Physics
Kinematics
Relative Velocity

Kinematics · JEE & NEET Physics

Relative Velocity: notes and previous year questions

How motion looks from something that is itself moving: in a line, in two dimensions, in the rain, across a river and in a wind.

16 JEE Main questions (2004–2026)
5 NEET questions (2000–2025)

Relative Velocity in short

The velocity of A as seen from B is $\vec{v}_A - \vec{v}_B$ : subtract the velocity of the one watching.
$\vec{v}_{BA} = -\vec{v}_{AB}$ : each sees the other at the same speed, in opposite directions.
Relative velocities chain: $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ .
Along one line, subtract speeds for the same direction and add them for opposite directions.

1The rule

Relative velocity is the velocity of one object as seen from another, moving, object. From the roadside, car A does 80 km/h and car B 60 km/h, both east. From inside B, A creeps past at only 20 km/h, and the roadside trees rush backward at 60 km/h.

\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

The velocity of A as seen from B: subtract the velocity of the one watching.

$\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = -\vec{v}_{AB}$ : same size, opposite direction.
$\vec{v}_{AB} + \vec{v}_{BA} = 0$ .
Chain rule: $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ .

Example: A at 80 km/h and B at 60 km/h, both east. $v_{AB} = +20$ km/h (A seems to move forward); $v_{BA} = -20$ km/h (B seems to move backward, west).

2In one line

Motion	Relative speed
Same direction	subtract: $\|v_A - v_B\|$
Opposite directions	add: $v_A + v_B$

Two trains approaching at 60 km/h each close in at $60 + 60 = 120$ km/h; in signs, $v_{AB} = 60 - (-60) = 120$ km/h. That is why head-on collisions are so violent.

3In two dimensions

When the velocities point in different directions, subtract them as vectors: add $-\vec{v}_B$ to $\vec{v}_A$ . For an angle $\theta$ between them:

|\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 - 2v_Av_B\cos\theta}

$\theta = 0°$ (same direction): $|v_A - v_B|$ .
$\theta = 180°$ (opposite): $v_A + v_B$ .
$\theta = 90°$ (perpendicular): $\sqrt{v_A^2 + v_B^2}$ .

4Rain and an umbrella

Rain falls straight down; a person walks. The rain's velocity relative to the walker is

\vec{v}_{rm} = \vec{v}_r - \vec{v}_m

For vertical rain $v_r$ and a horizontal walk $v_m$ , the rain seems to come from the front:

|\vec{v}_{rm}| = \sqrt{v_r^2 + v_m^2}

\tan\alpha = \frac{v_m}{v_r}

α is measured from the vertical; tilt the umbrella forward by α.

5Crossing a river

A boat with speed $v_b$ in still water crosses a river of width $d$ flowing at $v_r$ . Its velocity over the ground is the boat's velocity plus the river's.

Goal	Heading	Time	Drift
Shortest time	straight across	$d/v_b$	$v_r d/v_b$
Shortest path	upstream at $\sin^{-1}(v_r/v_b)$	$d/\sqrt{v_b^2 - v_r^2}$	0

Shortest time: all of the boat's speed goes into crossing, but the current carries it downstream.
Shortest path: head upstream so that $v_b\sin\theta = v_r$ cancels the current; the crossing speed is then $\sqrt{v_b^2 - v_r^2}$ .

6Flying in a wind

An aircraft with air velocity $\vec{v}_a$ in a wind $\vec{v}_w$ moves over the ground at

\vec{v}_g = \vec{v}_a + \vec{v}_w

To reach a destination directly, the pilot heads so that the resultant points at the destination, exactly like the boat aiming upstream.

Summary

Key ideas

The velocity of A as seen from B is $\vec{v}_A - \vec{v}_B$ : subtract the velocity of the one watching.
$\vec{v}_{BA} = -\vec{v}_{AB}$ : each sees the other at the same speed, in opposite directions.
Relative velocities chain: $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ .
Along one line, subtract speeds for the same direction and add them for opposite directions.
Meeting and chase problems: time = gap ÷ relative speed.
In two dimensions, subtract velocities as vectors; perpendicular velocities give $\sqrt{v_A^2 + v_B^2}$ .
Walking in vertical rain, the rain seems to come from the front: tilt the umbrella forward by $\alpha$ with $\tan\alpha = v_m / v_r$ .
To make slanted rain look vertical, walk at the rain's horizontal velocity.
Crossing a river in the shortest time: head straight across and accept a drift of $v_r d / v_b$ .
Crossing by the shortest path: head upstream with $v_b\sin\theta = v_r$ ; possible only if $v_b > v_r$ .
A plane's ground velocity is its air velocity plus the wind velocity.

Every equation

Relative velocity: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$
Reverse: $\vec{v}_{BA} = -\vec{v}_{AB}$
Chain: $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$
Any angle θ: $|\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 - 2v_Av_B\cos\theta}$
Rain for the walker: $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$
Its speed: $|\vec{v}_{rm}| = \sqrt{v_r^2 + v_m^2}$
Umbrella angle: $\tan\alpha = v_m / v_r$
River: shortest time: $t = d / v_b$
River: drift: $x = v_r d / v_b$
River: shortest path heading: $\sin\theta = v_r / v_b$
River: shortest path time: $t = \frac{d}{\sqrt{v_b^2 - v_r^2}}$
Plane in a wind: $\vec{v}_g = \vec{v}_a + \vec{v}_w$

Previous year questions with solutions

Real JEE and NEET questions on relative velocity. Try each one before you open the solution.

Q1JEE Main 2026One correct option

Two cars $A$ and $B$ are moving in the same direction along a straight line with speeds $100\ \mathrm{km/h}$ and $80\ \mathrm{km/h}$ respectively, such that car $A$ is moving ahead of car $B$ . A person in car $B$ throws a stone with a speed $v$ so that it hits car $A$ with a speed of $5\ \mathrm{m/s}$ . The value of $v$ is __________ $\mathrm{km/h}$ .

A $18$
B $28$
C $38$
D $48$

Show answer and solution

Answer: Option C

First make the units agree: $5\ \mathrm{m/s} = 5 \times 3.6 = 18\ \mathrm{km/h}$ , and that is the stone's speed relative to $A$ . The stone is thrown at $v$ relative to $B$ , so its ground speed is $80+v$ and its speed relative to $A$ is $(80+v)-100$ . Setting that to $18$ gives $v = 38\ \mathrm{km/h}$ .

Q2NEET 2025One correct option

Two cities $X$ and $Y$ are connected by a regular bus service with a bus leaving in either direction every $T$ minutes. A girl driving a scooty at $60\ \mathrm{km/h}$ in the direction $X$ to $Y$ notices that a bus goes past her every $30$ minutes in the direction of her motion, and every $10$ minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed of the buses.

A $10\ \mathrm{min}$ , $90\ \mathrm{km/h}$
B $15\ \mathrm{min}$ , $120\ \mathrm{km/h}$
C $9\ \mathrm{min}$ , $40\ \mathrm{km/h}$
D $25\ \mathrm{min}$ , $100\ \mathrm{km/h}$

Show answer and solution

Answer: Option B

Consecutive buses are a fixed distance $vT$ apart. Catching her from behind, they close at $v-60$ and take $30$ minutes: $(v-60)(30) = vT$ . Coming the other way they close at $v+60$ and take $10$ : $(v+60)(10) = vT$ . Equating gives $30v-1800 = 10v+600$ , so $v = 120\ \mathrm{km/h}$ , and then $T = 15$ minutes.

Q3JEE Main 2026One correct option

A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ______ and ______ respectively.

A20 s and 100 m
B40 s and 100 m
C40 s and 200 m
D40 s and 0 m

Show answer and solution

Answer: Option C

Minimum time means pointing the boat straight across on both legs. In metres per second, $18 \times \tfrac{5}{18} = 5\ \mathrm{m/s}$ for the river and $36 \times \tfrac{5}{18} = 10\ \mathrm{m/s}$ for the boat.

Each leg takes $\dfrac{200}{10} = 20\ \mathrm{s}$ , so the round trip takes $40\ \mathrm{s}$ . Option A gives the time for one leg only.

In each of those legs the current carries the boat $5 \times 20 = 100\ \mathrm{m}$ east. The boat's own velocity reverses for the return leg but the river's does not, so the second $100\ \mathrm{m}$ is east as well, and the displacement along the bank is $200\ \mathrm{m}$ .

Option D is the answer you reach by imagining the drift cancels itself on the way back, and it is the commonest mistake in this whole section.

Practice questions, easy to hard

Three questions from the relative velocity practice ladder: one easy, one medium, one hard.

Q4One or more correct options

Two bodies $A$ and $B$ move in a plane, each with its own velocity relative to the ground.

Select every statement that is true.

A $\vec{v}_{AB}$ and $\vec{v}_{BA}$ have the same magnitude and point opposite ways
BIf $A$ and $B$ have equal velocities, each sees the other at rest
CIf $A$ moves east at $3\ \mathrm{m/s}$ and $B$ moves north at $4\ \mathrm{m/s}$ , then $A$ moves at $5\ \mathrm{m/s}$ as seen from $B$
DThe speed of $A$ relative to $B$ can never be larger than the speed of $A$ relative to the ground

Show answer and solution

Answer: Options A, B, C

Swapping the frame reverses the subtraction, which reverses the vector and leaves its length alone — that is A. Equal velocities subtract to zero, so the gap between them never changes, which is what being at rest relative to somebody means — that is B. In C the two parts are $3$ east and $4$ south, closing at $\sqrt{3^{2}+4^{2}} = 5\ \mathrm{m/s}$ , and that one line also kills D: $5\ \mathrm{m/s}$ is larger than the $3\ \mathrm{m/s}$ the ground measures for $A$ .

Q5One or more correct options

A river of width $d$ flows at $v_r$ , and a boat does $v_b$ in still water, with $v_b > v_r$ .

Select every statement that is true.

AThe shortest crossing time is $\dfrac{d}{v_b}$ , whatever the current
BLanding directly opposite always takes longer than that shortest crossing time
CSteering upstream shortens the crossing time
DOn the quickest crossing the boat lands $\dfrac{v_r d}{v_b}$ downstream of the point opposite

Show answer and solution

Answer: Options A, B, D

The current has no component across the river, so it cannot touch the crossing time, and that time is smallest when the whole of $v_b$ is aimed across — A. Landing directly opposite spends part of $v_b$ on cancelling the current, leaving only $\sqrt{v_b^{2} - v_r^{2}}$ to go across, so it is always the slower crossing — B, and C is that same fact denied. D is the drift: the current's speed multiplied by the time the boat spends in the water.