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.
Accepted Solution
A:
Answer:The speed of the boat in still water is 18.03 mph.Step-by-step explanation:Let's call [tex]v[/tex] the speed of the boat in still water and [tex]u[/tex] the speed of the current.When the boat is heading upstream its absolute speed will be [tex]v-u[/tex].When the boat is heading downstream its absolute speed will be [tex]v+u[/tex].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[tex]distance=speed*time[/tex] (equation 1)So, when the boat is heading upstream, the equation 1 will be:[tex]30mi=(v-5mph)*t[/tex] (equation 2)And when the boat is heading downstream, the equation will be:[tex]30mi=(v+5mph)*(t-1h)[/tex] (equation 3)Equaling equations 2 and 3 we may find the value of v[tex](v-5)*t=(v+5)(t-1)[/tex][tex]vt-5t=vt-v+5t-5[/tex][tex]t=\frac{1}{10}(v+5)[/tex] (equation 4)Replacing equation 4 in equation 2:[tex]30=(v-5)*\frac{1}{10}(v+5)[/tex][tex]300=v^2-25[/tex][tex]v=\sqrt{325}[/tex][tex]v=18.03 mph[/tex]Then, the speed of the boat in still water is 18.03 mph.