# Speed of Boat in Still Water

• Dec 4th 2008, 02:36 PM
magentarita
Speed of Boat in Still Water
Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?

MY WORK:

Let A = against the current

Let W = with the current

The boat will travel faster with the current and slower against the current. Is this correct?

Here is my set up:

...........rate..........time........distance
A.......(x+4).....(x + 20)........(x+4)(x+20)
W.......(x-4).......(x)..............(x-4)(x)

Since the distance she travelled both ways is the same, I set up the following equation:

(x+4)(x+20) = (x-4)(x)

The letter x here represents the speed of the boat in still water.

Is any of this correct?

If not, what did I do wrong?

• Dec 4th 2008, 03:58 PM
nzmathman
$\displaystyle distance = velocity \times time$

Let t = time taken to travel downstream.
Let v = the velocity of the boat in still water.

using $\displaystyle d = vt$,

$\displaystyle 5 = (v - 4) \times (t + 20)$

$\displaystyle 5 = (v + 4) \times t$

We want 'v' on its own so we can solve, so divide the 'v parts' across to the other side.

$\displaystyle \frac{5}{v-4} = t + 20$

$\displaystyle \frac{5}{v+4} = t$

$\displaystyle \frac{5}{v - 4} = \frac{5}{v+4} + 20$

Multiply whole expression by $\displaystyle (v + 4)(v - 4)$

$\displaystyle 5(v + 4) = 5(v - 4) + 20(v + 4)(v - 4)$

Now simplify and solve the quadratic formed.

Don't hesitate to ask for further help...
• Dec 4th 2008, 07:16 PM
magentarita
ok..........
Quote:

Originally Posted by nzmathman
$\displaystyle distance = velocity \times time$

Let t = time taken to travel downstream.
Let v = the velocity of the boat in still water.

using $\displaystyle d = vt$,

$\displaystyle 5 = (v - 4) \times (t + 20)$

$\displaystyle 5 = (v + 4) \times t$

We want 'v' on its own so we can solve, so divide the 'v parts' across to the other side.

$\displaystyle \frac{5}{v-4} = t + 20$

$\displaystyle \frac{5}{v+4} = t$

$\displaystyle \frac{5}{v - 4} = \frac{5}{v+4} + 20$

Multiply whole expression by $\displaystyle (v + 4)(v - 4)$

$\displaystyle 5(v + 4) = 5(v - 4) + 20(v + 4)(v - 4)$

Now simplify and solve the quadratic formed.

Don't hesitate to ask for further help...

Thank you. I can take it from here.