Question

A boat heads upstream a distance of 30 miles on the Mississippi​ river, whose current is running at 5 miles per hour. If the trip back takes an hour​ less, what was the speed of the boat in still​ water? Give the answer rounded to two decimal​ places, if necessary.

Answer 1

The speed of the boat in still water is 18.03 mph.

Explanation:

Let's call
`v` the speed of the boat in still water and
`u` the speed of the current.

When the boat is heading upstream its absolute speed will be
`v-u`.

When the boat is heading downstream its absolute speed will be
`v+u`.

In any case, knowing the absolute speed of the boat (the speed with respect of the land), we can calculate the distance traveled during a given time

`distance=speed*time` (equation 1)

So, when the boat is heading upstream, the equation 1 will be:

`30mi=(v-5mph)*t` (equation 2)

And when the boat is heading downstream, the equation will be:

`30mi=(v+5mph)*(t-1h)` (equation 3)

Equaling equations 2 and 3 we may find the value of v

`(v-5)*t=(v+5)(t-1)`

`vt-5t=vt-v+5t-5`

`t=(1)/(10)(v+5)` (equation 4)

Replacing equation 4 in equation 2:

`30=(v-5)*(1)/(10)(v+5)`

`300=v^2-25`

`v=√(325)`

`v=18.03 mph`

Then, the speed of the boat in still water is 18.03 mph.