A cross country team race is about to begin and team Delta 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 2.8v. In addition, while Ann is riding the horse, she moves at a rate of 10.4v and Bob (since he is slightly overweight) rides at a rate of 7.1v. (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.
Too long.
onobnik
To solve this problem, we can start by setting up the equations for the time it takes for Ann and Bob to complete the race, given their respective speeds while walking and riding.
Let's assume the total race distance is D.
When Ann is walking, her speed is v, so the time it takes for her to cover the distance D is given by:
t_ann_walk = D / v
When Ann is riding the horse, her speed is 10.4v, so the time it takes for her to cover the distance D is given by:
t_ann_ride = D / (10.4v)
Similarly, when Bob is walking, his speed is 2.8v, so the time it takes for him to cover the distance D is given by:
t_bob_walk = D / (2.8v)
When Bob is riding the horse, his speed is 7.1v, so the time it takes for him to cover the distance D is given by:
t_bob_ride = D / (7.1v)
Now, let's consider how Bob and Ann alternate riding and walking.
If Bob rides for a fraction f of the race distance, then Ann will ride for (1-f) of the race distance. The total time it takes for the team to complete the race is the sum of the individual times for Ann and Bob.
Let's calculate the total time, T, as a function of f:
T = f * t_bob_ride + (1-f) * t_ann_ride + f * t_bob_walk + (1 - f) * t_ann_walk
= f * (D / (7.1v)) + (1-f) * (D / (10.4v)) + f * (D / (2.8v)) + (1 - f) * (D / v)
Simplifying this expression, we get:
T = D * [f/(7.1v) + (1-f)/(10.4v) + f/(2.8v) + (1-f)/v]
To minimize the total race time, we want to find the value of f that minimizes T.
Let's take the derivative of T with respect to f and set it equal to zero to find the critical points:
dT/df = D * [1/(7.1v) - 1/(10.4v) + 1/(2.8v) - 1/v] = 0
Simplifying this equation, we get:
1/7.1 - 1/10.4 + 1/2.8 - 1 = 0
Now, we can solve for f to find the fraction of the total race distance that Bob should ride to minimize the total race time.