# Chases and Escapes: The Mathematics of Pursuit and Evasion

By Paul J. Nahin | Go to book overview

Appendix A
Solution to the Challenge Problems of Section 1.1

Our first question is easy to answer. To calculate the distance sailed by the pirate ship until it captures the merchant vessel (n < 1), simply recall from (1.1.12) that capture occurs at (xo, n/(1 − n2)xo), i.e., the merchant has traveled a distance of n/(1 − n2)xo. Since the pirate ship travels 1/n times faster than does the merchant, the pirate travels 1/n times as far, that is, the pirate ship travels a total distance of 1/(1 –n2)xo.

To answer the second question, i.e., to determine the distance the pirate ship lags behind the merchant vessel after a long time has passed (for n = 1), refer to figure A. There we see the pirate ship at point (x,y), while the merchant vessel is at (xo, ym). Note, carefully, that this is for any arbitrary time t. The distance separating the pirate ship and the merchant vessel is D, where

Now, here’s the crucial observation: the line joining (x,y) to (xo, ym) is the tangent to the pirate’s pursuit curve, precisely because the chase

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