Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed. [Hint: Consider (??)=??(??)-h(??), ?where ?? ?and h ?are the position functions of the two runners.]
The Correct Answer and Explanation is:
We are given that two runners start a race at the same time and finish in a tie, and we are asked to prove that at some point during the race, they must have the same speed. The hint suggests using the function: f(t)=g(t)−h(t)f(t) = g(t) – h(t)
where:
- g(t)g(t) = position function of the first runner
- h(t)h(t) = position function of the second runner
- f(t)f(t) = difference in position between the two runners at time tt
Step-by-step Proof (Using the Mean Value Theorem):
Since both runners start at the same point and finish at the same point, we have: g(0)=h(0)andg(T)=h(T)g(0) = h(0) \quad \text{and} \quad g(T) = h(T)
This implies: f(0)=g(0)−h(0)=0andf(T)=g(T)−h(T)=0f(0) = g(0) – h(0) = 0 \quad \text{and} \quad f(T) = g(T) – h(T) = 0
So the function f(t)f(t) satisfies:
- ff is continuous on the closed interval [0,T][0, T]
- ff is differentiable on the open interval (0,T)(0, T)
- f(0)=f(T)=0f(0) = f(T) = 0
Now, apply Rolle’s Theorem:
If a function is continuous on [a,b][a, b], differentiable on (a,b)(a, b), and f(a)=f(b)f(a) = f(b), then there is at least one point c∈(a,b)c \in (a, b) such that f′(c)=0f'(c) = 0.
Since f(t)f(t) meets all the conditions, there exists a time c∈(0,T)c \in (0, T) such that: f′(c)=0f'(c) = 0
But recall: f′(t)=g′(t)−h′(t)f'(t) = g'(t) – h'(t)
So at time cc: g′(c)=h′(c)g'(c) = h'(c)
That means the instantaneous speed (velocity) of both runners is equal at some moment during the race.
Conclusion:
At some point during the race, both runners must have had the same speed. This result is guaranteed by Rolle’s Theorem applied to the difference of their position functions. It doesn’t matter how they individually sped up or slowed down—as long as they started and ended together, they had to be going the same speed at least once.