Thread: 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?

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

using ,





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







Multiply whole expression by



Now simplify and solve the quadratic formed.

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

Originally Posted by nzmathman

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

using ,





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







Multiply whole expression by



Now simplify and solve the quadratic formed.

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

Thank you. I can take it from here.