The next four questions refer to the situation below. A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i. e. in the direction of the current) at a constant speed, vS , with respect to the water. The swimmer travels a distance D in a time tOut . The swimmer then changes direction to swim upstream (i. e. against the direction of the current) at a constant speed, vS , with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn .
What is tOut in terms of vR, vS, and D, as needed?

 t_{out} = t_{in},      t_{out} =

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

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

            = D /

           D = v_{sg1}  t_{out}

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

          = D /

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

          t_{out} =  t_{in}

           t_{out} = t_{in}

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

            t_{in}= D /

we substitute

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

            t_{out} =

gas and liquid