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

Answer:

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

Step-by-step 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

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

$vt-5t=vt-v+5t-5$

$t=\frac{1}{10}(v+5)$ (equation 4)

Replacing equation 4 in equation 2:

$30=(v-5)*\frac{1}{10}(v+5)$

$300=v^2-25$

$v=\sqrt{325}$

$v=18.03 mph$

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