# Homework Help: Swimming vector problem

1. Aug 21, 2010

### kudoushinichi88

1. The problem statement, all variables and given/known data
The water in a river flows uniformly at a constant speed of 2.50m/s between two parallel banks 80.0m apart. You are to deliver a package directly across the river, but you can only swim at 1.5m/s.

If you choose to minimize the distance downstream that the water carries you, in what direction should you head?

2. Relevant equations
x=x_o+v_xt

3. The attempt at a solution
Let's say the river is flowing towards the east and taking that as the x-axis, you must swim at some angle, x from the x-axis in the opposite direction of the flow of the river.
To have a minimum distance downstream, I figure that we must have zero displacement on the x-axis and a displacement of 80m on the y-axis.

So I came up with this right angled vector triangle with hypotenus 1.5(cos x)t and sides 80 and 2.5t. Using the Pythagoras theorem, I came up with

t = sqrt(6400/(2.25Cos^2 (x)-6.25))

but now i'm stuck! cos I think i'm conceptually flawed right from the start.

2. Aug 21, 2010

### kuruman

Your reasoning is flawed. Think of it this way. Even if you were to angle yourself so that you swim directly upstream, you would still be carried downstream by the current at the speed of 1.00 m/s. To move directly across, you need to be able to swim faster than 2.50 m/s relative to the water and adjust your direction so that your upstream component is 2.50 m/s to cancel the effect of current. There is no way to make it straight across from where you are (zero displacement on the x-axis).

However, the problem is asking you to minimize the distance downstream where you land, not how to make it straight across. What is your thinking on that?

3. Aug 21, 2010

### Terocamo

-------------------------

--->2.5m/s
-------------------------
Try imagine thats a river...
Let v be the overall velocity of the man
vx=2.5-1.5sinA
vy=1.5cosA

Let time required for him to cross the river be t
t=80/vy=80/1.5cosA

Let S be the distance cross downstream
S=(2.5-1.5sinA)t

solving for S

S=(133-80sinA)/cosA

The problem is... I dont know how to find the minimum S for the equation, but i try to plot a graph and the min S is when A about 37.2 degree

4. Aug 21, 2010

### kuruman

Have you had calculus? Do you know how to minimize S with respect to A?

5. Aug 21, 2010

### Terocamo

Is it nessesary to do any differenciation work?

6. Aug 21, 2010

### kuruman

It is.

7. Aug 21, 2010

### kudoushinichi88

Ahhh!!! Two years away from physics has caused me to forget that in the vector diagram, the resultant vector and its components are actually independent of each other...

Taking the river flow going towards the positive direction of the x-axis,

$v_y=1.5\sin\theta$
$v_x=2.5-1.5\cos\theta$

$y=y_0+v_yt$
$80=0+(1.5\sin\theta)t$
$t= \frac{160}{3\sin\theta}$

$x=x_0+v_xt$
$x=(2.5-1.5\cos\theta)\left(\frac{160}{3\sin\theta}\right)$
$=\left(\frac{400}{3}\right)\csc\theta-80cot\theta$

$\frac{dx}{d\theta}=-\left(\frac{400}{3}\right)\cot\theta\csc\theta-80\csc^2\theta$
$0=-80\csc\theta(5/3\cot\theta+8\csc\theta)$

One of the solutions for \theta;

$\frac{5\cos\theta}{3sin\theta}=-\frac{1}{\sin\theta}$
$\cos\theta=-3/5$
$\theta=126.9^o$

hence the direction is 53.1 degrees from the x-axis! XDD Thanks!

8. Aug 21, 2010

### aq1q

lol who are you thanking? You really shouldn't just do the problem out like that. Nonetheless, Terocame you are very close to solving the problem (if you havent done it already). You need to find ds/d(theta) and minimize S with respect to theta.

P.S. The theta you found is correct. However, be careful what this theta actually represents

Last edited: Aug 21, 2010