extremly difficult problem

[isky van persie]

A cross country team race is about to begin and team Gamma decides to use their one-dimensional kinematics knowledge . Each team consists of two people and one horse. The clock is started when the team crosses the starting line with one of the members (let's call her "Ann") riding the horse and the other (we'll call him "Bob") on foot.

The rider will of course leave the walker behind. Ann may tether the horse and continue on foot. When Bob reaches the horse, he mounts it, rides past Ann and tethers the horse further on and continues walking. When Ann reaches the horse, she mounts it and rides past Bob. This alternation of riding/walking continues until the entire team (Ann, Bob and the horse) cross the finish line.

Suppose that Ann walks at a rate of v and Bob (owing to the fact that he has longer legs) walks at a rate of 4.9v. In addition, while Ann is riding the horse, she moves at a rate of 12.8v and Bob (since he is slightly overweight) rides at a rate of 7.2v. (Note also that Bob moves faster on the horse than on foot). Find the fraction of the total race distance that Bob should ride in order to minimize the total race time. Assume that the race is sufficiently long that Ann and Bob can alternate on the horse many times.

 

[galactus]

I have seen this problem before. It is in Physics for Scientists and Engineers, Vol. 2 by Lawrence Lerner.

This book is full of wacky, off-the-wall problems.

The solution they have (which they just present, but do not derive) is rather sparse, and is as follows:

\(\displaystyle D_{B}=\frac{\frac{1}{b}-\frac{1}{c}}{1+\frac{1}{b}-\frac{1}{c}-\frac{1}{e}}\)

\(\displaystyle e=\text{ratio of Bob's riding speed to Ann's riding speed}, \;\ D_{B}=\text{fraction of distance traveled by Bob on horse}\)

They do not define b and c for some reason. I believe b and c may represent Ann and Bob's walking speeds.

The ratio of their riding speeds would be \(\displaystyle e=\frac{\text{Bob}}{\text{Ann}}:\frac{\frac{36}{5}v}{\frac{64}{5}v}=\frac{9}{16}\)