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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.

User ChrisBe
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1 Answer

15 votes
15 votes

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}

A boat crossing a 153.0 m wide river is directed so that it will cross the river as-example-1
User Ugavetheroses
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