I am trying to understand fully the following question:

A boat is traveling west at 100m/s and is partially helped by a strong wind blowing at 20m/s 30 degrees north of west. What speed could the boat travel at in still water.

As I understand it I must resolve the wind into horizontal and vertical components subtract the horizontal component from 100 to get my answer.

So what I should do is: $$100-[(cos(30))(20)]=82.67949\space m/s$$

Is my understanding of the question correct or am I missing something?

This question is NOT from a homework assignment, rather a question I found elsewhere that I am attempting.

  • $\begingroup$ Where did you find this ridiculous question? Please provide a reference so we can all have a good laugh. $\endgroup$ – Olin Lathrop Nov 24 '13 at 22:28
  • $\begingroup$ The effective wind direction, moves forward with speed. Freeboard has a large effect. West=270°, plus 30° North of West would make it 300°. Wind direction is usually referred to as "out of" (where it is blowing from). Not enough information to form an answer. $\endgroup$ – Optionparty Nov 24 '13 at 23:03

This question can't be answered because important information is not given. We have no idea how much influence, if any, the wind has on the boat's velocity. In the limiting case, if the boat's wind resistance is 0, then the wind has no effect.

The question also asks what the boat would do in still water, but we were never told how still or not the water is in the original case.

To summarize, in the test case we know what the wind is doing but not the water, but are asked to find the speed knowing only what the water is doing and not the wind. In addition, we have no idea of the wind's effect on the speed. All around, this is a badly designed question that can't be answered.

  • $\begingroup$ Not to mention that the boat is doing $360 \frac{\text{km}}{\text{hr}}$! (wind aided) $\endgroup$ – DJohnM Nov 24 '13 at 15:52
  • $\begingroup$ Good point, I hadn't noticed that. That comes out to 224 miles/hour, which is faster than many small airplanes. Hmm, this whole question would make a lot more sense if it was about a airplane instead of a boat. $\endgroup$ – Olin Lathrop Nov 24 '13 at 22:27

Changing the "wind" to a "current" in an attempt to save the question, the OP's solution is still not quite correct.

By subtracting the current's west component from the boat's overall, west velocity over the ground, the OP has correctly found the west component of the boat's velocity through the water.

However, he has forgotten that the boat must cancel the current's north component of $20\cdot \sin(30^\circ)$, by having a matching south component to the boat's velocity through the water. The boat's overall velocity through the water is the resultant of these west and south components.

  • $\begingroup$ So the north component has an effect on the westward speed? $\endgroup$ – Hasan Nov 24 '13 at 16:35
  • $\begingroup$ In the sense that the boat must use some of its overall velocity to counteract the current's north velocity, and thus not use it to go west, yes... $\endgroup$ – DJohnM Nov 24 '13 at 17:01

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