Relative Velocity — Revision Notes

⚡ 30-Second Revision

1D Relative Velocity: — $v_{AB} = v_A - v_B$
2D Relative Velocity (Vector Form): — $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$
Relative Acceleration: — $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$
Opposite Directions (1D): — Relative speed = $v_A + v_B$
Same Direction (1D): — Relative speed = $|v_A - v_B|$
Rain-Man (Angle with vertical $\theta$): — $\tan\theta = \frac{v_M}{v_R}$
Boat-River (Resultant velocity): — $\vec{v}_{BG} = \vec{v}_B + \vec{v}_R$
Boat-River (Shortest Time): — Head perpendicular to river. $t = \frac{W}{v_B}$. Drift $x = v_R t$.
Boat-River (Shortest Path): — Head upstream at angle $\theta = \sin^{-1}(\frac{v_R}{v_B})$. Resultant speed $v_{BG} = \sqrt{v_B^2 - v_R^2}$. Time $t = \frac{W}{v_{BG}}$.

2-Minute Revision

Relative velocity is how one object's motion appears from another moving object's perspective. It's a vector quantity, calculated by subtracting the observer's velocity from the observed object's velocity: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.

This principle extends to relative acceleration: $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$. In one dimension, simply use signs for direction; if objects move in the same direction, subtract speeds; if opposite, add speeds.

For two dimensions, vector subtraction is crucial. This often involves resolving velocities into components (x and y) and subtracting them separately, or using the triangle law of vector subtraction (adding $\vec{v}_A$ to $-\vec{v}_B$).

Key applications include 'rain-man' problems, where the angle to hold an umbrella depends on the rain's velocity relative to the walking person, and 'boat-river' problems, which differentiate between crossing a river in the shortest time (heading perpendicular to the current) versus crossing along the shortest path (heading upstream to counteract the current).

Always ensure consistent units and correct vector operations.

5-Minute Revision

Relative velocity is a core concept in kinematics, defining the velocity of an object as seen from a moving frame of reference. The fundamental formula is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$, where $\vec{v}_A$ and $\vec{v}_B$ are velocities with respect to a common stationary frame (e.g., ground). This means the velocity of object A as observed by an observer in object B. The same logic applies to relative acceleration: $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$.

One-Dimensional Motion:

If two objects move along a straight line:

Same direction: — Relative speed is $|v_A - v_B|$. Example: Car A at $60,\text{km/h}$, Car B at $40,\text{km/h}$ in same direction. $v_{AB} = 60 - 40 = 20,\text{km/h}$.
Opposite directions: — Relative speed is $v_A + v_B$. Example: Car A at $60,\text{km/h}$ east, Car B at $40,\text{km/h}$ west. $v_{AB} = 60 - (-40) = 100,\text{km/h}$.

Two-Dimensional Motion:

Requires vector subtraction. You can use component method or graphical method.

Component Method: — If $\vec{v}_A = v_{Ax}\hat{i} + v_{Ay}\hat{j}$ and $\vec{v}_B = v_{Bx}\hat{i} + v_{By}\hat{j}$, then $\vec{v}_{AB} = (v_{Ax} - v_{Bx})\hat{i} + (v_{Ay} - v_{By})\hat{j}$.
Graphical Method: — Draw $\vec{v}_A$, then draw $-\vec{v}_B$ (same magnitude as $\vec{v}_B$ but opposite direction) from the head of $\vec{v}_A$. The resultant vector from the tail of $\vec{v}_A$ to the head of $-\vec{v}_B$ is $\vec{v}_{AB}$.

Key Applications:

1
Rain-Man Problems: — If rain falls vertically ($\vec{v}_R$) and a man walks horizontally ($\vec{v}_M$), the velocity of rain relative to the man is $\vec{v}_{RM} = \vec{v}_R - \vec{v}_M$. The angle $\theta$ with the vertical at which the umbrella should be held is given by $\tan\theta = \frac{|\vec{v}_M|}{|\vec{v}_R|}$.

*Mini-Example:* Rain $4,\text{m/s}$ down, man $3,\text{m/s}$ horizontal. $\tan\theta = 3/4$. $\theta = \tan^{-1}(3/4)$.

1
Boat-River Problems: — Velocity of boat relative to ground ($\vec{v}_{BG}$) is $\vec{v}_{BG} = \vec{v}_B + \vec{v}_R$, where $\vec{v}_B$ is boat's velocity in still water and $\vec{v}_R$ is river current velocity.

* Shortest Time to Cross: Boat heads perpendicular to river flow. Time $t = \frac{\text{width}}{v_B}$. Drift occurs. * Shortest Path to Cross (Directly opposite): Boat heads upstream at an angle to counteract current. Resultant velocity is perpendicular to current. $v_{BG} = \sqrt{v_B^2 - v_R^2}$. Time $t = \frac{\text{width}}{v_{BG}}$. Requires $v_B > v_R$.

Common Mistakes: Incorrect vector subtraction, sign errors in 1D, confusing shortest time vs. shortest path, and not converting units.

Prelims Revision Notes

Relative velocity is the velocity of an object with respect to an observer. It's a vector quantity. The formula for velocity of A relative to B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. Remember, $\vec{v}_A$ and $\vec{v}_B$ must be with respect to a common, usually stationary, frame (e.g., ground).

1. One-Dimensional Motion:

Same Direction: — Relative speed = $|v_A - v_B|$. Example: Car A at $v_A$, Car B at $v_B$. If $v_A > v_B$, A approaches B at $v_A - v_B$. If $v_B > v_A$, B approaches A at $v_B - v_A$.
Opposite Directions: — Relative speed = $v_A + v_B$. Example: Two trains approaching each other. Their speeds add up.
Total Distance for Crossing: — For two objects of lengths $L_1$ and $L_2$ to completely cross each other, the total distance is $L_1 + L_2$.

2. Two-Dimensional Motion:

Vector Subtraction: — $\vec{v}_{AB} = \vec{v}_A + (-\vec{v}_B)$. Resolve vectors into components ($\hat{i}, \hat{j}$) and subtract corresponding components.
Rain-Man Problems:

* $\vec{v}_{RM} = \vec{v}_R - \vec{v}_M$. * If rain falls vertically ($\vec{v}_R = -v_R\hat{j}$) and man walks horizontally ($\vec{v}_M = v_M\hat{i}$), then $\vec{v}_{RM} = -v_R\hat{j} - v_M\hat{i}$. * Angle $\theta$ with vertical for umbrella: $\tan\theta = \frac{v_M}{v_R}$.

Boat-River Problems:

* Velocity of boat relative to ground: $\vec{v}_{BG} = \vec{v}_B + \vec{v}_R$ (where $\vec{v}_B$ is boat's velocity in still water, $\vec{v}_R$ is river velocity). * Shortest Time to Cross: Boat heads perpendicular to river flow.

Time $t = \frac{\text{River Width}}{v_B}$. Drift downstream $x = v_R \times t$. * Shortest Path to Cross (Directly Opposite): Boat heads upstream at an angle such that its resultant velocity is perpendicular to the river flow.

Required $v_B > v_R$. Resultant speed $v_{BG} = \sqrt{v_B^2 - v_R^2}$. Time $t = \frac{\text{River Width}}{v_{BG}}$. Angle upstream $\theta = \sin^{-1}(\frac{v_R}{v_B})$.

3. Relative Acceleration: $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$. This is applicable when both frames are inertial or one is accelerating relative to the other.

Key Points for NEET:

Always convert units (e.g., $\text{km/h}$ to $\text{m/s}$).
Pay attention to the 'observer' and 'observed' in the question.
Master vector addition/subtraction, including component resolution and graphical methods.
Practice the specific scenarios of rain-man and boat-river problems extensively.

Vyyuha Quick Recall

Really Velocious Animals Subtract Observer's Velocity.

Really Velocious Animals: Reminds you of Relative Velocity and Acceleration.
Subtract Observer's Velocity: The core rule: $\vec{v}_{observed} - \vec{v}_{observer}$.

For Rain-Man problems, think: Umbrella Man Rain. $\tan\theta = \frac{v_M}{v_R}$ (Man's speed over Rain's speed for angle with vertical).