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Old 11-19-2008, 12:12 AM
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Default Speed of the Boat in Still Water

The speed of a stream is 3 mph. A boat travels 5 miles upstream in the same time it takes to travel 11 miles downstream. What is the speed of the boat in still water?
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Old 11-19-2008, 07:30 AM
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Speed up stream = x -3 (-3 because the stream movement will be negative in relation to the boat) and the boat will travel 5 miles

x-3=5

x=5+3

x=8

You can check using the other data.
Speed down stream = x+3 (because the stream and boat movement will add) and the boat will travel 11 miles

x+3=11

x=11-3

x=8

Now you see, the boat -on still water- travels at 8 mph.
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Old 11-19-2008, 10:31 AM
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Default perfectly done............

Quote:
Originally Posted by Alienis Back View Post
Speed up stream = x -3 (-3 because the stream movement will be negative in relation to the boat) and the boat will travel 5 miles

x-3=5

x=5+3

x=8

You can check using the other data.
Speed down stream = x+3 (because the stream and boat movement will add) and the boat will travel 11 miles

x+3=11

x=11-3

x=8

Now you see, the boat -on still water- travels at 8 mph.
Your answer is wonderfully simple.
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Old 11-19-2008, 10:48 AM
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Hello, magentarita!

Another approach . . .

We'll use: .\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}


Quote:
The speed of a stream is 3 mph.
A boat travels 5 miles upstream in the same time it takes to travel 11 miles downstream.
What is the speed of the boat in still water?
Let x = boat's speed in still water.


Going upstream, the current works against the boat. .The boat's speed is x - 3 mph.
. . It went 5 miles at x-3 mph. .This took: .\frac{5}{x-3} hours.

Going downstream, the current works with the boat. .The boat's speed is x+3 mph.
. . It went 11 miles at x+3 mph. .This took: .\frac{11}{x+3} hours.


These two times are equal: .\frac{5}{x-3} \:=\:\frac{11}{x+3}\quad\hdots\quad There!

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Old 11-19-2008, 05:37 PM
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Default ok............

Quote:
Originally Posted by Soroban View Post
Hello, magentarita!

Another approach . . .

We'll use: .\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}

Let x = boat's speed in still water.


Going upstream, the current works against the boat. .The boat's speed is x - 3 mph.
. . It went 5 miles at x-3 mph. .This took: .\frac{5}{x-3} hours.

Going downstream, the current works with the boat. .The boat's speed is x+3 mph.
. . It went 11 miles at x+3 mph. .This took: .\frac{11}{x+3} hours.


These two times are equal: .\frac{5}{x-3} \:=\:\frac{11}{x+3}\quad\hdots\quad There!
What can I say? Your replies indicate that you are a teacher or were a teacher at one time.
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