Question

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of 5.10 m/s in still water and the river flows uniformly at 3.70 m/s. Calculate the total distance the boat will travel to reach the opposite shore.

Answer 1

We have the relation

`\vec v_(B \mid E) = \vec v_(B \mid R) + \vec v_(R \mid E)`

where
`v_(A \mid B)` denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

`v_(B \mid R) = 5.10 (\rm m)/(\rm s)`

`v_(R \mid E) = 3.70 (\rm m)/(\rm s)`

Let's assume the river flows South-to-North, so that

`\vec v_(R \mid E) = v_(R \mid E) \, \vec\jmath`

and let
`-90^\circ < \theta < 90^\circ` be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

`\vec v_(B \mid R) = v_(B \mid R) \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)`

Then the velocity of the boat relative to the Earth is

`\vec v_(B\mid E) = v_(B \mid R) \cos(\theta) \, \vec\imath + \left(v_(B \mid R) \sin(\theta) + v_(R \mid E)\right) \,\vec\jmath`

The crossing is 153.0 m wide, so that for some time
`t` we have

`153.0\,\mathrm m = v_(B\mid R) \cos(\theta) t \implies t = (153.0\,\rm m)/(\left(5.10(\rm m)/(\rm s)\right) \cos(\theta)) = 30.0 \sec(\theta) \, \mathrm s`

which is minimized when
`\theta=0^\circ` so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

`\vec v_(B \mid E) = v_(B \mid R) \,\vec\imath + \vec v_(R \mid E) \,\vec\jmath \\\\ \implies v_(B \mid E) = \sqrt{\left(5.10(\rm m)/(\rm s)\right)^2 + \left(3.70(\rm m)/(\rm s)\right)^2} \approx 6.30 (\rm m)/(\rm s)`

The boat's position
`\vec x` at time
`t` is

`\vec x = \vec v_(B\mid E) t`

so that after 30.0 s, the boat's final position on the other side of the river is

`\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath`

and the boat would have traveled a total distance of

`\|\vec x(30.0\,\mathrm s)\| = √((153\,\mathrm m)^2 + (111\,\mathrm m)^2) \approx \boxed{189\,\mathrm m}`