Vector Analysis

Relative velocity

Relative velocity is a fundamental vector concept defining the velocity of one moving body with respect to another as the vector difference of their respective velocities. This topic covers this topic.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Relative Velocity

Let $\vec{v}_A$ and $\vec{v}_B$ be the velocity vectors of two particles $A$ and $B$ with respect to a common inertial frame of reference $S$. The velocity of particle $A$ relative to particle $B$, denoted by $\vec{v}_{AB}$, is defined as the vector difference: $$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$

📐 Geometric Interpretation

Geometrically, $\vec{v}_{AB}$ is represented by closing the vector triangle formed by $\vec{v}_A$ and $\vec{v}_B$, pointing from the tip of $\vec{v}_B$ to the tip of $\vec{v}_A$. It signifies the apparent motion of $A$ when an observer is rigidly attached to $B$ (treating $B$ as stationary).

⚙️ Algebraic & Structural Meaning

Algebraically, relative velocity decomposes motion into independent orthogonal components (e.g., $x$ and $y$ axes) obeying linear vector spaces, simplifying complex multi-body intercept and minimum-distance problems into standard parametric equations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Inertial Frame of Reference	`Constant velocity reference frame`	Strictly Mandatory / Necessary	If the reference frame accelerates, fictitious forces (such as centrifugal or Coriolis forces) must be incorporated, invalidating standard vector subtraction.
Simultaneous Measurement	`Same time coordinate $t$`	Strictly Mandatory	Comparing velocity vectors at different instants violates the instantaneous definition of kinematics in classical mechanics.
Uniformity of Velocity Vectors	`Constant or instantaneously differentiable vectors`	Essential for trajectory prediction	Discontinuous velocity jumps (infinite acceleration) render relative acceleration undefined and make intercept calculations indeterminate.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Minimum Distance Between Two Moving Particles

$$d_{\min} = \frac{|\vec{r}_0 \times \vec{v}_{rel}|}{|\vec{v}_{rel}|}$$

Rule: The minimum distance between two particles with initial position vector difference $\vec{r}_0 = \vec{r}_{A0} - \vec{r}_{B0}$ and relative velocity $\vec{v}_{rel} = \vec{v}_A - \vec{v}_B$ is given by the perpendicular distance from the origin of relative motion.

Application: Frequently asked in UP PGT/TGT and KVS to find the closest approach of two ships or particles.

Time to Reach Minimum Distance

$$t = -\frac{\vec{r}_0 \cdot \vec{v}_{rel}}{|\vec{v}_{rel}|^2}$$

Rule: Direct time formula to attain the closest point of approach without expanding quadratic position equations.

Rain-Man Umbrella Angle

$$\tan \theta = \frac{v_m}{v_r}$$

Rule: If rain falls vertically with velocity $v_r$ and a man runs horizontally with velocity $v_m$, the angle $\theta$ with the vertical at which he must hold his umbrella is given by $\tan \theta = \frac{v_m}{v_r}$.

River Crossing - Minimum Time Path

$$t_{\min} = \frac{d}{v_{sw}}$$

Rule: To cross a river of width $d$ in minimum time, swim perpendicular to the river flow. The downstream drift is $x = v_r \cdot t_{\min}$.

Application: Direct formula substitution tested in TGT/PGT vector applications.

River Crossing - Shortest Path (Perpendicular Arrival)

$$\sin \theta = \frac{v_r}{v_{sw}}, \quad t = \d\frac{d}{\sqrt{v_{sw}^2 - v_r^2}}$$

Rule: To reach the exact opposite point, swim upstream at angle $\theta$ with the line perpendicular to the bank, valid only if swimmer speed $v_{sw} >$ river speed $v_r$.

Rapid Elimination via Magnitude Bounds

$$|\vec{v}_A - \vec{v}_B| \le |\vec{v}_A| + |\vec{v}_B|$$

Rule: Eliminate options where relative speed exceeds the sum of individual speeds or falls below their difference.

Application: Enables instant elimination of 2 absurd options in multiple-choice questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing $\vec{v}_{AB}$ with $\vec{v}_{BA}$

Common Pitfall: Reversing the vector order in subtraction, leading to a sign error in direction.

Counterexample: If $\vec{v}_A = 3\hat{i}$ and $\vec{v}_B = 4\hat{i}$, then $\vec{v}_{AB} = -\hat{i}$ whereas $\vec{v}_{BA} = \hat{i}$.

Why it fails: Vector subtraction is anti-commutative: $\vec{v}_{AB} = -\vec{v}_{BA}$. Ignoring this flips direction vectors.

🛡️ Exam Defense: Always write subscript expansion: velocity of $A$ relative to $B$ means observer is at $B$, so compute $\vec{v}_A - \vec{v}_B$.

⚠️ COMMON ERROR #02

Shortest Path vs Minimum Time Condition

Common Pitfall: Applying the shortest path formula when swimmer speed is less than river flow speed.

Counterexample: If $v_{sw} = 3$ m/s and $v_r = 5$ m/s, setting $\sin \theta = \frac{5}{3}$ gives no real angle.

Why it fails: It is physically impossible to reach directly opposite if river current exceeds swimming speed.

🛡️ Exam Defense: Check if $v_{sw} > v_r$ before attempting shortest path calculations.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Relative Velocity (General)	Inertial frames	$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$	Vector triangle closing side	Velocity of train/car/boat w.r.t another
Minimum Distance	Constant relative velocity	$d_{\min} = \frac{\|\vec{r}_0 \times \vec{v}_{rel}\|}{\|\vec{v}_{rel}\|}$	Perpendicular distance to relative line	Closest approach / collision avoidance
River Crossing (Min Time)	Uniform stream	$t = \frac{d}{v_{sw}}$	Swimming across perpendicular to flow	Find minimum time to cross river
River Crossing (Shortest Path)	$v_{sw} > v_r$	$\sin \theta = \frac{v_r}{v_{sw}}$	Upstream angle compensating drift	Reach directly opposite point

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Two particles $A$ and $B$ are moving with velocities $\vec{v}_A = 3\hat{i} + 4\hat{j}$ m/s and $\vec{v}_B = 6\hat{i} + 8\hat{j}$ m/s. Find the magnitude of the velocity of $A$ relative to $B$.

(A) $5$ m/s(B) $9$ m/s(C) $25$ m/s(D) $1$ m/s

📚 Method A: Formal Textbook Derivation

Step 1: Compute $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (3-6)\hat{i} + (4-8)\hat{j} = -3\hat{i} - 4\hat{j}$. Step 2: Find magnitude $|\vec{v}_{AB}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = 5$ m/s.

⚡ Method B: 25-Second Exam Shortcut

Notice $\vec{v}_B = 2\vec{v}_A$, so $\vec{v}_{AB} = -\vec{v}_A$, magnitude is $|- (3\hat{i}+4\hat{j})| = 5$.

💡 Tip: Always check for scalar multiples between velocity vectors to compute relative speed instantly.

CANONICAL PROBLEM 02 Dual-Method Breakdown

A boat can swim at $5$ m/s in still water. It crosses a river $120$ m wide flowing at $3$ m/s. Find the minimum time taken to cross the river.

(A) $20$ s(B) $24$ s(C) $30$ s(D) $40$ s

📚 Method A: Formal Textbook Derivation

Step 1: For minimum time, swim perpendicular to current. $t = \frac{d}{v_{sw}} = \frac{120}{5} = 24$ seconds.

⚡ Method B: 25-Second Exam Shortcut

Direct formula $t = \frac{d}{v_{sw}}$ ignores river velocity for minimum time calculation.

💡 Tip: Do not mix up minimum time condition with shortest path condition.

Practice questions

Choose an answer to check it and see the solution:

If $\vec{v}_A = \hat{i} + \hat{j}$ and $\vec{v}_B = \hat{i} - \hat{j}$, then the relative velocity $\vec{v}_{AB}$ is:

A $2\hat{j}$

B $-2\hat{j}$

C $2\hat{i}$

D 0

Derivation & Analysis:

$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (1-1)\hat{i} + (1-(-1))\hat{j} = 2\hat{j}$.

Rain is falling vertically with a speed of $30$ m/s. A woman rides a bicycle with a speed of $10$ m/s in the east to west direction. What is the relative velocity of rain with respect to the woman?

A $10\hat{i} - 30\hat{j}$

B $-10\hat{i} - 30\hat{j}$

C $30\hat{i} - 10\hat{j}$

D $-30\hat{i} - 10\hat{j}$

Derivation & Analysis:

Let rain velocity be $\vec{v}_r = -30\hat{j}$ and bicycle velocity be $\vec{v}_m = -10\hat{i}$. Then $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m = -30\hat{j} - (-10\hat{i}) = 10\hat{i} - 30\hat{j}$.

The minimum distance of approach between two particles moving with constant relative velocity $\vec{v}_{rel}$ and initial position difference $\vec{r}_0$ is:

A $\frac{|\vec{r}_0 \times \vec{v}_{rel}|}{|\vec{v}_{rel}|}$

B $\frac{\vec{r}_0 \cdot \vec{v}_{rel}}{|\vec{v}_{rel}|}$

C $|\vec{r}_0| \sin \theta$

D Both (A) and (C)

Derivation & Analysis:

Both vector cross product formula and geometric projection $|\vec{r}_0| \sin \theta$ yield the correct minimum perpendicular distance.

A swimmer crosses a river of width $d$ with speed $v$ relative to water. If the river flow speed is $v$, the angle with the normal at which he must swim to reach the directly opposite point is:

A $\frac{\pi}{6}$

B $\frac{\pi}{4}$

C $\frac{\pi}{3}$

D $\frac{\pi}{2}$

Derivation & Analysis:

$\sin \theta = \frac{v_r}{v_{sw}} = \frac{v}{v} = 1 \implies \theta = \frac{\pi}{4} = 45^\circ$.

If the angle between two velocity vectors $\vec{v}_1$ and $\vec{v}_2$ is $\theta$, the magnitude of their relative velocity is:

A $\sqrt{v_1^2 + v_2^2 - 2v_1v_2 \cos \theta}$

B $\sqrt{v_1^2 + v_2^2 + 2v_1v_2 \cos \theta}$

C $v_1 - v_2$

D $v_1 + v_2$

Derivation & Analysis:

By vector subtraction magnitude law, $|\vec{v}_1 - \vec{v}_2| = \sqrt{v_1^2 + v_2^2 - 2v_1v_2 \cos \theta}$.

Two trains $100$ m and $120$ m long are moving in opposite directions with velocities $20$ m/s and $15$ m/s. The time taken by them to cross each other is:

A $\frac{22}{7} \text{ s}$

B $\frac{44}{7} \text{ s}$

C $\frac{22} {35} \text{ s}$

D $\frac{44}{35} \text{ s}$

Derivation & Analysis:

Total distance = $100 + 120 = 220$ m. Relative speed = $20 + 15 = 35$ m/s. Time $t = \frac{220}{35} = \frac{44}{7}$ s.

If a boat's speed in still water is less than the river flow speed ($v_{sw} < v_r$), then:

A It can cross the river perpendicularly

B It can never reach the directly opposite point

C It crosses in minimum time only upstream

D Drift is zero

Derivation & Analysis:

Since $\sin \theta = \frac{v_r}{v_{sw}} > 1$, no real angle exists to counteract the river current, making direct perpendicular crossing impossible.

The concept of relative velocity in classical mechanics is valid in:

A Inertial frames only

B Non-inertial frames only

C Both inertial and non-inertial frames without modification

D Only rotating frames

Derivation & Analysis:

Standard vector subtraction for relative velocity assumes inertial frames unless fictitious acceleration terms are added.

If $\vec{v}_{AB} = -\vec{v}_{BA}$, this property demonstrates that relative velocity vector subtraction is:

A Commutative

B Anti-commutative

C Associative

D Distributive

Derivation & Analysis:

The negative sign upon interchanging operands defines anti-commutativity.

Q10

A bird flies east with $10$ m/s and wind blows north with $10$ m/s. The magnitude and direction of velocity of bird relative to wind (if wind is considered moving frame) or vice versa is:

A $10\sqrt{2} \text{ m/s}, 45^\circ \text{ East of North}$

B $10 \text{ m/s}, 90^\circ$

C $20 \text{ m/s}, 30^\circ$

D $10\sqrt{2} \text{ m/s}, \text{South}$

Derivation & Analysis:

Magnitude = $\sqrt{10^2 + 10^2} = 10\sqrt{2}$ m/s at $45^\circ$.

Common questions

Subtraction arises from the definition of position difference vector $\vec{r}_{AB} = \vec{r}_A - \vec{r}_B$. Differentiating with respect to time yields $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. Adding them would calculate velocity relative to a hypothetical moving center of mass.

Confusing 'time to cross' with 'position of arrival'. Minimum time requires swimming straight across (ignoring drift), whereas shortest path requires angling upstream against the current.

Memorize standard shortcut formulas for minimum distance ($d = \frac{|\vec{r}_0 \times \vec{v}_{rel}|}{|\vec{v}_{rel}|}$) and leverage symmetry or vector component isolation.

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