Question

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?

Answer 1

The time taken is
`t_(out) = (D)/(v__(R)) + v__(S)}}`

Step-by-step explanation:

From the question we are told that

The speed of the current is
`v__(R)}`

The speed of the swimmer in direction of current is
`v__(S)}`

The distance traveled by the swimmer is
`D`

The time taken to travel this distance is
`t_(out)`

The speed of the swimmer against direction of current is
`v__(s)}`

The resultant speed for downstream current is

`V_(r) = v__(S)} +v__(R)}`

The time taken can be mathematically represented as

`t_(out) = (D)/(V_(r))`

`t_(out) = (D)/(v__(R)) + v__(S)}}`