Crossing The River

Let the width of the river be ' $d$ '. Let the time taken by the swimmer to cross the river in minimum time be ${t}_{1}$ and that to cross the river via the shortest path is ${t}_{2}$ . Let his horizontal drift for the first case be $x$ . Furthermore, let the speed of the swimmer in still water be ${v}_{m}$ and that of the river itself, be ${v}_{r}$ . We'll derive a general result and just plug in the given information to get our answer.

Case 1: Minimum time

For the swimmer to cross the river in minimum time, he must always swim perpendicular to the direction of flow of river. Now, look at the following diagram.

The time taken by man to reach $B$ from $O$ horizontally is only because of the river's velocity, or in other words:

${v}_{r} = \frac {x} {{t}_{1}}$

and for vertical motion, it is clear that the distance $d$ and the swimmer's velocity ${v}_{s}$ can be related as:

${v}_{m} = \frac {d} {{t}_{1}}$

Case 2: Shortest distance

For the swimmer to cross the river via the shortest path, he should swim at an angle $\theta$ with the river velocity, so as to cancel out the river's horizontal velocity, and allow himself to always travel in the vertical direction.

From the above diagram, it is clear, that to travel in the vertical direction, the following relation must always hold:

${v}_{m} \cos {\theta} = {v}_{r}$

and for vertical motion:

${v}_{m} \sin {\theta} = \frac {d} {{t}_{2}}$

Now, squaring and adding the last two equations, so as to eliminate $\theta$ , and using up the value of ${v}_{m}$ and ${v}_{r}$ from the equations formed earlier, we get:

$\frac {{d}^{2}} {{{t}_{1}}^{2}} =\frac {{d}^{2}} {{{t}_{2}}^{2}} + \frac {{x}^{2}} {{{t}_{1}}^{2}}$

or by simplifying:

${ d }^{ 2 }=\frac { { x }^{ 2 } }{ { { t }_{ 1 } }^{ 2 } } \frac { { { t }_{ 1 } }^{ 2 }{ { t }_{ 2 } }^{ 2 } }{ { { t }_{ 2 } }^{ 2 }-{ { t }_{ 1 } }^{ 2 } }$

and finally:

${ d }=\frac { x{ t }_{ 2 } }{ \sqrt { ({ t }_{ 2 }-{ t }_{ 1 })({ t }_{ 1 }+{ t }_{ 2 }) } }$

Now, inserting the given values, i.e. $x= 120m$ , ${t}_{1} = 10$ minutes and ${t}_{2} = 12.5$ minutes, we get $d = 200m$