Question

The next four questions refer to the situation below.

Answer 1

Answer:

 t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in},      t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         $v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           $v_{sg1}$ = D / $t_{out}$

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        $v_{sg 2} = v_{sr} - v_{rg}$

         $v_{sg 2}$ = D / $t_{in}$

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           $v_{sg1} t_{out} = v_{sg2} t_{in}$

          t_{out} =  t_{in}

           t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / $v_{sg2}$

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = $\frac{D}{v_s +v_r}$