One subtraction, six quantities deleted. Trains, chases, lifts, escalators — and the result that gravity cancels between any two falling bodies.
NCERT 2.5Priority 221 worked questions3 animated figuresExercises 2.13–2.15
1 · The concept in plain language
The one idea
Velocity is only ever measured with respect to something. Change what you measure it against and
the number changes — by simple subtraction.
Everything in this topic follows from one line:
vAB = vA − vB. Read it aloud as
“the velocity of A as seen from B”. In one dimension there are no components to resolve,
so it really is nothing more than a subtraction of two signed numbers — and the whole skill is
choosing your signs once and refusing to change them.
The reason the subtraction sometimes looks like an addition is the sign, not the physics. If B moves
the other way its velocity is negative, and subtracting a negative adds:
animated Same direction: the gap between the two cars closes slowly, at the difference of the speeds. Opposite directions: it closes at the sum. Both pictures come from the one subtraction — only the sign of the second velocity has changed.
Why change frames at all?
Because it deletes information you do not need. A question about two cars has two positions, two
velocities and possibly two accelerations — six quantities. Step into one car and three of them
become zero. What is left is a one-body problem, and you already know how to solve those.
NCERT sets this up in Exercise 2.14, which is the single best question in the chapter for showing what
the change of frame buys you:
animated Sit in car A and you no longer care how fast anything is moving over the ground. C is coming at 25 m/s through 1 km, so the deadline is 40 s. B is gaining at only 5 m/s and must gain 1000 m. One equation gives a = 1 m/s².
The result you will use most: gravity cancels
If two bodies are both in free fall, their relative acceleration is
g − g = 0, whatever their masses, whatever their initial velocities, whichever
way they were thrown. In each other’s frame they move in a straight line at constant speed.
That converts most two-body gravity questions into
time = separation ÷ relative speed — a single division, where the
direct method would need two quadratics.
Reading it off a graph
On a position–time plot, relative velocity is the difference of the two slopes and the
vertical gap between the curves is the relative displacement. The bodies meet where the curves cross.
animated Two objects on one x–t plot. The brass bar is the gap between them. It shrinks at a steady 12 m s⁻¹ — the difference of the slopes — reaches zero at t = 10 s where the lines cross, and then reopens on the other side as P pulls ahead.
A note on this section and your NCERT copy
The contents page of the 2026–27 reprint lists §2.5 Relative velocity, but the running
text jumps from Example 2.7 straight to the Summary — the section body is missing from that printing.
Exercises 2.13, 2.14 and 2.15 still require it, and NEET still asks it, which is why it is here as a
Priority 2 topic rather than dropped. Nothing on this page depends on the missing pages.
2 · Formula sheet
Definition
vAB = vA − vB
Read as ‘velocity of A as seen from B’. In 1D these are signed numbers.
Reversal
vBA = −vAB
Same magnitude, opposite sign. Always.
Same direction
|vA − vB|
The speeds subtract.
Opposite directions
vA + vB
The speeds add.
Relative position
xAB = xA − xB
They meet when this is zero.
Relative acceleration
aAB = aA − aB
Zero for any two bodies under gravity alone — the single most useful line on this sheet.
Kinematics in the relative frame
srel = urelt + ½arelt²
The three equations work unchanged, provided arel is constant.
Time to meet (approaching)
t = separation / relative speed
Only when arel = 0.
Train crossing a pole or a man
t = L / vrel
One length — the observer is a point.
Train crossing a train or a platform
t = (L₁ + L₂) / vrel
Both lengths.
Chained frames
vA,ground = vA,train + vtrain,ground
Inner labels match and cancel. Extends to any number of frames.
Two bodies dropped Δt apart
separation = g·Δt·(t − Δt/2)
The gap opens at the constant rate g·Δt.
Boat and stream
vdown = v + vs ; vup = v − vs
Round trip average speed = (v² − vs²)/v, always less than v.
Escalator / walkway
1/t = 1/t₁ + 1/t₂
Add reciprocals when the velocities add over the same distance.
Lift
arel = g + alift (lift going up)
An upward-accelerating lift acts like stronger gravity.
3 · Exceptions and traps
Trap — equal velocity is not the same as meeting
When one body chases another, the moment their velocities match is the moment the gap stops
growing. The chaser is at its furthest behind right then. Catching up means equal positions. Papers
put both times in the options.
Trap — opposite directions
“Opposite directions” means the speeds add, because you are subtracting a negative
number. Every year some candidates subtract the magnitudes instead and get an answer seven times too small.
Trap — the train has a length
Crossing a man or a pole: the train covers one length. Crossing a platform, a
bridge or another train: it covers the sum of the two lengths. Read the object being crossed
before choosing.
Trap — muzzle speed and launch speed are relative
A bullet’s muzzle speed is measured relative to the gun, and a ball dropped from a moving balloon
starts with the balloon’s velocity. Convert to the ground frame before you use any kinematic equation.
Exception — relative acceleration is only zero without air resistance
The elegant ‘gravity cancels’ result assumes free fall. Once drag matters it depends on shape,
speed and mass, so two bodies no longer share an acceleration and the relative motion stops being uniform.
Every NEET question of this type says ‘neglect air resistance’ — and it matters that it does.
Exception — this is Galilean, not universal
Simple subtraction of velocities is an approximation that holds when speeds are far below the speed of
light. NCERT flags this in Chapter 2’s Points to Ponder. It never affects a NEET numerical, but it
can appear as a statement in a conceptual question.
4 · Numbers to remember
Situation
Result
Two bodies under gravity alone
arel = 0 — relative motion is uniform
NCERT Exercise 2.14 (cars A, B, C)
deadline 40 s, amin = 1 m s⁻²
Same trains, same direction vs opposite
70 s vs 10 s
Escalator 60 s and 90 s together
36 s
Boat 10 km h⁻¹ in a 2 km h⁻¹ stream, round trip
9.6 km h⁻¹
Two stones dropped 1 s apart, g = 10
gap opens at a steady 10 m s⁻¹
Body from rest (a) chasing one at constant v
they meet again at t = 2v/a
Lift accelerating up at a
effective gravity g + a
km h⁻¹ → m s⁻¹
× 5/18 (54→15, 72→20, 90→25, 192→53.3)
5 · Scientists NEET names
Galileo Galilei (1564–1642)
The idea that the laws of motion look the same in every uniformly moving frame — and that you
cannot tell, from inside a smoothly sailing ship, whether you are moving — is Galilean
relativity, argued in his Dialogue Concerning the Two Chief World Systems (1632). The simple
subtraction of velocities you use here is called the Galilean transformation for that reason.
Albert Einstein (1879–1955)
Named only for the boundary: in 1905 he showed that velocity subtraction fails as speeds approach
c = 3 × 10⁸ m s⁻¹, where relative speeds combine so that
nothing ever exceeds c. NEET asks this only as a statement, never as a calculation.
6 · Twenty-one worked questions
These follow NCERT Exercises 2.13–2.15 and the standard
NEET treatment of the topic. No previous-year attributions are claimed here — the year labels for
relative-velocity items vary too much between question banks to state confidently.
Q1NEET-style
On a two-lane road, car A travels at 36 km h⁻¹. Two cars B and C approach A from opposite directions, each at 54 km h⁻¹. At an instant when AB = AC = 1 km, B decides to overtake A before C does. The minimum acceleration of B required to avoid an accident is
(a) 0.5 m s⁻²
(b) 2 m s⁻²
(c) 4 m s⁻²
(d) 1 m s⁻²
Given
vA = 10 m s⁻¹, vB = vC = 15 m s⁻¹, AB = AC = 1000 m; B is behind A, C is ahead coming the other way
Asked
minimum acceleration of B
Concept to use
Sit inside car A and let everything else move relative to you. C’s arrival fixes a deadline; B must cover 1000 m of relative displacement inside that deadline.
Formula or rule
t = d/(vC + vA) ; then srel = urelt + ½a t²
Baby steps
C approaches A head-on, so relative speed = 15 + 10 = 25 m s⁻¹. Deadline t = 1000/25 = 40 s.
B chases A in the same direction, so urel = 15 − 10 = 5 m s⁻¹.
B must gain 1000 m relative to A: 1000 = 5(40) + ½a(40)².
1000 = 200 + 800a ⇒ a = 1 m s⁻².
Answer
1 m s⁻²
Shortcut
Two relative speeds, one deadline. Head-on adds (25), same-direction subtracts (5). Everything after that is one equation.
Trap
Using ground speeds and asking ‘where is B when C reaches A?’ — it works, but takes three times as long and invites sign errors. Also note only B accelerates; A and C stay uniform. This is NCERT Exercise 2.14.
Q2NEET-style
Two trains, 150 m and 200 m long, run on parallel tracks in the same direction at 72 km h⁻¹ and 54 km h⁻¹. The time the faster train takes to completely overtake the slower one is
(a) 70 s
(b) 10 s
(c) 14 s
(d) 140 s
Given
L₁ = 150 m, L₂ = 200 m, v₁ = 20 m s⁻¹, v₂ = 15 m s⁻¹, same direction
Asked
time to completely overtake
Concept to use
‘Completely overtake’ means the front of the faster train travels from the rear of the slower train to a point one combined length ahead. In the slower train’s frame it must cover L₁ + L₂.
Formula or rule
t = (L₁ + L₂)/|v₁ − v₂| for the same direction
Baby steps
Convert: 72 × 5/18 = 20, 54 × 5/18 = 15 m s⁻¹.
Relative speed = 20 − 15 = 5 m s⁻¹.
Relative distance = 150 + 200 = 350 m.
t = 350/5 = 70 s.
Answer
70 s
Shortcut
Crossing problems are always (sum of lengths)/(relative speed). Only the relative speed changes between the two cases: subtract for same direction, add for opposite.
Trap
Using one length instead of the sum. The train is not a point — its own length has to clear too. Note also that 70 s is seven times the opposite-direction answer, which is why same-direction overtaking on a highway is so dangerous.
Q3NEET-style
The same two trains (150 m and 200 m, at 72 km h⁻¹ and 54 km h⁻¹) now approach each other on parallel tracks. The time they take to cross each other completely is
(a) 14 s
(b) 35 s
(c) 10 s
(d) 70 s
Given
L₁ = 150 m, L₂ = 200 m, v₁ = 20 m s⁻¹, v₂ = 15 m s⁻¹, opposite directions
Asked
crossing time
Concept to use
Opposite directions means the velocities have opposite signs, so subtracting them adds the speeds.
Formula or rule
t = (L₁ + L₂)/(v₁ + v₂) for opposite directions
Baby steps
Relative speed = 20 − (−15) = 35 m s⁻¹.
Relative distance = 350 m.
t = 350/35 = 10 s.
Answer
10 s
Shortcut
Same numbers as the previous question, one sign flipped. Solve them as a pair and the contrast sticks: 70 s versus 10 s.
Trap
Averaging the two speeds instead of adding them. Relative velocity is a difference of vectors, never an average.
Q4NEET-style
A ball is thrown vertically upward from the ground at 20 m s⁻¹. At the same instant another ball is released from rest from a height of 40 m directly above. Taking g = 10 m s⁻², they meet after
(a) 2 s
(b) 1 s
(c) 2.5 s
(d) 4 s
Given
uA = +20 m s⁻¹ (up), uB = 0, initial separation 40 m, both under gravity
Asked
time at which they meet
Concept to use
Both bodies have the same acceleration −g, so the relative acceleration is zero. In each other’s frame they move with constant velocity, and the problem collapses to distance ÷ speed.
Formula or rule
aAB = aA − aB = 0 ; t = srel/urel
Baby steps
uAB = 20 − 0 = 20 m s⁻¹, directed upward.
aAB = (−g) − (−g) = 0 — gravity cancels out.
So A approaches B at a steady 20 m s⁻¹ through a gap of 40 m.
t = 40/20 = 2 s.
Check the height: A is at 20(2) − 5(2)² = 20 m; B is at 40 − 5(2)² = 20 m. ✓
Answer
2 s
Shortcut
Whenever two bodies are in free fall together, delete gravity. Relative motion is uniform, so t = separation/relative speed in one line.
Trap
Writing two full quadratics and equating them. It gives the same answer but the ½gt² terms were always going to cancel — recognising that in advance is the whole point of the topic.
Q5NEET-style
Two stones are dropped from the top of a tower, the second one exactly 1 s after the first. Taking g = 10 m s⁻², the separation between them 3 s after the first stone is released is
(a) 5 m
(b) 10 m
(c) 20 m
(d) 25 m
Given
both released from rest, time lag T = 1 s, t = 3 s after the first
Asked
separation at t = 3 s
Concept to use
Relative acceleration is zero, so after the second stone is released the gap grows at a constant rate equal to the speed the first stone had already built up in the lag.
Formula or rule
Δs = ½gt² − ½g(t − T)² = gT(t − T/2)
Baby steps
First stone: ½(10)(3)² = 45 m.
Second stone has fallen for 2 s: ½(10)(2)² = 20 m.
Separation = 45 − 20 = 25 m.
Or directly: gT(t − T/2) = 10(1)(2.5) = 25 m.
Answer
25 m
Shortcut
The gap opens at a constant rate gT = 10 m s⁻¹ once both are falling — because their relative acceleration is zero. So the separation is just 10 × (time since the second was dropped) + 5 m.
Trap
Assuming the separation grows faster and faster because both are speeding up. They are, but equally, so the gap opens uniformly. Papers plant options such as 45 m and 20 m — the two individual distances — to catch a half-finished solution.
Q6NEET-style
A man can climb a stationary escalator in 90 s. The same escalator, when moving, carries a standing man to the top in 60 s. The time taken if he walks up the moving escalator is
(a) 36 s
(b) 30 s
(c) 45 s
(d) 75 s
Given
tman = 90 s, tescalator = 60 s, same flight of steps
Asked
time when both act together
Concept to use
Velocities relative to the ground add. Since the distance is the same in all three cases, work with rates — fractions of the escalator covered per second.
Let the escalator length be L. Man’s speed = L/90; escalator’s speed = L/60.
Walking on the moving escalator, his ground speed is the sum: L/90 + L/60 = L(2 + 3)/180 = L/36.
t = L ÷ (L/36) = 36 s.
Answer
36 s
Shortcut
Add reciprocals, never times. The combined time must be shorter than both individual times — 36 s < 60 s, so the answer is plausible before you even check it.
Trap
Averaging 60 and 90 to get 75 s. That option is always present. Adding times gives 150 s, which is absurd — help arriving cannot make a journey longer. If he walked down the up-escalator you would subtract instead: 1/60 − 1/90 gives 180 s.
Q7NEET-style
A motorboat whose speed in still water is 10 km h⁻¹ travels 24 km downstream and then returns. The stream flows at 2 km h⁻¹. The average speed of the boat for the whole trip is
(a) 8 km h⁻¹
(b) 10 km h⁻¹
(c) 12 km h⁻¹
(d) 9.6 km h⁻¹
Given
vboat = 10 km h⁻¹, vstream = 2 km h⁻¹, d = 24 km each way
Asked
average speed for the round trip
Concept to use
Downstream the velocities add; upstream they subtract. Equal distances at two different speeds means the average speed is the harmonic mean, not the arithmetic one.
Formula or rule
vdown = v + vs, vup = v − vs, v̄ = 2vdownvup/(vdown + vup)
Baby steps
Downstream = 12 km h⁻¹, upstream = 8 km h⁻¹.
Times: 24/12 = 2 h and 24/8 = 3 h.
Total distance 48 km in 5 h.
v̄ = 48/5 = 9.6 km h⁻¹.
Answer
9.6 km h⁻¹
Shortcut
v̄ = (v² − vs²)/v = (100 − 4)/10 = 9.6. The distance never enters — it cancels.
Trap
Answering 10 km h⁻¹, on the reasoning that the current helps as much as it hinders. It does not: you spend longer in the slow leg, so the slow speed carries more weight. The average is always below the still-water speed.
Q8NEET-style
A thief’s car passes a police station at a steady 10 m s⁻¹. A police jeep starts from rest 7.5 s later with a constant acceleration of 2 m s⁻². Measured from the moment the jeep starts, the thief is caught after
(a) 5 s
(b) 10 s
(c) 20 s
(d) 15 s
Given
vthief = 10 m s⁻¹ constant, jeep: u = 0, a = 2 m s⁻², head start 7.5 s
Asked
time to catch, measured from the jeep’s start
Concept to use
‘Caught’ means equal positions, not equal velocities. Convert the time lead into a distance lead first, then set the two displacements equal.
Formula or rule
½a t² = d₀ + v t
Baby steps
Head start in distance: 10 × 7.5 = 75 m.
Jeep: x = ½(2)t² = t². Thief: x = 75 + 10t.
t² − 10t − 75 = 0 ⇒ (t − 15)(t + 5) = 0.
t = 15 s. Both have then covered 225 m. ✓
Answer
15 s
Shortcut
In the thief’s frame the jeep starts 75 m behind at −10 m s⁻¹ with a = 2 m s⁻²: 75 = −10t + t² — the same quadratic, reached without writing two position functions.
Trap
Setting the velocities equal. The jeep matches the thief’s speed at t = 5 s, and at that moment it is still 100 m behind — that is when the gap stops growing, not when it closes. Papers put 5 s in the options for exactly this reason.
Q9NEET-style
Two cars are 100 km apart on a straight road and drive towards each other at 50 km h⁻¹ each. A bird starts from one car at the same instant, flies at 80 km h⁻¹ to the other car, turns instantly, and keeps shuttling until the cars meet. The total distance flown by the bird is
(a) 40 km
(b) 80 km
(c) 60 km
(d) 160 km
Given
d = 100 km, both cars 50 km h⁻¹ towards each other, bird 80 km h⁻¹
Asked
total path length flown by the bird
Concept to use
Do not sum the infinite series of legs. The bird flies for exactly as long as the cars take to meet, and its speed never changes — so distance = speed × time.
Formula or rule
t = d/(v₁ + v₂) ; s = vbird t
Baby steps
Cars approach at 50 + 50 = 100 km h⁻¹.
t = 100/100 = 1 h.
Bird flies for that whole hour at 80 km h⁻¹.
s = 80 × 1 = 80 km.
Answer
80 km
Shortcut
Ask ‘how long?’ before ‘how far?’. The number of turns is irrelevant, and each turn is instantaneous, so the speed is 80 km h⁻¹ for the entire hour.
Trap
Trying to add the geometric series of shorter and shorter legs. It converges to the same 80 km, but it is a five-minute detour in an exam where you have roughly one minute. The story about von Neumann solving this by summing the series is worth remembering as a warning.
Q10NEET-style
A coin is dropped from rest inside a lift that is accelerating upward at 2 m s⁻². The coin falls 2 m to the floor of the lift. Taking g = 10 m s⁻², the time it takes is nearest to
(a) 0.58 s
(b) 0.45 s
(c) 0.63 s
(d) 0.71 s
Given
h = 2 m relative to the lift floor, alift = +2 m s⁻² (up), g = 10 m s⁻²
Asked
time for the coin to reach the lift floor
Concept to use
The question is asked in the lift’s frame, so use the relative acceleration of the coin with respect to the lift. Taking down as positive, the floor accelerates upward to meet the coin.
Relative to the lift the coin starts from rest, so 2 = ½(12)t².
t² = 1/3 ⇒ t = 0.577 s.
Answer
0.58 s
Shortcut
An upward-accelerating lift behaves like stronger gravity, g + a; a downward-accelerating one like weaker gravity, g − a. In free fall (a = g) the coin never lands.
Trap
Using g = 10 alone, which gives 0.63 s — that option is deliberately placed. The 2 m is measured inside the lift, so the acceleration must be measured there too. Note the coin’s initial velocity relative to the lift is zero even though both are moving fast relative to the ground.
Q11NEET-style
A particle P starts from x = 20 m and moves at a constant 15 m s⁻¹. Another particle Q starts from x = 140 m at the same instant and moves at a constant 3 m s⁻¹ in the same direction. The velocity of P relative to Q, and the time at which P overtakes Q, are
(a) 12 m s⁻¹, 8 s
(b) 18 m s⁻¹, 10 s
(c) 12 m s⁻¹, 10 s
(d) 18 m s⁻¹, 6.7 s
Given
xP = 20 + 15t, xQ = 140 + 3t
Asked
vPQ and the overtaking time
Concept to use
On a position–time graph the relative velocity is the difference of the slopes, and overtaking is where the two lines intersect.
Formula or rule
vPQ = vP − vQ ; t = (xQ0 − xP0)/vPQ
Baby steps
vPQ = 15 − 3 = 12 m s⁻¹.
Initial gap = 140 − 20 = 120 m.
t = 120/12 = 10 s.
Check: xP = 20 + 150 = 170 m, xQ = 140 + 30 = 170 m. ✓
Answer
12 m s⁻¹ and 10 s
Shortcut
Relative position = 120 − 12t. Set it to zero. One line instead of two.
Trap
Adding the speeds to get 18 m s⁻¹. Both particles move the same way, so the velocities subtract. Adding only applies when the signs are opposite.
Q12NEET-style
Two stones are projected vertically upward from the same point with the same speed 20 m s⁻¹, the second one 2 s after the first. Taking g = 10 m s⁻², they meet at a height of
(a) 15 m
(b) 10 m
(c) 20 m
(d) 25 m
Given
u = 20 m s⁻¹ for both, lag = 2 s, g = 10 m s⁻²
Asked
height at which they cross
Concept to use
Both are under gravity alone, so relative acceleration is zero: the second stone sees the first moving away at a constant relative speed. Meeting means equal heights.
Formula or rule
y = ut − ½gt² for each, then equate
Baby steps
First stone at time t: 20t − 5t².
Second stone has been in flight for (t − 2): 20(t − 2) − 5(t − 2)².
Equate. The t² terms cancel, leaving 0 = −60 + 20t ⇒ t = 3 s.
Height = 20(3) − 5(9) = 60 − 45 = 15 m.
Answer
15 m
Shortcut
The relative velocity of the first with respect to the second is a constant g×2 = 20 m s⁻¹ downward once both are flying, and the head start in position is 20 m, giving 1 s after the second launch, i.e. t = 3 s.
Trap
Forgetting that the second stone’s clock reads (t − 2), not t. Also check that 3 s is before the first stone lands (it lands at 4 s) — otherwise the ‘meeting’ is fictional.
Q13NEET-style
A particle A starts from rest with a constant acceleration of 4 m s⁻². At the same instant and from the same point, particle B moves with a constant velocity of 10 m s⁻¹ in the same direction. They are together again at
(a) 2.5 s
(b) 10 s
(c) they never meet again
(d) 5 s
Given
A: u = 0, a = 4 m s⁻²; B: v = 10 m s⁻¹, a = 0; same start point and instant
Asked
time when they coincide again
Concept to use
Work in B’s frame: A starts from rest there too, but with the full relative acceleration. Relative displacement must return to zero.
Formula or rule
srel = urelt + ½arelt² = 0
Baby steps
uAB = 0 − 10 = −10 m s⁻¹; aAB = 4 − 0 = 4 m s⁻².
0 = −10t + ½(4)t² = t(2t − 10).
t = 0 (the start) or t = 5 s.
Answer
5 s
Shortcut
For a body starting from rest against one at constant speed v, they meet again at t = 2v/a — because A’s average velocity over the interval must equal v, and its average is half its final value.
Trap
Choosing 2.5 s, which is when their velocities match (4t = 10). At that instant A is at its furthest behind. Equal velocity is the turning point of the gap; equal position is the meeting.
Q14NEET-style
The velocity of body A with respect to body B is 12 m s⁻¹ due east. The velocity of B with respect to A is
(a) 12 m s⁻¹ due east
(b) 12 m s⁻¹ due west
(c) 24 m s⁻¹ due west
(d) zero
Given
vAB = 12 m s⁻¹ east
Asked
vBA
Concept to use
Relative velocity is a difference of vectors, so swapping the two bodies reverses the sign but keeps the magnitude.
Formula or rule
vBA = −vAB
Baby steps
vAB = vA − vB.
vBA = vB − vA = −(vA − vB).
Same size, opposite direction: 12 m s⁻¹ due west.
Answer
12 m s⁻¹ due west
Shortcut
Think of two people on a train: each sees the other receding at the same rate, in opposite directions. The magnitudes must agree or the two observers would disagree about how fast the gap changes.
Trap
Doubling the magnitude to 24. Nothing is being added here — only re-labelled. The same reversal applies to relative acceleration and relative position.
Q15NEET-style
A man running at 5 m s⁻¹ finds that raindrops appear to fall vertically. Another man standing still finds the raindrops falling at 13 m s⁻¹ along the vertical. The horizontal component of the rain’s actual velocity is
(a) 5 m s⁻¹ in the direction of the man’s motion
(b) zero
(c) 5 m s⁻¹ opposite to the man’s motion
(d) 13 m s⁻¹ horizontally
Given
vman = 5 m s⁻¹ horizontal; rain appears vertical to him
Asked
horizontal component of the rain’s true velocity
Concept to use
‘Appears vertical’ means the horizontal part of vrain − vman is zero. So the rain must already carry the man’s horizontal velocity.
Formula or rule
vrain,man = vrain − vman; set the horizontal component to zero
Baby steps
Horizontally: vrain,x − 5 = 0.
So vrain,x = 5 m s⁻¹, in the same direction he runs.
The standing man sees only the vertical component, 13 m s⁻¹, unchanged by the other man’s motion.
Answer
5 m s⁻¹ in the direction of the man’s motion
Shortcut
If a moving observer sees something fall straight down, that thing shares the observer’s horizontal velocity exactly.
Trap
Reversing the direction. The classical umbrella problem — where you tilt the umbrella forwards — is the case where the rain falls vertically to the ground observer. Read carefully which observer sees the vertical fall; the two setups give opposite answers.
Q16NEET-style
A police van moving at 30 m s⁻¹ fires a bullet at a thief’s car speeding away at 192 km h⁻¹. The muzzle speed of the bullet is 150 m s⁻¹. The speed with which the bullet strikes the thief’s car is
(a) 105 m s⁻¹
(b) 150 m s⁻¹
(c) 127 m s⁻¹
(d) 180 m s⁻¹
Given
vvan = 30 m s⁻¹, muzzle speed = 150 m s⁻¹ (relative to the van), vthief = 192 km h⁻¹
Asked
speed of the bullet relative to the thief’s car
Concept to use
Muzzle speed is measured relative to the gun. Add the van’s velocity to get the ground speed, then subtract the car’s velocity to get the impact speed.
Bullet relative to the ground: 150 + 30 = 180 m s⁻¹.
Relative to the car: 180 − 53.33 = 126.7 ≈ 127 m s⁻¹.
Answer
127 m s⁻¹
Shortcut
Two steps, both additions of vectors: gun’s frame → ground → target’s frame. Equivalently 150 + 30 − 53.3 in one line.
Trap
Forgetting to add the van’s 30 m s⁻¹ because ‘the bullet leaves at 150’. It leaves at 150 relative to the van. This is NCERT Exercise 2.13 in substance.
Q17NEET-style
Two parallel rail tracks run north–south. Train A moves north at 54 km h⁻¹ and train B moves south at 90 km h⁻¹. A monkey runs on the roof of A, against its motion, at 18 km h⁻¹ relative to A. The velocity of the monkey as observed by a man standing on the ground is
(a) 10 m s⁻¹ south
(b) 10 m s⁻¹ north
(c) 20 m s⁻¹ north
(d) 35 m s⁻¹ south
Given
vA = +15 m s⁻¹ (north), monkey at 5 m s⁻¹ relative to A, against A’s motion
Asked
monkey’s ground velocity
Concept to use
The monkey’s velocity relative to the ground is its velocity relative to the train plus the train’s velocity relative to the ground. Train B is a distractor — it never touches this part.
Formula or rule
vmonkey,ground = vmonkey,A + vA,ground
Baby steps
Take north as positive. vA = 54 × 5/18 = +15 m s⁻¹.
Monkey runs against A: vmonkey,A = −5 m s⁻¹.
vmonkey,ground = −5 + 15 = +10 m s⁻¹, i.e. north.
Answer
10 m s⁻¹ north
Shortcut
Chain the subscripts: monkey→A, A→ground. The inner labels match and cancel, leaving monkey→ground. The same trick works for any number of nested frames.
Trap
Including train B’s 25 m s⁻¹ somewhere. NEET routinely supplies a third body that is irrelevant to the part being asked — identify the two frames named in the question and ignore the rest. This is NCERT Exercise 2.15 in substance.
Q18NEET-style
Two bodies are held at the same height and released, one 1 s after the other. As time passes, the relative velocity of the first with respect to the second
(a) increases linearly with time
(b) decreases with time
(c) stays constant
(d) increases as the square of the time
Given
both in free fall, released 1 s apart
Asked
behaviour of their relative velocity
Concept to use
Relative acceleration is the difference of two identical accelerations, hence zero. Zero relative acceleration means constant relative velocity.
Formula or rule
a12 = a₁ − a₂ = g − g = 0
Baby steps
At any moment after both are falling, v₁ = gt and v₂ = g(t − 1).
v₁ − v₂ = g(1) = 9.8 m s⁻¹ — no t in it.
So the relative velocity is constant, while the relative displacement grows linearly.
Answer
stays constant
Shortcut
Free fall wipes out relative acceleration for every pair of bodies, regardless of mass, initial speed or direction. This single fact solves most two-body gravity questions in one step.
Trap
Confusing relative velocity (constant) with relative displacement (growing linearly). Both individual speeds do increase — but equally, so the difference does not. Note this holds only while air resistance is neglected; with drag the heavier or faster body behaves differently and the symmetry breaks.
Q19NEET-style
A train 120 m long is moving at 20 m s⁻¹. A boy on the platform notes that it takes 8 s for the train to pass him completely. The boy is
(a) stationary
(b) walking at 5 m s⁻¹ opposite to the train
(c) walking at 5 m s⁻¹ in the direction of the train
(d) running at 15 m s⁻¹ in the direction of the train
Given
L = 120 m, vtrain = 20 m s⁻¹, observed crossing time = 8 s
Asked
the boy’s state of motion
Concept to use
For a point observer, the train must cover only its own length relative to the observer. The observed time therefore fixes the relative speed, and from that the observer’s velocity.
Formula or rule
t = L/vrel
Baby steps
vrel = 120/8 = 15 m s⁻¹.
But the train’s ground speed is 20 m s⁻¹, larger than 15.
So the boy must be moving with the train: 20 − vboy = 15 ⇒ vboy = 5 m s⁻¹, in the train’s direction.
Answer
walking at 5 m s⁻¹ in the direction of the train
Shortcut
A relative speed smaller than the train’s own speed means the observer is moving the same way; larger means the opposite way. Check that inequality before doing any algebra.
Trap
Assuming the boy is stationary because ‘he is on a platform’. A stationary observer would have measured 120/20 = 6 s, not 8 s. The extra 2 s is the clue the question is built on.
Q20NEET-style
Car A moves at a constant 20 m s⁻¹. Car B, 200 m behind, is moving at 10 m s⁻¹ and accelerates uniformly. To just catch A after 20 s, B’s acceleration must be
(a) 1 m s⁻²
(b) 2 m s⁻²
(c) 1.5 m s⁻²
(d) 3 m s⁻²
Given
gap = 200 m, vA = 20 m s⁻¹ constant, uB = 10 m s⁻¹, t = 20 s
Asked
acceleration of B
Concept to use
In A’s frame, B must close 200 m in 20 s, starting with a relative velocity of 10 − 20 = −10 m s⁻¹ — it is actually falling further behind at first.
Formula or rule
srel = urelt + ½arelt²
Baby steps
Take B’s direction as positive. Relative displacement needed = +200 m.
urel = 10 − 20 = −10 m s⁻¹, arel = a.
200 = (−10)(20) + ½a(400).
200 = −200 + 200a ⇒ a = 2 m s⁻².
Answer
2 m s⁻²
Shortcut
Relative frame turns a two-car problem into a one-body problem: a body 200 m from its target, thrown the wrong way at 10 m s⁻¹, needing acceleration a to arrive in 20 s.
Trap
Dropping the minus sign on urel, which gives 0 and then a = 1. B is slower than A at the start, so the gap widens before it closes — the acceleration has to pay back that loss as well as cover the original 200 m.
Q21NEET-style
Which of the following statements about relative velocity in one dimension is incorrect?
(a) If two bodies move with equal velocities, their relative velocity is zero
(b) If two bodies move in opposite directions, their relative speed is the difference of their speeds
(c) The relative velocity of A with respect to B has the same magnitude as that of B with respect to A
(d) Two bodies in free fall have zero relative acceleration
Given
four statements about one-dimensional relative motion
Asked
the incorrect one
Concept to use
Relative velocity is a vector difference. Opposite directions carry opposite signs, so subtracting them adds the magnitudes.
Formula or rule
vAB = vA − vB
Baby steps
Equal velocities give vA − vB = 0 — correct, and this is why passengers on the same train appear at rest to one another.
Swapping the bodies only flips the sign — correct.
g − g = 0 — correct.
Opposite directions: vA − (−vB) = vA + vB, a sum. So this statement is the incorrect one.
Answer
If two bodies move in opposite directions, their relative speed is the difference of their speeds
Shortcut
Same direction → subtract. Opposite directions → add. The words in the option have them the wrong way round.
Trap
Reading ‘incorrect’ as ‘correct’ and picking the first true statement you meet. Circle the word incorrect in the question paper before you look at the options — NEET sets several of these per paper.
Answer key
dQ1
aQ2
cQ3
aQ4
dQ5
aQ6
dQ7
dQ8
bQ9
aQ10
cQ11
aQ12
dQ13
bQ14
aQ15
cQ16
bQ17
cQ18
cQ19
bQ20
bQ21
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.