Fraction of a distance (1D Kinematics)

• chatnay

Homework Statement

A cross country team race is about to begin and team Beta decides to use their one-dimensional kinematics knowledge that they learned from their first year physics course. 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 3.4v. In addition, while Ann is riding the horse, she moves at a rate of 11.6v and Bob (since he is slightly overweight) rides at a rate of 8.1v. (Note also that Bob moves faster on the horse than on foot). Find the fraction of the total race distance that Ann 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.

s = u +at

The Attempt at a Solution

I can't resolve this exercise because it seems that my answer is wrong (0.464).

I've set up the following the equations in order to equal them and find the fraction of the distance that Anna rides:

3.4vx + (1-x)8.1v = (1-x)v + 11.6vx

Then I solved for x and i got as a result x = 71/153, which is ≈ 0.464. I've plugged in the value to the equation in order to see if the distance was the same, and it was. Possibly I have committed a mistake in setting up the equation. Is the procedure that I've done appropriate?

Thank you very much PF.