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?

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moya1316 moya1316 · Answer 1 · 2021-02-16T13:36:33+01:00

Answer:

t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}, t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

[tex]v_{sg 1} = v_{sr} + v_{rg}[/tex]

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

[tex]v_{sg1}[/tex] = D / [tex]t_{out}[/tex]

D = v_{sg1} t_{out}

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

[tex]v_{sg 2} = v_{sr} - v_{rg}[/tex]

[tex]v_{sg 2}[/tex] = D / [tex]t_{in}[/tex]

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

[tex]v_{sg1} t_{out} = v_{sg2} t_{in}[/tex]

t_{out} = t_{in}

t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}

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

t_{in}= D / [tex]v_{sg2}[/tex]

we substitute

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

t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]