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**Hint:**These problems are called river problems in one dimensional motion. In these problems we have to calculate the absolute velocity of the swimmer across all three edges with respect to absolute velocity of the river. As the swimmer has constant speed in swimming this will be taken as the velocity of the swimmer in still water.

**Complete step by step answer:**

Let’s look at the diagram. Let’s assume the flow of water is along a positive x-axis.

Let $v$ be the velocity of the swimmer in still water, ${v_w}$ be the velocity of water flow and ${v_a}$ be the absolute velocity of the swimmer with respect to the flow of the river.

Thus the final velocity or absolute velocity of the swimmer will be the vector sum of its velocity in still water and the absolute velocity of the river.

Or \[\overrightarrow {{v_a}} = \overrightarrow v + \overrightarrow {{v_w}} \]

The magnitude of the resultant velocity vector is given by,

$ \Rightarrow {v_a} = {v^2} + {v_w}^2 + 2v{v_w}\cos \theta $

Where $\theta $ is the angle between the vectors.

From this we can conclude that the value of absolute velocity of the swimmer will be maximum for $\theta = 0^\circ $ and minimum for $\theta = 180^\circ $ .

Let’s now analyse the options.

If the swimmer now swims from A to B or along the axis CA, the angles between the velocity of the river and velocity of the swimmer will be $60^\circ $ .

But when the swimmer is travelling from B point to C point, the angle between the vectors will be $180^\circ $ , so the velocity will be minimum.

Thus the time taken by the swimmer will be maximum when he swims along BC.

**The correct answer is (B), from B to C.**

**Note:**We can also use vector resolving on these problems, but as it was a simpler problem, vector resolving was not necessary. If the triangle was inverted and the examiner asked about the shortest time interval the swimmer would take, the answer would have been from point C to point B, as the angle between the vectors of velocities of river and swimmer is $0^\circ $.

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