# Math Help - Speed of Boat in Still Water

1. ## 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?

2. $distance = velocity \times time$

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

using $d = vt$,

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

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

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

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

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

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

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

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

Now simplify and solve the quadratic formed.

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

3. ## ok..........

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

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

using $d = vt$,

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

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

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

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

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

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

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

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

Now simplify and solve the quadratic formed.

Don't hesitate to ask for further help...
Thank you. I can take it from here.