Answer:
Speed of the boat in still water = 6.125 miles/hour
Step-by-step explanation:
We are given that a boat travels 33 miles downstream in 4 hours and the return trip takes the boat 7 hours.
We are to find the speed of the boat in the still water.
Assuming [tex]S_b[/tex] to be the speed of the boat in still water and [tex]S_w[/tex] to be the speed of the water.
The speeds of the boat add up when the boat and water travel in the same direction.
[tex]Speed = \frac{distance}{time}[/tex]
[tex]S_b+S_w=\frac{d}{t_1}=\frac{33 miles}{4 hours} [/tex]
And the speed of the water is subtracted from the speed of the boat when the boat is moving upstream.
[tex]S_b-S_w=\frac{d}{t_2}=\frac{33 miles}{7 hours} [/tex]
Adding the two equations to get:
[tex]S_b+S_w=\frac{d}{t_1}[/tex]
+ [tex]S_b-S_w=\frac{d}{t_2} [/tex]
___________________________
[tex]2S_b=\frac{d}{t_1} +\frac{d}{t_2}[/tex]
Solving this equation for [tex]S_b[/tex] and substituting the given values for [tex]d,t_1, t_2[/tex]:
[tex]S_b=\frac{(t_1+t_2)d}{2t_1t_2}[/tex]
[tex]S_b=\frac{(4 hour + 7hour)33 mi}{2(4hour)(7hour)}[/tex]
[tex]S_b=\frac{(11 hour)(33mi)}{56hour^2}[/tex]
[tex]S_b=6.125 mi/hr[/tex]
Therefore, the speed of the boat in still water is 6.125 miles/hour.
Answer:
[tex]6.48\frac{mi}{h}[/tex]
Step-by-step explanation:
Let' call "b" the speed of the boat and "c" the speed of the river.
We know that:
[tex]V=\frac{d}{t}[/tex]
Where "V" is the speed, "d" is the distance and "t" is the time.
Then:
[tex]d=V*t[/tex]
We know that distance traveled downstream is 33 miles and the time is 4 hours. Then, we set up the folllowing equation:
[tex]4(b+c)=33[/tex]
For the return trip:
[tex]7(b-c)=33[/tex] (Remember that in the return trip the speed of the river is opposite to the boat)
By solving thr system of equations, we get:
- Make both equations equal to each other and solve for "c".
[tex]4(b+c)=7(b-c)\\\\4b+4c=7b-7c\\\\4c+7c=7b-4b\\\\11c=3b\\\\c=\frac{3b}{11}[/tex]
- Substitute "c" into any original equation and solve for "b":
[tex]4b+\frac{3b}{11} =33\\\\4b+\frac{12b}{11}=33\\\\\frac{56b}{11}=33\\\\b=6.48\frac{mi}{h}[/tex]
