α + β)
Total distance = ½vmaxt.
3 · Exceptions and traps
The governing condition
All three equations require the acceleration to be constant in both magnitude and direction
throughout the interval. NCERT states this in Points to Ponder 6: the definitions of v and a are always
exact, the kinematic equations are not. The moment a changes — a car that accelerates then brakes,
a ball that bounces — you must break the motion into phases and start fresh.
Trap — ‘in n seconds’ vs ‘in the nth second’
One is a total, the other is a one-second slice. From rest they differ by a factor of
n²/(2n − 1), which is 1 at n = 1 but already 9/5 at n = 3.
Papers put both values in the options.
Trap — mid-point of distance vs mid-point of time
Mid-time velocity is (u+v)/2. Mid-point velocity is
√[(u²+v²)/2], which is always larger. The words
‘mid-point of the path’ and ‘half-way in time’ select different formulas.
Trap — km h⁻¹ left unconverted
Multiply by 5/18. Every stopping-distance question in NEET states the speed in
km h⁻¹ precisely because the conversion is the easiest mark to drop.
Trap — catching up means equal positions, not equal velocities
When a body chases another, set the displacements equal. The instant of equal velocity is when
the gap stops growing — the chaser is still behind at that moment.
Exception — signs are algebraic, not decorative
NCERT Points to Ponder 5: the quantities in these equations may be positive or negative, and the
equations work in every case provided you substitute with proper signs. Fix your positive direction
before you write a single number, and never change it mid-problem.
4 · Numbers to remember
Fact
Value
km h⁻¹ → m s⁻¹
× 5/18
Common conversions
36 → 10, 54 → 15, 72 → 20, 90 → 25, 126 → 35
Distances from rest in 1, 2, 3, 4 s
1 : 4 : 9 : 16
Distances in the 1st, 2nd, 3rd, 4th second from rest
1 : 3 : 5 : 7
Distance in the last second before stopping
a/2
Stopping time for a body brought to rest over distance s
t = 2s/u
NCERT braking data (10, 20, 34, 50 m)
at 11, 15, 20, 25 m s⁻¹
NCERT Exercise 2.5 answers
a = 3.06 m s⁻², t = 11.4 s
Accelerate at α, brake at β, total time t
vmax = αβt/(α+β)
5 · Scientists NEET names
Galileo Galilei (1564–1642)
Named twice in this chapter. He established the law of odd numbers: the distances traversed in
equal successive intervals by a body falling from rest are in the ratio 1 : 3 : 5 : 7 … NCERT calls
him the first person to make quantitative studies of free fall. Remember the dates — assertion-reason
items sometimes attach the law to the wrong name.
6 · Twenty worked questions
Red left-border = previous-year question.
Q1NEET-style
A car moving along a straight highway with a speed of 126 km h⁻¹ is brought to a stop within a distance of 200 m. The retardation of the car (assumed uniform) and the time it takes to stop are
(a) 3.06 m s⁻², 11.4 s
(b) 3.06 m s⁻², 5.7 s
(c) 6.12 m s⁻², 11.4 s
(d) 1.53 m s⁻², 22.8 s
Given
u = 126 km h⁻¹, v = 0, s = 200 m
Asked
retardation and time
Concept to use
Convert to SI first, then use the equation that has no time in it, then the one that has no distance.
Formula or rule
v² = u² + 2as ; v = u + at
Baby steps
u = 126 × 5/18 = 35 m s⁻¹.
0 = 35² + 2a(200) ⇒ a = −1225/400 = −3.06 m s⁻².
0 = 35 + (−3.06)t ⇒ t = 35/3.0625 = 11.43 s.
Retardation = 3.06 m s⁻², time = 11.4 s.
Answer
3.06 m s⁻² and 11.4 s
Shortcut
For a body brought to rest, t = 2s/u = 400/35 = 11.4 s directly, because the average velocity is u/2.
Trap
Forgetting the 5/18 conversion. Using 126 raw gives a retardation near 40 m s⁻² — physically absurd, and a sign you skipped the conversion. This is NCERT Exercise 2.5.
Q2NEET-style
A body starts from rest and moves with a uniform acceleration of 2 m s⁻². The distance it covers in the 5th second is
(a) 5 m
(b) 10 m
(c) 25 m
(d) 9 m
Given
u = 0, a = 2 m s⁻², n = 5
Asked
distance in the 5th second
Concept to use
‘Distance in the nth second’ means the distance covered between t = n − 1 and t = n — a one-second slice, not the total up to n seconds.
Formula or rule
Sn = u + (a/2)(2n − 1)
Baby steps
S₅ = 0 + (2/2)(2×5 − 1).
= 1 × 9 = 9 m.
Check by subtraction: s(5) − s(4) = 25 − 16 = 9 m. ✓
Answer
9 m
Shortcut
From rest, the distance in the nth second is simply (a/2)(2n − 1), so the slices run 1a/2, 3a/2, 5a/2 … — the odd numbers again.
Trap
Using s = ½at² with t = 5 and answering 25 m. That is the total distance in five seconds, not the fifth slice. Note the units of Sn look like m s⁻¹ but it really is a distance — the ‘per second’ is hidden.
Q3PYQ · AIPMT 2009
A body covers 200 cm in the first 2 s and 220 cm in the next 4 s under constant acceleration. Its velocity at the end of the 7th second is
(a) 5 cm s⁻¹
(b) 15 cm s⁻¹
(c) 10 cm s⁻¹
(d) 20 cm s⁻¹
Given
s(0–2 s) = 200 cm = 2 m; s(2–6 s) = 220 cm = 2.2 m
Asked
v at t = 7 s
Concept to use
Two unknowns (u and a) need two equations. Convert the second window into a total distance over 6 s before writing the second equation.
Formula or rule
s = ut + ½at², then v = u + at
Baby steps
First 2 s: 2 = 2u + 2a, so u + a = 1.
First 6 s (2.0 + 2.2 = 4.2 m): 4.2 = 6u + 18a.
Substitute u = 1 − a: 4.2 = 6 + 12a ⇒ a = −0.15 m s⁻².
So u = 1.15 m s⁻¹ and v = 1.15 − 0.15(7) = 0.10 m s⁻¹ = 10 cm s⁻¹.
Answer
10 cm s⁻¹
Shortcut
The body is slowing down (220 cm in 4 s is slower than 200 cm in 2 s), so expect a negative a before you compute. A positive a would mean you mis-set the second window.
Trap
Treating ‘the next 4 s’ as a fresh journey starting from rest, or writing 2.2 = 4u + 8a with the original u. Always convert to totals measured from t = 0.
Q4NEET-style
A bullet loses 1/20 of its velocity while passing through a plank. The least number of such identical planks required to just stop the bullet is
(a) 10
(b) 20
(c) 11
(d) 39
Given
each plank removes 1/20 of the current velocity on the first plank; identical planks, so identical retarding force
Asked
least number of planks to bring the bullet to rest
Concept to use
Identical planks means identical work done per plank, so each plank removes the same amount of v², not the same amount of v. Work with v² throughout.
Formula or rule
v² = u² − 2a(nd)
Baby steps
After one plank: v = (19/20)u, so v² = (361/400)u².
Loss of v² per plank = u² − 361u²/400 = (39/400)u².
To lose all of u²: n = u² ÷ (39/400)u² = 400/39 = 10.26.
After 10 planks the bullet is still moving, so it needs an 11th. Least number = 11.
Answer
11 planks
Shortcut
Fraction f of velocity lost per plank ⇒ n = 1/[1 − (1 − f)²], rounded up. Here 1/(1 − 0.9025) = 10.26 → 11.
Trap
Two traps. First, assuming equal velocity loss per plank, which gives 20. Second, rounding 10.26 down to 10 — but after 10 planks some speed remains, so 10 planks do not stop it.
Q5PYQ · AIPMT 1994
A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, the maximum velocity acquired is
(a) (α + β)t / αβ
(b) αβt / (α + β)
(c) (α² − β²)t / αβ
(d) (α + β)t / 2
Given
rest → accelerate at α for t₁ → decelerate at β for t₂ → rest; t₁ + t₂ = t
Asked
maximum velocity v
Concept to use
The two phases share one quantity: the peak velocity v at the junction. Write t₁ and t₂ in terms of v and add.
Formula or rule
v = αt₁ and v = βt₂
Baby steps
t₁ = v/α (starting from rest).
t₂ = v/β (ending at rest).
v/α + v/β = t ⇒ v(α + β)/αβ = t.
v = αβt/(α + β).
Answer
αβt / (α + β)
Shortcut
Recognise the shape: v is the harmonic-type combination of α and β. The total distance in the same problem is ½vt = αβt²/2(α+β).
Trap
Dimension-check the options if you are unsure: only αβt/(α+β) reduces to LT⁻¹. Option (a) has the wrong dimensions entirely.
Q6NEET-style
A body moving with uniform acceleration covers 20 m in the 2nd second and 32 m in the 4th second. The distance it covers in the first 6 seconds is
(a) 144 m
(b) 156 m
(c) 174 m
(d) 192 m
Given
S₂ = 20 m, S₄ = 32 m
Asked
total distance in 6 s
Concept to use
Two nth-second equations give two linear equations in u and a. Subtract them — u cancels immediately.
Formula or rule
Sn = u + (a/2)(2n − 1) ; s = ut + ½at²
Baby steps
S₂ = u + 1.5a = 20.
S₄ = u + 3.5a = 32.
Subtract: 2a = 12 ⇒ a = 6 m s⁻², then u = 20 − 9 = 11 m s⁻¹.
s = 11(6) + ½(6)(36) = 66 + 108 = 174 m.
Answer
174 m
Shortcut
The nth-second distances form an arithmetic progression with common difference a. From S₂ to S₄ is two steps, so a = (32 − 20)/2 = 6 in one line.
Trap
Adding 20 + 32 and treating that as the distance in 6 s. Those are two isolated one-second slices, not a running total.
Q7NEET-style
A train decelerating uniformly slows from 20 m s⁻¹ to 10 m s⁻¹ over a distance of 100 m. How much further does it travel before stopping?
(a) 25.0 m
(b) 50.0 m
(c) 33.3 m
(d) 100 m
Given
u = 20 m s⁻¹, v = 10 m s⁻¹ after 100 m, then to rest
Asked
extra distance to stop
Concept to use
Find a from the first leg, then apply the same a to the second leg with u = 10 m s⁻¹.
Formula or rule
v² = u² + 2as
Baby steps
Leg 1: 100 = 400 + 2a(100) ⇒ a = −1.5 m s⁻².
Leg 2: 0 = 10² + 2(−1.5)s.
s = 100/3 = 33.3 m.
Answer
33.3 m
Shortcut
Distance to stop ∝ v². From 20 to 0 the total would be 400/3 = 133.3 m; minus the 100 m already used leaves 33.3 m.
Trap
Assuming the remaining distance is half of 100 m because the speed halved. Speed halving means v² falls to a quarter, so three quarters of the journey is already done.
Q8NEET-style
A body starting from rest moves with constant acceleration 4 m s⁻². The ratio of the distance covered in the first 3 seconds to the distance covered in the 3rd second is
(a) 3 : 1
(b) 5 : 9
(c) 2 : 1
(d) 9 : 5
Given
u = 0, a = 4 m s⁻²
Asked
s(3 s) : S₃
Concept to use
One is a total, the other is a one-second slice. Compute both from the same a and divide. The value of a will cancel.
Formula or rule
s = ½at² ; Sn = (a/2)(2n − 1)
Baby steps
s(3) = ½(4)(9) = 18 m.
S₃ = (4/2)(5) = 10 m.
Ratio = 18 : 10 = 9 : 5.
Answer
9 : 5
Shortcut
From rest the ratio is always n² : (2n − 1), independent of a. For n = 3 that is 9 : 5 without touching the number 4.
Trap
Reading ‘in 3 seconds’ and ‘in the 3rd second’ as the same phrase. They differ by a factor that grows with n.
Q9NEET-style
A bus starts from rest with a constant acceleration of 2 m s⁻². At the same instant a scooter moving with a constant speed of 20 m s⁻¹ overtakes it. The bus catches up with the scooter after
(a) 20 s, at 400 m
(b) 10 s, at 100 m
(c) 20 s, at 200 m
(d) 40 s, at 800 m
Given
bus: u = 0, a = 2 m s⁻²; scooter: constant 20 m s⁻¹; same start point and instant
Asked
time and distance at which they meet
Concept to use
They meet when their positions are equal, not their velocities. Set the two position expressions equal.
Formula or rule
xbus = ½at² ; xsc = vt
Baby steps
½(2)t² = 20t.
t² = 20t ⇒ t(t − 20) = 0.
t = 20 s (t = 0 is the starting instant).
Distance = 20 × 20 = 400 m.
Answer
20 s, at 400 m
Shortcut
From rest, a body catches a constant-speed rival at t = 2v/a and the meeting point is 2v²/a. Here 2(20)/2 = 20 s and 2(400)/2 = 400 m.
Trap
Solving for equal velocities, which happens at t = 10 s. At that instant the bus is still 100 m behind — it is only just matching speed, not catching up. Option (b) is that error.
Q10NEET-style
A particle moving with uniform acceleration passes point A with velocity 10 m s⁻¹ and point B with velocity 20 m s⁻¹. Its velocity at the mid-point of AB is
(a) 15.0 m s⁻¹
(b) 14.1 m s⁻¹
(c) 15.8 m s⁻¹
(d) 17.5 m s⁻¹
Given
u = 10 m s⁻¹ at A, v = 20 m s⁻¹ at B, uniform a
Asked
velocity at the mid-point of the distance
Concept to use
Mid-point of distance is not mid-point of time. Use v² = u² + 2as over half the distance — the velocity-squared, not the velocity, is what varies linearly with distance.
Formula or rule
vmid-point = √[(u² + v²)/2]
Baby steps
Over the full distance s: v² − u² = 2as.
Over half of it: vm² − u² = 2a(s/2) = (v² − u²)/2.
vm² = (u² + v²)/2 = (100 + 400)/2 = 250.
vm = √250 = 15.8 m s⁻¹.
Answer
15.8 m s⁻¹
Shortcut
Mid-time velocity is the plain average (u+v)/2 = 15; mid-point velocity is the root-mean-square √[(u²+v²)/2] = 15.8. The RMS value is always the larger of the two.
Trap
Answering 15 m s⁻¹. Option (a) is planted for exactly this. Read whether the question says mid-point of the path or mid-point of the time.
Q11NEET-style
A car travelling at 54 km h⁻¹ is brought to rest with a uniform retardation of 5 m s⁻². The stopping distance and stopping time are
(a) 45 m, 3 s
(b) 22.5 m, 3 s
(c) 22.5 m, 1.5 s
(d) 11.25 m, 3 s
Given
u = 54 km h⁻¹, a = −5 m s⁻², v = 0
Asked
stopping distance and time
Concept to use
Standard stopping-distance calculation. Convert units first, always.
Formula or rule
ds = u²/2a ; t = u/a
Baby steps
u = 54 × 5/18 = 15 m s⁻¹.
ds = 15²/(2×5) = 225/10 = 22.5 m.
t = 15/5 = 3 s.
Answer
22.5 m and 3 s
Shortcut
Cross-check with the average velocity: d = (u/2)t = 7.5 × 3 = 22.5 m. If the two routes disagree, one of them has an arithmetic slip.
Trap
Writing d = u²/a and losing the factor 2, giving 45 m. Option (a) is that dropped 2.
Q12NEET-style
A body moving with a uniform retardation comes to rest. If the retardation is 4 m s⁻², the distance covered in the last second of its motion is
(a) 2 m
(b) 1 m
(c) 4 m
(d) 8 m
Given
uniform retardation a = 4 m s⁻², final velocity zero
Asked
distance in the last second
Concept to use
Run the motion backwards. Reversed in time, the last second becomes the first second of a body starting from rest with acceleration 4 m s⁻².
Formula or rule
s = ½at² with t = 1 s
Baby steps
Reverse the film: the body starts from rest and accelerates at 4 m s⁻².
s = ½(4)(1)².
= 2 m.
Answer
2 m
Shortcut
Distance in the last second before stopping is always a/2, whatever the initial speed was. It is the single most reusable fact in this topic.
Trap
Trying to find u first. You do not need it — and the question does not give it, which is the hint that time reversal is intended.
Q13NEET-style
A body moving with uniform acceleration covers 12 m in the 3rd second and 20 m in the 5th second. The distance covered in 10 s from the start is
(a) 200 m
(b) 210 m
(c) 240 m
(d) 220 m
Given
S₃ = 12 m, S₅ = 20 m
Asked
distance in 10 s
Concept to use
Same two-equation structure as before. Solve for a and u, then use the standard displacement equation.
Formula or rule
Sn = u + (a/2)(2n − 1)
Baby steps
u + 2.5a = 12 and u + 4.5a = 20.
Subtract: 2a = 8 ⇒ a = 4 m s⁻².
u = 12 − 10 = 2 m s⁻¹.
s = 2(10) + ½(4)(100) = 20 + 200 = 220 m.
Answer
220 m
Shortcut
Common difference of the nth-second series is a. Two steps apart gives a = (20 − 12)/2 = 4 instantly.
Trap
Mixing up the (2n−1) factor as (2n+1). For n = 3 the multiplier is 5, not 7. Test it against a case you know: from rest the first second must give a/2.
Q14NEET-style
For a car of a particular make, the braking distances are 10 m, 20 m, 34 m and 50 m at speeds of 11, 15, 20 and 25 m s⁻¹. If the initial speed is doubled, the stopping distance becomes
(a) four times as large
(b) twice as large
(c) √2 times as large
(d) unchanged
Given
ds data from NCERT Example 2.6
Asked
effect of doubling the initial speed
Concept to use
Stopping distance depends on the square of the initial speed for a fixed braking deceleration.
Formula or rule
ds = −v₀²/2a
Baby steps
ds ∝ v₀² for fixed a.
Replace v₀ by 2v₀: (2v₀)² = 4v₀².
Check against the data: doubling 11 → 22 m s⁻¹ should give about 4 × 10 = 40 m, and the table indeed reads 34 m at 20 m s⁻¹ and 50 m at 25 m s⁻¹. ✓
Answer
four times as large
Shortcut
Any ‘what happens to stopping distance’ question is answered by the square law. This is exactly why school-zone speed limits are set low — NCERT makes that point explicitly.
Trap
Confusing stopping distance with stopping time. Time scales linearly with v₀ (doubling doubles it); distance scales quadratically.
Q15NEET-style
A body starts from rest with constant acceleration. The ratio of the distance covered in n seconds to the distance covered in the nth second is
(a) (2n − 1) : n²
(b) n² : (2n − 1)
(c) n : (2n − 1)
(d) (2n + 1) : n²
Given
u = 0, uniform a
Asked
s(n) : Sn
Concept to use
Write both quantities with the same a and divide. Do this once and memorise the result — it converts several questions into one-liners.
Formula or rule
s(n) = ½an² ; Sn = (a/2)(2n − 1)
Baby steps
s(n)/Sn = (½an²) ÷ [(a/2)(2n − 1)].
The a/2 cancels.
= n²/(2n − 1).
Answer
n² : (2n − 1)
Shortcut
Sanity-test with n = 1: the ratio must be 1 : 1, since the first second and ‘in 1 second’ are the same thing. Only option (b) gives 1 : 1.
Trap
Picking (2n−1) : n² by writing the ratio upside down. The total must be the larger quantity for n ≥ 2, so the bigger expression goes on top.
Q16NEET-style
A body accelerates uniformly from 5 m s⁻¹ to 25 m s⁻¹ in 4 s. The distance it covers in this interval is
(a) 40 m
(b) 50 m
(c) 80 m
(d) 60 m
Given
u = 5 m s⁻¹, v = 25 m s⁻¹, t = 4 s
Asked
distance covered
Concept to use
For constant acceleration the average velocity is the plain arithmetic mean of the initial and final velocities. This shortcut skips finding a altogether.
Formula or rule
x = [(u + v)/2] t
Baby steps
v̄ = (5 + 25)/2 = 15 m s⁻¹.
x = 15 × 4 = 60 m.
Cross-check: a = (25−5)/4 = 5, x = 5(4) + ½(5)(16) = 20 + 40 = 60 m. ✓
Answer
60 m
Shortcut
x = v̄t with v̄ = (u+v)/2 is the fastest route whenever both u and v are given. NCERT calls this Eq. 2.7a.
Trap
Using (u+v)/2 when acceleration is not constant. The arithmetic-mean rule is flagged ‘constant acceleration only’ in NCERT Eq. 2.7b.
Q17NEET-style
A body moving at 20 m s⁻¹ is brought to rest in 50 m under uniform retardation. The retardation and the time taken are
(a) 2 m s⁻², 10 s
(b) 4 m s⁻², 5 s
(c) 4 m s⁻², 2.5 s
(d) 8 m s⁻², 2.5 s
Given
u = 20 m s⁻¹, v = 0, s = 50 m
Asked
retardation and time
Concept to use
The no-time equation for a, then the no-distance equation for t.
Formula or rule
v² = u² + 2as ; t = u/|a|
Baby steps
0 = 400 + 2a(50) ⇒ a = −4 m s⁻².
t = 20/4 = 5 s.
Cross-check with average velocity: 50 = (20/2)t ⇒ t = 5 s. ✓
Answer
4 m s⁻² and 5 s
Shortcut
For a body brought to rest, t = 2s/u always. Here 2(50)/20 = 5 s with no need for a at all.
Trap
Using t = s/u = 2.5 s. That would be right only if the body kept its full speed the whole way; in reality it averages half the speed, so it takes twice as long.
Q18PYQ · NEET 2019
A person travelling in a straight line moves with a constant velocity v₁ for a certain distance x and with a constant velocity v₂ for the next equal distance. The average velocity v is given by
(a) 2/v = 1/v₁ + 1/v₂
(b) v = (v₁ + v₂)/2
(c) v = √(v₁v₂)
(d) 1/v = 1/v₁ + 1/v₂
Given
equal distances x at speeds v₁ and v₂
Asked
relation defining the average velocity
Concept to use
Average velocity = total displacement ÷ total time. When equal distances are involved the answer is a harmonic mean; when equal times are involved it is an arithmetic mean.
Formula or rule
v̄ = total distance / total time
Baby steps
t₁ = x/v₁, t₂ = x/v₂.
v = 2x/(t₁ + t₂) = 2x / (x/v₁ + x/v₂).
Cancel x: v = 2/(1/v₁ + 1/v₂).
Rearranged: 2/v = 1/v₁ + 1/v₂, i.e. v = 2v₁v₂/(v₁ + v₂).
Answer
2/v = 1/v₁ + 1/v₂
Shortcut
Equal distances → harmonic mean. Equal times → arithmetic mean. Two words in the question decide which one you write.
Trap
Answering (v₁ + v₂)/2. That is correct only if the two speeds are held for equal durations, not equal distances. The slower leg takes longer, so it must be weighted more heavily.
Q19NEET-style
A body starts from rest and moves with constant acceleration. The ratio of the time taken to cover the first half of a distance to the time taken to cover the second half is
(a) 1 : 1
(b) 1 : (√2 − 1)
(c) (√2 − 1) : 1
(d) 1 : 2
Given
u = 0, constant a, distance split into two equal halves
Asked
tfirst half : tsecond half
Concept to use
From rest, distance grows as t², so time grows as √(distance). Half the distance takes 1/√2 of the total time, not half of it.
Formula or rule
s = ½at² ⇒ t ∝ √s
Baby steps
Let the total time be T for total distance s.
For s/2: t₁ = T/√2.
t₂ = T − T/√2 = T(1 − 1/√2).
t₁:t₂ = (1/√2) : (1 − 1/√2) = 1 : (√2 − 1).
Answer
1 : (√2 − 1)
Shortcut
√2 − 1 ≈ 0.41, so the second half takes less than half the time — which must be true, because the body is faster there. Use this to eliminate options instantly.
Trap
Answering 1 : 1. Equal distances take equal times only at constant velocity. Under acceleration the later half is always quicker.
Q20NEET-style
A body starts from rest, accelerates uniformly at 4 m s⁻² for 5 s, then decelerates uniformly at 8 m s⁻² until it stops. The total distance travelled is
(a) 50 m
(b) 62.5 m
(c) 100 m
(d) 75 m
Given
phase 1: u = 0, a = 4 m s⁻², t = 5 s; phase 2: deceleration 8 m s⁻² to rest
Asked
total distance
Concept to use
Two-phase problem. The peak velocity at the junction is the bridge between them: it is the final velocity of phase 1 and the initial velocity of phase 2.
Formula or rule
v = u + at ; s = ½at² ; s = v²/2a
Baby steps
Phase 1 peak velocity: v = 4 × 5 = 20 m s⁻¹.
Phase 1 distance: ½(4)(25) = 50 m.
Phase 2 distance: 20²/(2×8) = 400/16 = 25 m.
Total = 50 + 25 = 75 m (total time 5 + 2.5 = 7.5 s).
Answer
75 m
Shortcut
Draw the v–t triangle: base 7.5 s, height 20 m s⁻¹, area ½(7.5)(20) = 75 m. One triangle, one answer.
Trap
Applying one set of kinematic equations across both phases. The equations assume a single constant a — the moment the acceleration changes, you must start a fresh calculation.
Answer key
aQ1
dQ2
cQ3
cQ4
bQ5
cQ6
cQ7
dQ8
aQ9
cQ10
bQ11
aQ12
dQ13
aQ14
bQ15
dQ16
bQ17
aQ18
bQ19
dQ20
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.