Relative Velocity Concept for Class 11, JEE and NEET: Formulas, River-Boat, Rain-Man and Closest Approach

 Relative velocity is part of kinematics. The one-dimensional case belongs to Motion in a Straight Line (Chapter 2 of NCERT Class 11 Physics), and the two-dimensional case uses the vectors of Motion in a Plane (Chapter 3). The JEE Main 2026 syllabus lists relative velocity in Unit 2, Kinematics. These notes give you the method once, and then show it in every problem type that exams use.

Relative Velocity Concept: Chapter Overview and Syllabus

Check the latest syllabi of CBSE, NTA (JEE Main) and NMC (NEET) for the exact lists.

Tip from the eSaral Physics team: For every relative motion problem, first name the observer (the object whose point of view you take), and then subtract the observer’s velocity from the velocity of the other object. If you do this one step first, the sign mistakes mostly disappear.

What Is Relative Velocity? in Relative Velocity Concept

Every velocity is measured with respect to a frame of reference. There is no absolute velocity. When we say “a car moves at 60 km/h”, we mean with respect to the ground. The relative velocity of A with respect to B is the velocity of A as seen by an observer who is moving with B:

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

where $\vec{v}_A$ and $\vec{v}_B$ are the velocities of A and B measured in the same frame (usually the ground). Two results follow:

• $\vec{v}_{BA} = -\vec{v}_{AB}$. The velocity of B as seen from A is the opposite of the velocity of A as seen from B.
• If you change to another frame that is not accelerating, both $\vec{v}_A$ and $\vec{v}_B$ change by the same amount, so $\vec{v}_{AB}$ stays the same in every such frame.

Everyday example: you sit in a train moving at 60 km/h, and another train passes at 80 km/h in the same direction. It seems to move at only $80 - 60 = 20$ km/h relative to you.

Relative Velocity in One Dimension in Relative Velocity Concept

Take one direction as positive and write each velocity with its sign.

Relative acceleration and displacement:

$$a_{AB} = a_A - a_B, \qquad x_{AB} = x_A - x_B$$

If the relative acceleration is constant, the equations of motion hold for the relative quantities: $v_{AB} = u_{AB} + a_{AB}t$ and $s_{AB} = u_{AB}t + \tfrac{1}{2}a_{AB}t^2$. For the equations themselves, see eSaral’s Motion in a Straight Line JEE notes.

Meeting, overtaking and crossing:

• Two bodies approaching from a distance $d$ meet after $t = \dfrac{d}{v_{rel}}$.
• A train of length $L_1$ crosses a train (or a pole, with $L_2 = 0$) of length $L_2$ in $t = \dfrac{L_1 + L_2}{v_{rel}}$, where $v_{rel}$ is the difference of the speeds (same direction) or the sum (opposite directions).

Relative Velocity in Two Dimensions in Relative Velocity Concept

In two dimensions, subtract the velocities as vectors. Take the components along the $x$ and $y$ axes:

$$\vec{v}_{AB} = (v_{Ax} - v_{Bx})\,\hat{i} + (v_{Ay} - v_{By})\,\hat{j}$$

If $\vec{v}_A$ and $\vec{v}_B$ have magnitudes $v_A$ and $v_B$ and the angle between them is $\theta$:

$$\lvert \vec{v}_{AB} \rvert = \sqrt{v_A^{2} + v_B^{2} - 2\,v_A v_B \cos\theta}$$

For $\theta = 0^\circ$ (the same direction), this gives $v_A - v_B$, and for $\theta = 180^\circ$ (opposite directions), it gives $v_A + v_B$. The direction of $\vec{v}_{AB}$ is found from its components. Revise vector subtraction in eSaral’s vectors and two-dimensional motion notes if you need it.

Chain rule for relative velocities: for three objects (or frames) A, B and C:

$$\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$$

This is the equation behind river-boat and aeroplane-wind problems, because it adds the velocity of the object with respect to a medium to the velocity of the medium with respect to the ground.

River-Boat Problems in Relative Velocity Concept

Let $v_{br}$ be the speed of the boat in still water, $v_r$ the speed of the river and $d$ the width of the river. The velocity of the boat with respect to the ground is $\vec{v}_{bg} = \vec{v}_{br} + \vec{v}_{r}$. If the boat heads at an angle $\theta$ upstream from the perpendicular to the bank:

$$t = \frac{d}{v_{br}\cos\theta}, \qquad x_{\text{drift}} = \big(v_r - v_{br}\sin\theta\big)\,t$$

If $v_r > v_{br}$, the boat cannot reach the point directly opposite. The time to cross depends only on the component of the boat’s velocity perpendicular to the bank, so the river does not change it.

Rain-Man Problems in Relative Velocity Concept

The velocity of the rain with respect to the man is:

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

For vertical rain of speed $v_r$ and a man walking horizontally at $v_m$, the rain seems to come at an angle $\theta$ from the vertical, in the direction opposite to the man’s motion, so he must tilt the umbrella forward:

$$\tan\theta = \frac{v_m}{v_r}, \qquad \lvert \vec{v}_{rm} \rvert = \sqrt{v_r^{2} + v_m^{2}}$$

If the rain falls with speed $v_r$ at an angle $\alpha$ from the vertical, blown in the direction of the man’s motion, the horizontal and vertical components of $\vec{v}_{rm}$ are $v_r\sin\alpha - v_m$ and $v_r\cos\alpha$. Then:

$$\tan\theta = \frac{\lvert v_r\sin\alpha - v_m \rvert}{v_r\cos\alpha}$$

Aeroplane and Wind Problems in Relative Velocity Concept

Use the chain rule: $\vec{v}_{ag} = \vec{v}_{aw} + \vec{v}_{wg}$, where $\vec{v}_{aw}$ is the velocity of the aeroplane with respect to the air (its airspeed and heading), $\vec{v}_{wg}$ is the velocity of the wind with respect to the ground, and $\vec{v}_{ag}$ is the velocity of the aeroplane with respect to the ground. The method is the same as for a boat in a river.

Relative Motion of Two Projectiles in Relative Velocity Concept

Two projectiles that move under gravity alone have the same acceleration, $\vec{g}$. So their relative acceleration is zero, and their relative velocity is constant:

$$\vec{a}_{AB} = \vec{g} - \vec{g} = 0, \qquad \vec{v}_{AB} = \vec{u}_A - \vec{u}_B \ \text{(constant)}$$

When two projectiles are launched at the same instant, each moves in a straight line as seen from the other, with constant relative speed. Their separation after time $t$ is $\lvert \vec{u}_A - \vec{u}_B \rvert\,t$ plus their initial separation along the same line.

Closest Approach of Two Moving Bodies in Relative Velocity Concept

Two bodies move with constant velocities, and the distance between them at the start is $d$. Work in the frame of one body. The other body moves in a straight line with the relative velocity $\vec{v}_{rel}$. If $\theta$ is the angle between $\vec{v}_{rel}$ and the line that joins the bodies (taken towards the observer), then:

$$d_{min} = d\sin\theta, \qquad t_{min} = \frac{d\cos\theta}{v_{rel}}$$

If $\theta = 0$, the bodies collide. In vector form, with $\vec{r}$ the relative position, $d_{min} = \dfrac{\lvert \vec{r} \times \vec{v}_{rel} \rvert}{\lvert \vec{v}_{rel} \rvert}$.

Key Formulas in Relative Velocity Concept

Solved Examples from Relative Velocity Concept

Example 1 (overtaking): A train of length 100 m moves at 54 km/h and overtakes a train of length 150 m that moves at 36 km/h on a parallel track. How long does the overtaking take?

$v_{rel} = 54 - 36 = 18$ km/h $= 5$ m/s. The distance to cover is $100 + 150 = 250$ m, so $t = \dfrac{250}{5} = 50$ s.

Example 2 (opposite directions): Car A moves at 30 m/s east, and car B moves at 20 m/s west. They are 500 m apart and approaching each other. When do they meet?

$v_{rel} = 30 + 20 = 50$ m/s, so $t = \dfrac{500}{50} = 10$ s.

Example 3 (two dimensions): A moves at 3 m/s east, and B moves at 4 m/s north. Find $\vec{v}_{AB}$.

$\vec{v}_{AB} = (3, 0) - (0, 4) = (3, -4)$ m/s. The magnitude is $\sqrt{9 + 16} = 5$ m/s, and the direction is $\tan^{-1}\dfrac{4}{3} \approx 53^\circ$ south of east.

Example 4 (river, shortest time): A river is 100 m wide, and it flows at $v_r = 3$ m/s. A boat has $v_{br} = 5$ m/s. Find the shortest time to cross and the drift.

$t_{min} = \dfrac{100}{5} = 20$ s, and the drift is $x = 3 \times 20 = 60$ m.

Example 5 (river, shortest path): For the same river and boat, find the heading and the time to reach the point directly opposite.

$\sin\theta = \dfrac{3}{5}$, so $\theta \approx 37^\circ$ upstream from the perpendicular. The resultant speed is $\sqrt{5^2 - 3^2} = 4$ m/s, so $t = \dfrac{100}{4} = 25$ s.

Example 6 (rain-man): Rain falls vertically at 10 m/s, and a man walks at 5 m/s. At what angle should he tilt the umbrella, and how fast does the rain seem to fall?

$\tan\theta = \dfrac{5}{10} = 0.5$, so $\theta \approx 26.6^\circ$ from the vertical, tilted forward. The rain seems to fall at $\sqrt{10^2 + 5^2} \approx 11.2$ m/s.

Example 7 (aeroplane and wind): An aeroplane has an airspeed of 200 km/h and heads north. The wind blows towards the east at 50 km/h. Find its velocity with respect to the ground.

$\vec{v}_{ag} = \vec{v}_{aw} + \vec{v}_{wg}$ has components $(50, 200)$ km/h. The speed is $\sqrt{50^2 + 200^2} \approx 206$ km/h, and the direction is $\tan^{-1}\dfrac{50}{200} \approx 14^\circ$ east of north.

Example 8 (closest approach): Body A is at the origin and moves at 4 m/s east. Body B is at (0, 100) m and moves at 3 m/s south. Find the closest distance and the time at which it occurs.

$\vec{v}_{BA} = (0, -3) - (4, 0) = (-4, -3)$ m/s, so $v_{rel} = 5$ m/s. The angle between $\vec{v}_{BA}$ and the line from B to A satisfies $\cos\theta = \dfrac{3}{5}$, so $\sin\theta = 0.8$. Then $d_{min} = 100 \times 0.8 = 80$ m, and $t = \dfrac{100 \times 0.6}{5} = 12$ s.

Example 9 (two projectiles): Two balls are thrown from the same point at the same instant, each with a speed of 20 m/s, at angles of 30° and 60° to the horizontal. Find their separation after 2 s.

$\vec{u}_1 - \vec{u}_2 = (20\cos 30^\circ - 20\cos 60^\circ,\ 20\sin 30^\circ - 20\sin 60^\circ) \approx (7.32, -7.32)$ m/s. The relative speed is about 10.35 m/s, and it is constant, so the separation after 2 s is about 20.7 m.

Relative Velocity Class 11 – Formula, Examples 

Common Mistakes in Relative Velocity Concept Problems

How to Study Relative Velocity Concept

1. Revise one-dimensional kinematics with eSaral’s Motion in a Straight Line JEE notes, and vector subtraction with the vectors and two-dimensional motion notes.
2. For every problem, write the observer, draw the vector triangle, and then calculate.
3. Practise one problem type each day: trains, river-boat, rain-man, aeroplane-wind, projectiles and closest approach.
4. Revise the Kinematics unit with eSaral’s JEE Main Physics revision notes, and test yourself with eSaral’s JEE Main chapter-wise PYQs and NEET chapter-wise PYQs.