Relative Velocity — JEE Advanced Physics

Frames of Reference and Relative Velocity

Observer, Frame of Reference, and Relative Motion

Consider a person sitting inside a moving train. To another passenger sitting in the same train, the person is at rest. However, to an observer standing on the railway platform, the person is moving along with the train at high speed.

This demonstrates a fundamental principle of physics: rest and motion are relative terms. There is no such thing as absolute rest or absolute motion in the universe. The state of motion of an object depends entirely on the reference frame from which it is observed.

To describe and measure motion quantitatively, we define a frame of reference:

Observer: A person or instrument that measures the position, displacement, and time of an event.
Frame of Reference: A coordinate system (such as $x, y, z$ axes) equipped with a clock, attached to an observer.

Changing the reference frame changes the measured velocity of an individual object. However, the relative velocity between any two objects is identical in all inertial reference frames.

The same passenger described from two frames of reference: Carriage Frame vs Platform Frame.

Key Takeaway: Motion must always be specified relative to a chosen reference frame. Always write the reference frame clearly (for example, $v_{C,G}$ denotes the velocity of cart $C$ with respect to ground $G$ ).

You are sitting still inside a moving train reading a book. Someone standing on the ground claims you are moving at $300\,\text{km/h}$ . Is this claim correct?

Two cars travel side-by-side on a straight highway with the exact same velocity relative to the ground. What does the driver of Car A observe about Car B?

Relative Position Vector

Let $G$ represent a fixed ground reference frame. At any given instant, let the position vectors of two objects $A$ and $B$ measured from the ground origin be $\vec{r}_{A,G}$ and $\vec{r}_{B,G}$ respectively:

\vec{r}_{A,G} \qquad \text{and} \qquad \vec{r}_{B,G}

The position vector of object $A$ relative to object $B$ (denoted as $\vec{r}_{A,B}$ ) is the vector pointing from $B$ to $A$ . By vector subtraction, this is given by:

\vec{r}_{A,B} = \vec{r}_{A,G} - \vec{r}_{B,G}

Press play. The purple arrows are the positions of A and B measured from the ground; the orange arrow is the position of A as B records it, and it equals their difference at every instant.

The position changes (displacements) of drone A and drone B in 1 second measured from the ground are $\vec{s}_{A,G} = (3,\,4)\,\text{m}$ and $\vec{s}_{B,G} = (3,\,1)\,\text{m}$ . What is the displacement of A relative to B?

Two cars move along a straight road. Which of the following quantities will be measured to have the EXACT same value by all observers in different inertial frames?

Definition of Relative Velocity

The relative velocity of object A with respect to object B (written as $\vec{v}_{A,B}$ ) is defined as the time rate of change of the position of A as observed from frame B.

Mathematically, it is calculated as the vector difference of their velocities measured in the same reference frame (such as ground frame $G$ ):

\vec{v}_{A,B} = \vec{v}_{A,G} - \vec{v}_{B,G}

Physical Intuition: If car A moves forward at $10\,\text{m/s}$ and car B moves forward at $6\,\text{m/s}$ , the driver of B observes car A pulling ahead at $10 - 6 = 4\,\text{m/s}$ . Relative velocity isolates this difference in motion.

Counting on the left, measuring on the right. In both columns the bottom row is the only part the second observer notices.

Subscript Convention: Always write subscript labels clearly. $\vec{v}_{A,B}$ means velocity of Object A with respect to Observer B. The first subscript is the object being observed; the second subscript is the reference frame or observer.

Note that reversing the order of subscripts reverses the direction of the vector.

Relative Displacement

Similarly, the displacement of object A relative to object B ( $\vec{s}_{A,B}$ ) is the vector difference of their individual displacements relative to the ground:

\vec{s}_{A,B} = \vec{s}_{A,G} - \vec{s}_{B,G}

Ten metres and six metres from the ground, four metres between them. The difference is what B records.

Derivation by Differentiation

Differentiating the relative position relation with respect to time yields the relative velocity and relative acceleration equations:

1. Relative Velocity (First Derivative):

\frac{d}{dt}\vec{r}_{A,B} = \frac{d}{dt}\vec{r}_{A,G} - \frac{d}{dt}\vec{r}_{B,G} \qquad\Longrightarrow\qquad \vec{v}_{A,B} = \vec{v}_{A,G} - \vec{v}_{B,G}

2. Relative Acceleration (Second Derivative):

\vec{a}_{A,B} = \vec{a}_{A,G} - \vec{a}_{B,G}

Conclusion: Every relative physical quantity (position, velocity, acceleration) is obtained by subtracting the corresponding ground-frame quantity of the observer from that of the object.

A cyclist rides east at $6\,\text{m/s}$ and a jogger runs east at $2\,\text{m/s}$ . What is the relative velocity of the jogger with respect to the cyclist ( $\vec{v}_{\text{jogger, cyclist}}$ )?

Four Fundamental Properties of Relative Velocity

The following four properties directly follow from the definition of relative velocity:

Self-Relative Velocity is Zero: An object is always at rest relative to itself: $\vec{v}_{A,A} = \vec{v}_{A,G} - \vec{v}_{A,G} = 0$ .
Anti-Symmetric Property: Swapping the object and observer reverses the sign/direction of relative velocity: $\vec{v}_{B,A} = -\vec{v}_{A,B}$ .

The same pair read from each side: swapping the observer reverses the relative velocity direction while preserving magnitude.

Chain Rule (Frame Transformation): For any three frames A, B. C, relative velocities combine as: $\vec{v}_{A,C} = \vec{v}_{A,B} + \vec{v}_{B,C}$ . The intermediate frame B cancels out.

One walker, two frames. Seen from the ground the walker covers 12 m each second; switch to the train's frame and he covers only 2, while the ground streams backward at 10 m/s. The chain rule adds the two.

Principle of Relativity: The laws of mechanics have the exact same form in all inertial reference frames moving at constant velocity. Uniform linear motion cannot be detected by internal mechanical experiments.

To a passenger on a moving train, a car on an adjacent road appears to move backward at $12\,\text{m/s}$ . How does the train appear to move when observed from the car?

The Principle of Relativity

The Principle of Relativity states that the laws of physics are identical in all inertial reference frames (frames moving with constant velocity without acceleration).

Conceptual Example: Coin Toss inside a Moving Train

A train moves along a straight, smooth track at a constant velocity of $300\,\text{km/h}$ . A passenger tosses a coin vertically upward. Where will the coin land?

How does the outcome change if the train is accelerating or braking?

Show Explanation

Case 1: Constant Velocity (Inertial Frame)
The coin lands back in the passenger's hand. In the train frame, the coin has zero horizontal velocity initially, and no horizontal force acts on it. Therefore, it moves straight up and down.

From the ground frame, the coin follows a parabolic projectile path with a horizontal speed of $300\,\text{km/h}$ , landing in the hand which moves forward by the exact same distance in that time.

Two descriptions of one coin flip. They must agree, because the landing is a physical event and the descriptions are bookkeeping.

Case 2: Train Braking (Non-Inertial Frame)
If the train brakes while the coin is in mid-air, the train slows down but the coin maintains its forward speed. As a result, the coin lands ahead of the passenger's hand. Acceleration of a frame can be detected from within the frame.

Which of the following experiments performed inside a completely sealed windowless cabin could reveal that the cabin is moving at a constant velocity?

Relative Velocity in One Dimension

One-Dimensional Motion and Sign Convention

In one-dimensional motion (motion along a straight line), vectors point either forward or backward along a single line. Therefore, vector equations simplify to signed scalar equations.

First, choose a positive direction (for example, rightward/eastward as positive). Write each velocity with its appropriate sign. The relative velocity of object A with respect to object B is calculated as:

v_{A,B} = v_{A,G} - v_{B,G}

Important Rule: Always use the single formula $v_{A,B} = v_{A,G} - v_{B,G}$ along with appropriate signs. Do not memorize separate addition and subtraction rules.

Same direction: Both velocities have the same sign. $v_{A,B} = v_{A,G} - v_{B,G}$ .
Opposite directions: One velocity is positive and the other is negative, so subtracting a negative number yields an addition: $v_{A,B} = v_{A,G} - (-v_{B,G}) = v_{A,G} + v_{B,G}$ .

Fill in the last column and check it. The same subtraction covers all four sign combinations.

Two trains approach each other on parallel tracks at $25\,\text{m/s}$ and $15\,\text{m/s}$ . What is their relative speed?

Separation Distance and Rate of Approach

If object A is at position $x_A$ and object B is at position $x_B$ on a straight line, the separation distance between them is $x_A - x_B$ . Differentiating this distance with respect to time gives the rate of change of separation:

\frac{d}{dt}(x_A - x_B) = v_{A,G} - v_{B,G} = v_{A,B}

Relative velocity represents the exact rate at which the separation distance between two objects increases or decreases.

Time of Meeting / Overtaking:

t = \frac{\text{Initial separation distance}}{\text{Relative speed}}

Two position-time curves on the left, their difference on the right. The gap falls along a straight line and reaches zero exactly once.

On a position-time graph, the point of meeting corresponds to the instant when the relative displacement $x_A - x_B = 0$ . Note that relative displacement curves are independent of uniform motion of the observer's frame.

Car A is $100\,\text{m}$ behind Car B on a straight road. Car A travels at $30\,\text{m/s}$ and Car B travels at $20\,\text{m/s}$ in the same direction. How long will Car A take to catch Car B?

Worked Example: Bird Flying Along a Moving Train

Problem Statement

A train of length $120\,\text{m}$ moves along a straight track at a speed of $20\,\text{m/s}$ . A bird flies parallel to the train in the opposite direction at a speed of $10\,\text{m/s}$ relative to the ground. Calculate the time taken by the bird to cross the train from front to rear.

Press play, then read the same crossing from the train. The train stops moving and the bird alone covers the whole 120 m.

Show Solution

Step 1: Choose the Reference Frame of the Train.
In the train's reference frame, the train is stationary. The bird moves with relative velocity:

v_{\text{bird, train}} = v_{\text{bird, ground}} - v_{\text{train, ground}} = (-10) - (20) = -30\,\text{m/s}

The magnitude of relative speed is $30\,\text{m/s}$ .

Step 2: Calculate the Time Taken.
Distance to be covered by the bird relative to the train is equal to the length of the train ( $120\,\text{m}$ ):

t = \frac{\text{Length of train}}{\text{Relative speed}} = \frac{120\,\text{m}}{30\,\text{m/s}} = 4.0\,\text{s}

Ground Frame Verification: In $4.0\,\text{s}$ , the bird covers $10 \times 4 = 40\,\text{m}$ backward, while the rear of the train covers $20 \times 4 = 80\,\text{m}$ forward. Total combined distance $40 + 80 = 120\,\text{m}$ .

Two cyclists move in the same direction along a straight road at $8\,\text{m/s}$ and $5\,\text{m/s}$ relative to the ground. A car driving at $20\,\text{m/s}$ in the opposite direction records their motion. What is the rate of separation between the two cyclists as measured by the driver of the car?

Worked Example: Two Trains Crossing Each Other

Problem Statement

Train A has a length of $150\,\text{m}$ and travels at $25\,\text{m/s}$ . Train B has a length of $100\,\text{m}$ and travels at $15\,\text{m/s}$ in the opposite direction on a parallel track. Find the time taken for the two trains to completely cross each other after their front ends meet.

The same passing read from the ground and then from train A. From A, train A stands still and B covers the whole 250 m on its own.

Show Solution

Step 1: Calculate Relative Velocity.
Working in the reference frame of Train A (taking Train A's direction as positive):

v_{B,A} = v_{B,G} - v_{A,G} = (-15) - (25) = -40\,\text{m/s}

The relative speed between the trains is $40\,\text{m/s}$ .

Step 2: Calculate Total Distance to Clear.
To completely cross each other, the relative displacement required is the sum of their lengths:

d = L_A + L_B = 150 + 100 = 250\,\text{m}

Step 3: Calculate Time.

t = \frac{d}{v_{B,A}} = \frac{250\,\text{m}}{40\,\text{m/s}} = 6.25\,\text{s}

Problem Solving Tip: Whenever a problem involves two moving objects and asks for collision time, meeting time, or crossing time, shift to the reference frame of one object so that it becomes stationary.

Consider a $120\,\text{m}$ long train moving at $20\,\text{m/s}$ . A bird flies in the SAME direction as the train at $10\,\text{m/s}$ relative to the ground. How long does the train take to completely overtake the bird?

Worked Example: Police Chase and Bullet Firing

Problem Statement

A police van travelling on a straight highway at $30\,\text{km/h}$ fires a bullet at a thief's car speeding away in the same direction at $192\,\text{km/h}$ . If the muzzle speed of the bullet is $150\,\text{m/s}$ (speed of bullet relative to the gun), with what speed does the bullet strike the thief's car?

The muzzle speed is quoted relative to the van, not to the ground.

Show Solution

Step 1: Convert Speeds into SI Units (m/s).

v_{p,G} = 30\,\text{km/h} = 30 \times \frac{5}{18} = \frac{25}{3}\,\text{m/s}

v_{t,G} = 192\,\text{km/h} = 192 \times \frac{5}{18} = \frac{160}{3}\,\text{m/s}

Step 2: Determine Muzzle Velocity in Ground Frame.
Muzzle speed is the bullet velocity relative to the police van ( $v_{b,p} = 150\,\text{m/s}$ ). By the chain rule:

v_{b,G} = v_{b,p} + v_{p,G} = 150 + \frac{25}{3} = \frac{475}{3}\,\text{m/s}

Step 3: Calculate Bullet Velocity Relative to Thief's Car.

v_{b,t} = v_{b,G} - v_{t,G} = \frac{475}{3} - \frac{160}{3} = \frac{315}{3} = 105\,\text{m/s}

Alternative One-Line Method:

v_{b,t} = v_{b,p} + v_{p,t} = 150 + (30 - 192)\text{ km/h} = 150 - 45\,\text{m/s} = 105\,\text{m/s}

In the police chase example above, what would be the striking speed of the bullet if the thief's car were driving TOWARDS the police van at $192\,\text{km/h}$ ?

Relative Acceleration and Free-Fall Motion

Differentiating relative velocity gives relative acceleration:

\vec{a}_{A,B} = \vec{a}_{A,G} - \vec{a}_{B,G}

Special Case: Freely Falling Bodies
If two objects A and B are both moving under gravity alone (free fall), their ground-frame accelerations are identical: $\vec{a}_{A,G} = \vec{a}_{B,G} = \vec{g}$ (downward). Thus, their relative acceleration is zero:

\vec{a}_{A,B} = \vec{g} - \vec{g} = 0

Physical Meaning: In the reference frame of a freely falling body, any other freely falling body moves with zero acceleration (constant relative velocity) in a straight line.

Two stones dropped a second apart, second by second. The difference of the speeds never changes, and the gap between them opens by the same 10 m every second.

An elevator accelerates upward at $2\,\text{m/s}^2$ . A passenger inside drops a coin. Taking acceleration due to gravity $g = 10\,\text{m/s}^2$ downward, what is the acceleration of the coin RELATIVE TO THE ELEVATOR?

Applications of Relative Acceleration

Conceptual Problem: The Hunter and the Monkey

A monkey hangs from a tree branch. A dart is aimed directly along the line of sight at the monkey and fired. At the exact instant the dart is fired, the monkey lets go and drops freely under gravity. Will the dart hit the monkey?

Show Solution

Yes, the dart hits the monkey regardless of the firing speed.

Explanation using Relative Motion:
Once released, both the monkey and the dart are in free fall under gravity. Consequently, $\vec{a}_{\text{dart, monkey}} = \vec{g} - \vec{g} = 0$ .
In the monkey's reference frame, the dart has zero acceleration and travels along a straight line pointing directly at the monkey. Hence, it is guaranteed to hit the monkey.

Equal drops in equal times, so the straight aim is right after all.

Ground Frame Explanation of Hunter-Monkey Problem

From the ground reference frame, in time $t$ , the dart falls by a vertical distance $\frac{1}{2}gt^2$ below its initial straight line path. In the same time $t$ , the monkey drops by the exact same distance $\frac{1}{2}gt^2$ from the branch. Because both drop by equal vertical distances in equal times, their paths intersect.

Equal accelerations mean the gap opens along a straight line, not a curve.

Two stones are dropped from the same height, with the second stone dropped 1 second after the first stone. As they fall, what happens to the separation distance between them?

Relative Velocity in Two Dimensions

Vector Subtraction in Two Dimensions

The fundamental definition of relative velocity remains unchanged in two dimensions:

\vec{v}_{A,B} = \vec{v}_{A,G} - \vec{v}_{B,G}

Geometrically, subtracting a vector $\vec{v}_{B,G}$ is equivalent to adding its opposite vector $-\vec{v}_{B,G}$ :

\vec{v}_{A,B} = \vec{v}_{A,G} + (-\vec{v}_{B,G})

To construct $\vec{v}_{A,B}$ graphically: draw $\vec{v}_{A,G}$ , place the tail of $-\vec{v}_{B,G}$ at the head of $\vec{v}_{A,G}$ . Draw the resultant vector from the tail of $\vec{v}_{A,G}$ to the head of $-\vec{v}_{B,G}$ . The three vectors form a closed triangle.

Concept Test: Suppose two objects move with equal speeds $v$ . At what angle between their velocity vectors is their relative speed equal to $v$ ?

Use the interactive controls below to verify your answer.

The relative velocity closes the triangle. Its size depends on the angle between the two velocities as much as on the speeds themselves.

Key Insight: Relative velocity has both a magnitude and a direction. The relative speed depends on both the individual speeds and the angle between their directions of motion.

Two aircraft fly at the exact same speed $v$ on straight courses at an angle of $90^\circ$ to each other. What is their relative speed?

Relative Velocity of Two Ships

Consider two ships A and B moving on perpendicular courses. In the ground reference frame, both ships move along straight paths. However, in the reference frame of Ship A, Ship A is at rest at the origin. Ship B moves along a single relative velocity straight line.

A top view. Press play, then change frames. The distance between the two ships is the same number in both, only the tracks change.

If Ship A moves east at $24\,\text{km/h}$ and Ship B moves south at $18\,\text{km/h}$ , the relative velocity of B with respect to A is:

\vec{v}_{B,A} = \vec{v}_{B,G} - \vec{v}_{A,G} = (0,\,-18) - (24,\,0) = (-24,\,-18)\,\text{km/h}

The magnitude of relative velocity is $|\vec{v}_{B,A}| = \sqrt{24^2 + 18^2} = 30\,\text{km/h}$ in the south-west direction.

Two ships move along straight lines on courses $90^\circ$ apart. Under what condition will the navigator on Ship A observe Ship B at a CONSTANT COMPASS BEARING (constant line-of-sight angle)?

Magnitude of Relative Velocity (Law of Cosines)

If two objects A and B have speeds $v_{A,G}$ and $v_{B,G}$ . The angle between their velocity vectors is $\phi$ , the magnitude of relative velocity is calculated using the Law of Cosines:

|\vec{v}_{A,B}| = \sqrt{v_{A,G}^{2} + v_{B,G}^{2} - 2v_{A,G}v_{B,G}\cos\phi}

Three Important Special Cases:

1. Motion in Same Direction ( $\phi = 0^\circ$ ): $\cos 0^\circ = 1$ , so $|\vec{v}_{A,B}| = |v_{A,G} - v_{B,G}|$ (Minimum relative speed).
2. Perpendicular Motion ( $\phi = 90^\circ$ ): $\cos 90^\circ = 0$ , so $|\vec{v}_{A,B}| = \sqrt{v_{A,G}^2 + v_{B,G}^2}$ (Pythagoras Theorem).
3. Opposite Direction / Head-on ( $\phi = 180^\circ$ ): $\cos 180^\circ = -1$ , so $|\vec{v}_{A,B}| = v_{A,G} + v_{B,G}$ (Maximum relative speed).

Verification: At $\phi = 0^\circ$ and $\phi = 180^\circ$ , the formula correctly reduces to standard one-dimensional results.

The three angles that come up most often, and the expression each one collapses to.

Two ships have speeds of $12\,\text{km/h}$ and $5\,\text{km/h}$ relative to the water. Which of the following relative speeds is IMPOSSIBLE regardless of their directions of motion?

Chain Rule for Frame Transformations (Triangle Law)

The relative velocity of object A with respect to frame C can be expressed using an intermediate reference frame B as:

\vec{v}_{A,C} = \vec{v}_{A,B} + \vec{v}_{B,C}

Proof: Substituting definitions: $(\vec{v}_{A,G} - \vec{v}_{B,G}) + (\vec{v}_{B,G} - \vec{v}_{C,G}) = \vec{v}_{A,G} - \vec{v}_{C,G} = \vec{v}_{A,C}$ . The intermediate frame B cancels out.

Two links laid head to tail, and the third side that closes them. The shared middle object cancels.

Subscript Memory Rule: In the sum $\vec{v}_{A,\mathbf{B}} + \vec{v}_{\mathbf{B},C}$ , the inner subscripts ( $B$ ) match and cancel, leaving the outer subscripts $\vec{v}_{A,C}$ .

Three links laid end to end. The inner subscripts pair off and cancel, leaving the outermost two.

A passenger walks toward the front of a train at $1.5\,\text{m/s}$ relative to the train. The train moves forward at $30\,\text{m/s}$ relative to the ground. A second train moves in the opposite direction at $25\,\text{m/s}$ relative to the ground. What is the velocity of the passenger relative to the second train?

Worked Example: Distance of Closest Approach

Problem Statement

At 12:00 noon, Ship B is $20\,\text{km}$ due west of Ship A. Ship A travels due north at $20\,\text{km/h}$ . Ship B travels due east at $15\,\text{km/h}$ .
(a) Find the magnitude and direction of the velocity of Ship B relative to Ship A.
(b) Calculate the shortest distance between the two ships and the time when it occurs.

Show Solution to Part (a)

Step 1: Write Velocity Vectors in Component Form.
Let east be $+x$ and north be $+y$ :

\vec{v}_{A,G} = (0,\,20)\,\text{km/h}, \qquad \vec{v}_{B,G} = (15,\,0)\,\text{km/h}

\vec{v}_{B,A} = \vec{v}_{B,G} - \vec{v}_{A,G} = (15,\,-20)\,\text{km/h}

Step 2: Calculate Magnitude and Direction.

|\vec{v}_{B,A}| = \sqrt{15^2 + (-20)^2} = 25\,\text{km/h}

\tan\theta = \frac{15}{20} = 0.75 \quad\Longrightarrow\quad \theta = 36.9^\circ \text{ East of South}

The two courses, and the subtraction that turns them into one arrow. Reversing A's velocity and adding is all that happens.

Show Solution to Part (b)

Step 3: Work in Reference Frame of Ship A.
In Ship A's frame, Ship A is stationary at the origin $(0,0)$ . Ship B starts at position $\vec{r}_0 = (-20,\,0)\,\text{km}$ and moves in a straight line with constant velocity $\vec{v}_{B,A} = (15,\,-20)\,\text{km/h}$ .

Step 4: Find Time of Closest Approach ( $t^*$ ).
Distance is minimum when the relative position vector is perpendicular to the relative velocity vector ( $\vec{r} \cdot \vec{v}_{B,A} = 0$ ):

t^* = -\frac{\vec{r}_0 \cdot \vec{v}_{B,A}}{|\vec{v}_{B,A}|^2} = -\frac{(-20)(15) + (0)(-20)}{25^2} = \frac{300}{625} = 0.48\,\text{h} = 28.8\,\text{minutes}

Step 5: Calculate Minimum Separation Distance.
At $t^* = 0.48\,\text{h}$ , position of B relative to A is:

\vec{r} = (-20, 0) + 0.48(15, -20) = (-12.8, -9.6)\,\text{km}

d_{\text{min}} = \sqrt{(-12.8)^2 + (-9.6)^2} = 16\,\text{km}

The same two ships in A's frame. B runs a straight line, and the closest approach is just the perpendicular from A onto it.

Which of the following vector expressions is NOT mathematically equivalent to $\vec{v}_{A,C}$ ?

Relative Displacement in Two Dimensions

For uniform motion, the relative displacement of object A with respect to object B in time $t$ is given by:

\vec{s}_{A,B} = \vec{v}_{A,B}\,t

The distance between the two objects at time $t$ is the magnitude of this relative displacement vector: $|\vec{s}_{A,B}| = |\vec{v}_{A,B}|\,t$ .

Two ships leave the same port at the same instant. Ship A travels due north at $12\,\text{km/h}$ and Ship B travels due east at $16\,\text{km/h}$ . What is the distance between them after $30$ minutes ( $0.5\,\text{h}$ )?

River Crossings and Falling Rain

Standard Applications: River Crossing & Rain Problems

Two classic application problems frequently appear in examinations: a boat/swimmer crossing a flowing river, and a person walking through falling rain. Both situations are solved using the 2D vector addition equation:

\vec{v}_{\text{object, ground}} = \vec{v}_{\text{object, medium}} + \vec{v}_{\text{medium, ground}}

Important Reference Frame Distinction:

Swimmer/Boat speed in still water ( $v_{s,r}$ ) is velocity relative to the river water.
Rainfall velocity relative to ground is $v_{r,G}$ .
Walking speed relative to ground is $v_{m,G}$ .

Two problems that look nothing alike, drawn as the same three-sided figure.

River Crossing: General Vector Equation

Let a river of width $d$ flow with velocity $\vec{v}_{r,G}$ along the bank (say, $+x$ direction). A swimmer swims with speed $v_{s,r}$ relative to the water at an angle $\theta$ with the perpendicular across the river.

By the chain rule, the swimmer's velocity relative to the ground is:

\vec{v}_{s,G} = \vec{v}_{s,r} + \vec{v}_{r,G}

Component form:

Perpendicular to bank (across river): $v_{y} = v_{s,r}\cos\theta$
Parallel to bank (along river): $v_{x} = v_{r,G} - v_{s,r}\sin\theta$

Key Concept Test: To cross the river in the shortest possible time, should you steer straight across or angle upstream?

Use the interactive controls below to verify your answer.

Aiming straight across is quickest; aiming upstream lands you opposite. You cannot have both.

Result: The crossing time is given by $t = \frac{d}{v_y} = \frac{d}{v_{s,r}\cos\theta}$ . For shortest time, $\cos\theta = 1$ (i.e. $\theta = 0^\circ$ , steering straight across).

Note: The river current flow velocity $v_{r,G}$ has zero component across the river. Consequently, it has no effect on the time taken to cross the river. It only causes downstream drift.

Two identical swimmers attempt to cross a flowing river. Swimmer A steers straight across perpendicular to the bank, while Swimmer B angles upstream. Who reaches the opposite bank first?

Worked Example: Shortest Time Crossing

Problem Statement

A swimmer can swim at $4\,\text{km/h}$ in still water. A river $1\,\text{km}$ wide flows at $3\,\text{km/h}$ .
(a) Find the direction the swimmer must head to cross in the shortest possible time. Calculate this minimum time.
(b) Find the downstream drift distance when landing on the opposite bank.

Show Solution

(a) Shortest Time Crossing:
To minimize crossing time, steer straight across perpendicular to the bank ( $\theta = 0^\circ$ ).

t_{\text{min}} = \frac{d}{v_{s,r}} = \frac{1\,\text{km}}{4\,\text{km/h}} = 0.25\,\text{h} = 15\,\text{minutes}

(b) Downstream Drift:
During these 15 minutes, the river current carries the swimmer downstream:

x = v_{r,G} \times t = 3\,\text{km/h} \times 0.25\,\text{h} = 0.75\,\text{km} = 750\,\text{m}

Ground path length $s = \sqrt{1^2 + 0.75^2} = 1.25\,\text{km}$ at ground speed $v_{s,G} = \sqrt{4^2 + 3^2} = 5\,\text{km/h}$ ( $t = 1.25 / 5 = 0.25\,\text{h}$ ).

The answer put back on the picture: fifteen minutes across, 0.75 km downstream.

A swimmer heads straight across a river and reaches the far bank in 15 minutes. On the next day, the river current doubles in speed. If the swimmer again heads straight across with the same swimming speed, how long will the crossing take?

Worked Example: Shortest Path Crossing (Zero Drift)

Problem Statement

Using the same swimmer ( $v_{s,r} = 4\,\text{km/h}$ ) and river (width $d = 1\,\text{km}$ , current $v_{r,G} = 3\,\text{km/h}$ ), find the direction the swimmer must head to land at the point directly opposite the starting point (zero drift). Calculate the corresponding crossing time.

Show Solution

Step 1: Zero Drift Condition.
For zero drift, the net velocity along the bank must be zero ( $v_x = 0$ ). The upstream component of swimming velocity must balance the river current:

v_{s,r}\sin\theta = v_{r,G} \quad\Longrightarrow\quad \sin\theta = \frac{v_{r,G}}{v_{s,r}} = \frac{3}{4} = 0.75 \quad\Longrightarrow\quad \theta = 48.6^\circ \text{ Upstream}

Step 2: Calculate Across-River Velocity Component.

v_y = v_{s,r}\cos\theta = \sqrt{v_{s,r}^2 - v_{r,G}^2} = \sqrt{4^2 - 3^2} = \sqrt{7} \approx 2.65\,\text{km/h}

Step 3: Calculate Crossing Time.

t = \frac{d}{v_y} = \frac{1\,\text{km}}{2.65\,\text{km/h}} = 0.378\,\text{h} \approx 22.7\,\text{minutes}

Aimed upstream at 48.6°, the along-bank velocity cancels and the ground path is straight across.

Condition for Possibility of Zero Drift:
Since $\sin\theta = v_{r,G} / v_{s,r} \le 1$ , zero drift is possible only if $v_{s,r} \ge v_{r,G}$ (swimmer speed in still water is greater than or equal to river current speed). If $v_{s,r} < v_{r,G}$ , zero drift is impossible.

A boat slower than the current. However far upstream it points, the resultant still carries it downstream.

Crossing time and drift against heading angle. One is smallest at 0°, the other at 48.6°, and no angle does both.

A motorboat has a speed of $3\,\text{m/s}$ in still water. The river flows at $5\,\text{m/s}$ . Can the motorboat reach the point directly opposite on the far bank?

Rain-Umbrella Problem

Suppose rain falls vertically downward with velocity $\vec{v}_{r,G}$ . A man walks horizontally forward with velocity $\vec{v}_{m,G}$ . To protect himself from the rain, he must hold his umbrella along the direction of the velocity of rain relative to the man ( $\vec{v}_{r,m}$ ).

By vector subtraction:

\vec{v}_{r,m} = \vec{v}_{r,G} - \vec{v}_{m,G} = \vec{v}_{r,G} + (-\vec{v}_{m,G})

The forward motion of the man adds a backward component ( $-\vec{v}_{m,G}$ ) to the rain's velocity relative to the man. Consequently, the rain appears to come from the front at an angle.

Concept Question: As you walk faster through vertical rain, should you tilt your umbrella further forward, further back, or keep it unchanged?

Use the interactive controls below to verify your answer.

The faster you walk, the more the apparent rain leans forward, and the further forward the umbrella has to go.

Umbrella Tilt Angle ( $\alpha$ from vertical):

\tan\alpha = \frac{v_{m,G}}{v_{r,G}} \quad\Longrightarrow\quad \alpha = \tan^{-1}\left(\frac{v_{m,G}}{v_{r,G}}\right)

Apparent Speed of Rain:

v_{r,m} = \sqrt{v_{m,G}^2 + v_{r,G}^2}

The apparent speed of the rain against walking speed. It starts at the falling speed and never stops rising.

While walking through vertical rain, you hold an umbrella tilted forward. If you now increase your walking speed, how should you adjust the umbrella angle, and what happens to the apparent speed of rain?

Common Misconception in Rain-Umbrella Problem

Conceptual Misconception

A student argues: "Since I am moving forward, the rain moves backward relative to me. 'Backward' means behind me. Therefore, I should tilt the umbrella backward." Explain why this reasoning is incorrect.

Show Explanation

Explanation:
The statement "the rain moves backward relative to me" means the velocity vector $\vec{v}_{r,m}$ has a backward horizontal component. A velocity pointing backward and downward means rain drops are travelling toward your front chest/face from ahead.
To block oncoming rain drops travelling backward toward you, the umbrella must be tilted forward into the direction of oncoming rain.

The two tilts side by side. Only one of them is between the walker and the oncoming rain.

Rain falls vertically at $4\,\text{m/s}$ . A man walking at $4\,\text{m/s}$ holds his umbrella at $45^\circ$ from the vertical. If he doubles his walking speed to $8\,\text{m/s}$ , what is the new umbrella tilt angle from vertical?