The major thing that we ignoring is the drag that the water places upon the
ship. Much of the power output of the destroyer's engines must be used to
overcome drag.
The force due to drag is proportional to the velocity, and works to oppose the
ship's motion. The constant of proportionality is called the drag
coefficient, and its exact value depends upon the density of the water and the
construction of the ship. For the purposes of this problem, we will assume a
drag coefficient of 3300 kg/sec.
The force that drag exerts on the destroyer is:
| drag := t -> 3.3e3 * velocity(t); |
Our improved power formula must include both the power required
to produce the desired acceleration (in the absence of drag) and
the power required to overcome the drag:
| power := t -> (force(t) + drag(t)) * velocity(t); |
Now what is the top speed to which the destroyer can accelerate? Here is a
function that will convert from meters/second to miles/hour:
| mps2mph := mps -> 2.237*mps; |
Click here for the answer
This little bit of assessment tells us that our model approximates reality.
Now we can consider generalizing it.
Joseph L. Zachary
Hamlet Project
Department of Computer Science
University of Utah