Relative Velocity

Prerequisites

Before studying this topic, make sure you understand:

Frame of Reference - reference frames concept
Motion in a Plane - vector addition

What is Relative Velocity?

Relative velocity is the velocity of one object as observed from another moving object.

Train Fight in Jawan!

Remember the epic train fight scene in Jawan (2023)? When SRK fights on top of a moving train, his velocity relative to the train is slow (he’s on it!), but relative to the ground, he’s zooming at 100+ km/h! If he jumps between train cars moving at the same speed, relative to them, it’s like a normal jump. That’s relative velocity in action!

Key Concept

Motion is always relative. When you say “a car is moving at 60 km/h,” you mean relative to the ground. But what if you observe it from another moving car?

The Formula

The velocity of object A relative to object B:

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

Similarly:

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

Important Properties

$\vec{v}_{AB} + \vec{v}_{BA} = 0$
$\vec{v}_{AB} = -\vec{v}_{BA}$ (equal magnitude, opposite direction)
$\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ (chain rule)

Cases in 1D Motion

Same Direction

If A and B move in the same direction:

$$v_{AB} = v_A - v_B$$

Example: Car A at 80 km/h, Car B at 60 km/h (same direction)

A relative to B: 80 - 60 = 20 km/h (A appears to move forward)
B relative to A: 60 - 80 = -20 km/h (B appears to move backward)

Opposite Directions

If A and B move in opposite directions:

$$|v_{AB}| = v_A + v_B$$

Example: Two trains approaching each other at 60 km/h each

Relative speed = 60 + 60 = 120 km/h

Fast & Furious Head-On Chase

In Fast X (2023), when two cars race TOWARD each other, the relative speed is the SUM of their speeds! That’s why head-on collisions in movies look so dramatic — if both cars are at 100 km/h, they approach each other at 200 km/h. Physics makes the stunts even more intense!

Quick Rule

Same direction: Subtract speeds
Opposite directions: Add speeds

Cases in 2D Motion

For 2D motion, use vector subtraction:

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

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

where $\theta$ is the angle between $\vec{v}_A$ and $\vec{v}_B$.

Special cases:

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

Interactive Demo: Visualize Vector Addition

See how relative velocity is calculated using vector subtraction.

Application 1: Rain and Man Problem

Bollywood Rain Scenes!

Every Bollywood romantic rain song — from Kabhi Khushi Kabhie Gham to Jawan — shows heroes walking in rain. Here’s the physics: when you walk, rain appears to come from the front! The faster you run, the more you need to tilt your umbrella forward. That’s $\vec{v}_{rain,you} = \vec{v}_{rain} - \vec{v}_{you}$!

Problem Setup

A man walks with velocity $\vec{v}_m$ while rain falls with velocity $\vec{v}_r$. What angle does the rain appear to fall?

Solution Approach

Velocity of rain relative to man:

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

The rain appears to come from the direction of $\vec{v}_{rm}$.

Common Scenarios

Case 1: Rain falling vertically, man walking horizontally

If rain falls at speed $v_r$ (vertical) and man walks at $v_m$ (horizontal):

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

Angle with vertical: $\tan\alpha = \frac{v_m}{v_r}$

To stay dry: The man should hold his umbrella at angle $\alpha$ from vertical, tilted forward.

Case 2: At what speed should man walk so rain appears vertical?

Man should walk with velocity equal to the horizontal component of rain velocity.

Application 2: River Crossing

A boat crosses a river of width $d$ with current velocity $\vec{v}_r$.

Problem Types

Type 1: Shortest Time

Point boat directly across (perpendicular to bank).

Time: $t = \frac{d}{v_b}$
Drift: $x = v_r \times t = \frac{v_r \cdot d}{v_b}$

Type 2: Shortest Path (No Drift)

Point boat upstream at angle $\theta$ such that:

$$v_b\sin\theta = v_r$$ $$\sin\theta = \frac{v_r}{v_b}$$

This requires $v_b > v_r$ (boat faster than current).

Resultant velocity: $v_{res} = \sqrt{v_b^2 - v_r^2}$
Time: $t = \frac{d}{\sqrt{v_b^2 - v_r^2}}$

Important Condition

For shortest path crossing, the boat speed must be greater than river speed ($v_b > v_r$). Otherwise, the boat cannot go straight across.

Summary Table

Objective	Boat Direction	Time	Drift
Shortest time	Perpendicular	$\frac{d}{v_b}$	$\frac{v_r d}{v_b}$
Shortest path	Upstream at $\sin^{-1}(\frac{v_r}{v_b})$	$\frac{d}{\sqrt{v_b^2 - v_r^2}}$	0

Application 3: Aircraft in Wind

For an aircraft with airspeed $\vec{v}_a$ flying in wind $\vec{v}_w$:

Ground velocity: $\vec{v}_g = \vec{v}_a + \vec{v}_w$

To reach destination directly

The pilot must head at an angle such that the resultant velocity points toward the destination.

Practice Problems

Problem 1

Two cars A and B move with velocities 20 m/s east and 15 m/s north respectively. Find velocity of A relative to B.

Solution: $\vec{v}_A = 20\hat{i}$ m/s, $\vec{v}_B = 15\hat{j}$ m/s

$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = 20\hat{i} - 15\hat{j}$ m/s

$|\vec{v}_{AB}| = \sqrt{20^2 + 15^2} = \sqrt{625} = $ 25 m/s

Direction: $\tan\theta = \frac{15}{20} = 0.75$, $\theta = 36.87°$ south of east

Problem 2

Rain falls vertically at 10 m/s. A man walks at 6 m/s. At what angle should he hold his umbrella?

Solution: $\tan\alpha = \frac{v_m}{v_r} = \frac{6}{10} = 0.6$

$\alpha = \tan^{-1}(0.6) ≈ $ 31° from vertical (tilted forward)

Problem 3

A river is 400 m wide with current 3 m/s. A boat has speed 5 m/s in still water. Find: (a) Time for shortest path (b) Time for shortest time

Solution: (a) $v_{res} = \sqrt{5^2 - 3^2} = 4$ m/s $t = \frac{400}{4} = $ 100 s

(b) $t = \frac{400}{5} = $ 80 s (with drift = $\frac{3 \times 400}{5} = 240$ m)