# Velocity of boat in river with varying current

1. Oct 13, 2011

### s_j_sawyer

1. The problem statement, all variables and given/known data

See attached.

2. Relevant equations

3. The attempt at a solution

I originally began by assuming the rate was constant, obtaining the equations

$V_x^2 + V_y^2 = 1$

$\tan{\sigma} = \frac{V_y}{V_x}$

with the assumptions

$V_x(\sigma) = f(\sigma)$
$V_y(\sigma) = r + g(\sigma)$

and used the above equations to solve for f and g, and ended up getting the correct result for f, i.e.

$f(\sigma) = \cos{\sigma}$

but did not get the right result for $g(\sigma).$

Then I realize that the path was CURVED, hence there had to be acceleration involved, and upon reading other parts of the question figured out that we were not supposed to make the assumption that r was constant. i.e. it turns out

$r = r(x)$

and I am unsure about how to deal with this to get the correct result

$V_x(\sigma) = \cos{\sigma}$
$V_y(\sigma) = r + \sin{\sigma}$

Any help would be greatly appreciated.

#### Attached Files:

• ###### boat question.png
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2. Oct 13, 2011

### verty

Sigma is not the steering angle, right? Because if it was just the angle, you would not know enough to determine the velocity of the boat. So I take it sigma is the velocity of the boat relative to the water.

And if sigma is the velocity, and if I understand correctly, there is nothing to calculate. For surely, if it was not necessary for the question to say that sigma is a velocity, it was expected for you to have some physics intuition. So you can use that same intuition to supply the formulas without proof. So I would move on to part B.

I think I don't understand what the intent of the question was.

Last edited: Oct 13, 2011
3. Oct 13, 2011

### s_j_sawyer

Well $\sigma$ is the steering angle... my professor told us just to assume the boat had a constant speed, which is why I introduced the formula

$V_x^2 + V_y^2 = 1^2$

and in thus doing so assumed the speed to simply be 1.

Can someone read the question and see if I'm interpreting something wrong here?

4. Oct 13, 2011

### s_j_sawyer

Nevermind I got it. I was overthinking it apparently...