Find the distance with the relativity problem

[Please Wait]

Homework Statement

Popeye was sailing his boat in a straight channel one half km wide. The steady current flows south at four km/h and a wind blows from the north, parallel to the current. Popeye starts from the west shore. He sets his sail and tiller so that he moves diagonally across and up the channel. The velocity of the boat with respect to the water is twelve km/h in a direction 45° East of North. When he gets very close to the east shore, he stops readjusts the sil and tiller, and establishes an identical upstream tack back toward the west bank (his direction through the water is now 45° West of north). Due to lack of spinach, he was effectively stopped in the water for one whole minute at the turn around. Calculate the distance measured north along the west bank between his start and finish points.

I have created a little description image by myself:
http://s1323.beta.photobucket.com/user/mommadaddy2/media/hhh_zpsa050c917.jpg.html
This image has a error the current is at the same side as wind

The Attempt at a Solution

okay this is how i decided to start this off.
V of current to water = Vcurrent to boat + V boat to water
V = 4 km/h [ S ] + 12 km/h [E 45 N]
V = 4 km/h + 8.485 km/h [ E] + 8.485 km/h [N]
V = 4.485 km/h [ N] + 8.485 km/h [ E]
i do not know what to do next

 

[TSny]

Can you use your eastward component of velocity to find the time to cross the channel from the west shore to the east shore?

 

[Please Wait]

Can you use your eastward component of velocity to find the time to cross the channel from the west shore to the east shore?

alright so the velocity is 12 km/h and the distance is 1.5 km
1.5 km/ 12 km/h = 0.125 h

 

[TSny]

alright so the velocity is 12 km/h and the distance is 1.5 km
1.5 km/ 12 km/h = 0.125 h

What distance is 1.5 km? I don't believe this is correct.
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Instead, try to answer the following two questions:

(a) How far east does the boat need to travel to get across the channel? The boat will also travel some in the north direction, but just think about how far the boat needs to move eastward to get across.

(b) What is the rate at which the boat moves eastward? (Hint: this is just another way of asking for the eastward component of velocity.) You've already calculated it in your first post.

Use the answers to (a) and (b) to find the time to cross the channel from the west side to the east side.

 

[Nickie]

.

I would approach this problem by first breaking down the given information into mathematical equations and variables. Based on the given information, we can define the following variables:

- d: distance measured north along the west bank between start and finish points
- t: time spent on the first leg of the journey (before Popeye stops to readjust)
- t2: time spent on the second leg of the journey (after Popeye stops to readjust)
- Vc: velocity of the current
- Vw: velocity of the wind
- θ: angle of the boat's direction relative to the current (45° in this case)
- Vb: velocity of the boat relative to the water

Using these variables, we can create the following equations:

On the first leg of the journey:
- d = Vc * t + Vw * t + Vb * t
- Vb = Vc cosθ + Vw sinθ

On the second leg of the journey:
- d = Vc * t2 + Vw * t2 + Vb * t2
- Vb = Vc cosθ - Vw sinθ

Since we know that Popeye was stopped in the water for one minute at the turn around point, we can set t2 = 1 minute.

We also know that the distance between the start and finish points is equal, so we can set the two d equations equal to each other and solve for d:

Vc * t + Vw * t + Vb * t = Vc * t2 + Vw * t2 + Vb * t2
Vc * t + Vw * t + Vb * t = Vc * 1 + Vw * 1 + Vb * 1
Vc * (t - 1) = Vw * (1 - t) + Vb * (1 - t)
Vc = Vw + Vb

Substituting this into the equation for Vb on the first leg of the journey, we get:

Vb = Vc cosθ + Vw sinθ
Vb = (Vw + Vb) cosθ + Vw sinθ
Vb = Vw cosθ + Vb cosθ + Vw sinθ
Vb - Vb cosθ = Vw cosθ + Vw sinθ
Vb (1 - cos