海水,淡水,湖水 邊一個介質可以行得較遠?

要解決這個問題，我們主要是比較三隻艇所受的阻力的分別。

這個阻力（可以稱為 friction, resistance, drag force 等）是取決於船的速度，船浸在水中的形狀，水的黏度（viscosity)等。

船浸在水中的形狀：
海水密度比淡水高約 3% ，所以船在海水中會浮得高一些，當船在海面上前行，它浸在海中的橫切面面積會少一些，阻力（或叫摩擦等等）會少一些（若水的黏度一樣）。

水的黏度：
水的黏度絕對會影響阻力，海水的黏度比淡水高大約8-9%，所以單純考慮水的黏度的話，海水的阻力是最大，但是水的黏度最終對阻力有多大影響是一個十分複雜的題目：例如，船身的「線」水油漆可以降低黏度的影響。

再者 F = bv 只是在低速時大概正確，但是高速時黏度的重要性會降低，b 也不是一個常數，它也取決於船浸在水中的形狀。

由於有兩個互補的因素，我們很難絕對確定哪船行得較遠，船的構造會十分影響結果，但我推算在高速航行時，水的黏度變得比較不重要，反而船隻浸在水中深一些（即是食水深一些）會大大增加它的阻力，所以船在海上行得較遠。

（註：若這問題是會考程度，那麼水的黏度的影響是可以忽略。）

For the fluids, the is also viscosity (黏性). You can see the rate of flow in the different in water and oil in daily life. But you neglected it here.

Assume the force to drive the ship to move is constant, F0
The net force add on the ship F=F0-f, f is the fiction force. Since F=ma, if the ships are the same mass.

Let fsea, flake and fwater be the fiction due to the sea, lake and water repectivily.

fsea > flake > fwater

F0 is contant, then a=(F0-f)/m, asea < alake < awater,
since F0-fsea

2007-06-15 10:30:31 補充：
since F0-fsea < F0-flake < F0-fwater,and v=at, then vsea

2007-06-16 12:21:09 補充：
Yes, I meanthe ship's situation are the same in 3 cases.

2007-06-16 12:22:28 補充：
You just need to know the method and make your own assumptions. My assumptions may be wrong, but the method can be apply to any cases.

2007-06-18 10:07:07 補充：
As I neglect viscosity here, in the real case, your speed is fixed due to viscosity, as by follow:

2007-06-18 10:07:24 補充：
Force on your ship = F - viscosity = F - bv, where b is a constant, depend on your ship's sharp and the viscosity of the fluid (may or may not be water), where F is the force output by your ship's engine.

2007-06-18 10:07:47 補充：
If your speed is high, bv is also large, then you will attend the max. speed vmax when F - bv = 0, v=F/b.