A swimmer desires to cross a 500 m wide river that is flowing at 5...

Let the width of the river be $d = 500 \text{ m} = 0.5 \text{ km}$.

The velocity of the swimmer relative to the water is $v_{s} = 3 \text{ km/h}$.

The velocity of the river flow is $v_{r} = 5 \text{ km/h}$.

To cross the river in the shortest possible time, the swimmer must head exactly perpendicular to the river flow. The component of the swimmer's velocity along the width of the river is maximum in this case and is equal to $v_{s}$.

The shortest time taken to cross the river is given by:

t_{\text\min} = \dfrac{d}{v_{s}}

t_{\text\min} = \dfrac{0.5}{3} \text{ hours}

Converting the time into minutes:

t_{\text\min} = \dfrac{0.5}{3} \times 60 \text{ minutes} = 10 \text{ minutes}