A river flows due east at a speed of 3m/s. A swimmer can swim in still water at a speed of 4m/s. Point A is on

NCERT Exemplar · Q34

3\ \text{m/s}

(a) If the swimmer starts swimming due north, find his resultant velocity (magnitude and direction).

(b) If he wants to start from A and reach the exactly opposite point B on the north bank,

(i) in which direction should he swim, and

(ii) what will be his resultant speed?

(a) and (b), in which case does he reach the opposite bank in a shorter time?

Bihar BsebLong· 5mImportance★★★★★est

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Concept understanding — Relative Velocity in Two Dimensions

Relative Velocity in Two Dimensions

You already know relative velocity in one dimension — the speed of a car measured from another moving car. Two dimensions is the same idea, but now velocities have both magnitude and direction, so subtraction becomes vector subtraction.

The Intuition

Imagine you're standing on a train platform. A train passes you at 15 m/s. From your frame, the train's velocity is 15 m/s forward. Now imagine you're on that train, walking toward the front at 2 m/s relative to the train. Your velocity relative to the platform is 15 + 2 = 17 m/s forward. That's one-dimensional relative velocity: just add or subtract numbers.

Now picture a boat crossing a river. The boat points straight across, but the river flows sideways. From the bank, the boat moves diagonally — its velocity is the vector sum of its own velocity (relative to water) and the water's velocity (relative to bank). To find how the boat moves relative to the bank, you add vectors.

The reverse problem: you know the boat's velocity relative to bank and the water's velocity. To find the boat's velocity relative to water, you subtract vectors. That subtraction is the core of relative velocity in two dimensions.

The Precise Statement

If object A has velocity $\vec{v}_A$ and object B has velocity $\vec{v}_B$ , both measured in the same frame (say, the ground), then the velocity of A relative to B is:

\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B

This is a vector equation. It says: to find how A appears to move from B's perspective, take A's velocity and subtract B's velocity (vector subtraction). Equivalently, $\vec{v}_{A/B}$ is the velocity you would measure if you were riding on B.

Important

The order matters: $\vec{v}_{A/B}$ means "velocity of A as seen from B". Always write the first object minus the second.

Why Vector Subtraction?

In one dimension, subtraction is just flipping a sign. In two dimensions, velocities have components. Subtracting vectors means subtracting corresponding components:

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}_{A/B} = (v_{Ax} - v_{Bx})\,\hat{i} + (v_{Ay} - v_{By})\,\hat{j}

Geometrically, you place the tails of $\vec{v}_A$ and $\vec{v}_B$ together. The vector from the tip of $\vec{v}_B$ to the tip of $\vec{v}_A$ is $\vec{v}_{A/B}$ .

The Two Classic Problems

Rain-Man Problem: A man walks in rain. Rain falls vertically downward at speed $v_r$ relative to ground. The man walks horizontally at speed $v_m$ . To him, the rain appears to come at an angle. Why? Because the rain's velocity relative to the man is $\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m$ . Since $\vec{v}_r$ is downward and $\vec{v}_m$ is horizontal, the relative velocity is diagonal — the man must tilt his umbrella forward. …

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Add the swimmer's velocity and the river's velocity as vectors. Swimming straight north gives a $3$ - $4$ - $5$ right triangle, so $5\ \text{m/s}$ at about $37^\circ$ east of north. To land exactly opposite, he must angle upstream so the westward part of his stroke cancels the current, leaving only $\sqrt7\ \text{m/s}$ across. Because the whole $4\ \text{m/s}$ goes across the river in case (a) but only $\sqrt7\ \text{m/s}$ in case (b), case (a) crosses faster.

(a) Swimming due north

The swim velocity is $4\ \text{m/s}$ north; the current is $3\ \text{m/s}$ east. These are perpendicular, so the resultant magnitude is

v = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\ \text{m/s}.

Its direction, measured from north toward east, is

\phi = \tan^{-1}\!\left(\frac{3}{4}\right) \approx 36.9^\circ \ \text{east of north}.

The swimmer drifts downstream and lands to the east of B.

(b) Reaching the point B directly opposite

To have zero net eastward drift, the westward (upstream) component of his swim must cancel the $3\ \text{m/s}$ current. If he swims at angle $\phi$ west of north,

4\sin\phi = 3 \;\Rightarrow\; \sin\phi = \frac{3}{4} \;\Rightarrow\; \phi = \sin^{-1}\!\left(\frac34\right)\approx 48.6^\circ \ \text{west of north}.

The resultant (purely northward) speed is the remaining across-river component:

v_{\perp} = 4\cos\phi = 4\sqrt{1 - \tfrac{9}{16}} = 4\cdot\frac{\sqrt7}{4} = \sqrt7 \approx 2.65\ \text{m/s (due north)}.

(c) Which is faster

Let the river width be $d$ . The crossing time depends only on the across-river (northward) speed.

Case (a): the northward component is the full $4\ \text{m/s}$ , so $t_a = d/4$ . …

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Concept: Solve for the Resultant's Magnitude First via Vector-Triangle Closure, Then Back Out the Angle

Method: Treat the Unknown Swim Direction as "Whatever Makes the Resultant Purely Northward," Solved by Pythagoras

The direct method finds the upstream angle first (from $\sin\phi=3/4$ ) and only then computes the resultant speed. This method reverses the order: it writes the swim vector itself as "resultant minus current" using the closure condition, and solves for the resultant's magnitude directly from the fixed swim speed via Pythagoras -- the angle is a by-product at the end.

Steps

State the vector closure condition for part (b). To land exactly at B, the resultant of swim velocity and current must point due north only, with unknown magnitude $R$ : $\vec v_{\text{result}} = (0,R)$ .
Isolate the (as yet unresolved) swim vector from the closure equation, using $\vec v_{\text{swim}} = \vec v_{\text{result}} - \vec v_{\text{current}}$ , with the current fixed at $\vec v_{\text{current}}=(3,0)$ :

\vec v_{\text{swim}} = (0,R) - (3,0) = (-3,\,R).

Use the one fact about the swim vector that IS fixed -- its magnitude is always $4\text{ m/s}$ -- to solve for $R$ via Pythagoras:

|\vec v_{\text{swim}}|^2 = (-3)^2+R^2 = 4^2 \implies 9+R^2=16 \implies R=\sqrt7\approx2.65\text{ m/s}.

Only now extract the swim direction, as the angle of $\vec v_{\text{swim}}=(-3,\sqrt7)$ from north:

\tan(\text{angle west of north}) = \frac{3}{\sqrt7} \implies \text{angle}\approx48.6^\circ \text{ west of north}.

(Equivalently, $\sin(\text{angle})=3/4$ from the horizontal component of the fixed-length swim vector -- the same relation the direct method starts from, reached here as a consequence rather than a starting assumption.) …

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Q.A river flows due east at a speed of 3 m/s3\ \text{m/s}3 m/s. A swimmer can swim in still water at a speed of 4 m/s4\ \text{m/s}4 m/s. Point A is on the south bank and point B is the point on the north bank directly opposite A (i.e. due north of A).

(a) Swimming due north

(b) Reaching the point B directly opposite

(c) Which is faster

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Q.A river flows due east at a speed of $3\ \text{m/s}$ . A swimmer can swim in still water at a speed of $4\ \text{m/s}$ . Point A is on the south bank and point B is the point on the north bank directly opposite A (i.e. due north of A).

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