# Thread: Speed of the Boat in Still Water

1. ## 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?

2. 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.

3. ## perfectly done............

Originally Posted by Alienis Back
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.

4. Hello, magentarita!

Another approach . . .

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

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: . $\frac{5}{x-3} \:=\:\frac{11}{x+3}\quad\hdots\quad There!$

5. ## ok............

Originally Posted by Soroban
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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