Answer:
Speed of the current was 10 mi/h
Step-by-step explanation:
Let's start by recalling the formula that relates speed, distance and time:
[tex]speed = \frac{distance}{time}[/tex]
From which we can derive the expression for time:
[tex]time = \frac{distance}{speed}[/tex]
Now let's analyze the composition of velocities of the motorboat ([tex]s_b[/tex]) and that of the current ([tex]s_c[/tex]):
When the motorboat goes upstream (against the current), its actual traveling speed is its speed relative to water minus the speed of the current [tex]s_b+s_c[/tex]
When the motorboat goes downstream (same direction as the current) its actual traveling speed is the addition of its speed relative to the water plus the speed of the current ([tex]s_b+s_c[/tex]).
We know the speed of the motorboat relative to the water ([tex]s_b=22\,\,\frac{mi}{h}[/tex]), while the speed of the current ([tex]s_c[/tex]) is our unknown.
Since another piece of information we have is the distance covered on each direction (48 miles going upstream, and also 48 miles going downstream), and the time it took the total trip (5.5 hours), we can set an equation for the total time as the addition of the time to travel downstream, plus the time to travel downstream:
[tex]Total \,\,time=time_{upstream}+time_{downstream}\\5.5 =\frac{48}{22-s_c} \,+\,\frac{48}{22+s_c}[/tex]
We can then solve for our unknown in this retional equation:
[tex]5.5 =\frac{48}{22-s_c} \,+\,\frac{48}{22+s_c}\\5.5 =\,\frac{48(22+s_c)+48(22-s_c))}{(22-s_c)(22+s_c)}\\5.5\,(22^2-s_c^2)=1056+48\,s_c+1056-48\,s_c\\2662-5.5s_c^2=2112\\2662-2112=5.5\,s_c\\550=5.5\,s_c^2\\100=s_c^2\\s_c=+/-\,10\,\, \frac{mi}{h}[/tex]
Since the speed of the current must be a positive number, we select 10 mi/h as the answer
