Consider two athletes running at variable speeds  𝑣1(𝑡)  and  𝑣2(𝑡).  The runners start and finish a race at exactly the same time. Explain why the two runners must be going the same speed at some point.

This scenario is a classic application of the **Mean Value Theorem (MVT)** from calculus. The Mean Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \in (a, b) \) where:

\[
f'(c) = \frac{f(b) - f(a)}{b - a}.
\]

### Applying the Mean Value Theorem to this Problem
1. **Set the stage**: Let \( s_1(t) \) and \( s_2(t) \) represent the positions of the two athletes as functions of time \( t \). These functions describe how far each athlete has traveled at time \( t \). Their speeds are the derivatives of these functions: \( v_1(t) = s_1'(t) \) and \( v_2(t) = s_2'(t) \).

2. **Key observations**:
   - The runners start and finish at the same time. This means:
     \[
     s_1(a) = s_2(a) \quad \text{and} \quad s_1(b) = s_2(b),
     \]
     where \( a \) is the start time and \( b \) is the finish time.
   - Because \( s_1(t) \) and \( s_2(t) \) are both continuous (athletes can't teleport) and differentiable (their motion is smooth), the conditions of the Mean Value Theorem are satisfied for both position functions.

3. **Difference in positions**:
   Define \( \Delta s(t) = s_1(t) - s_2(t) \). The function \( \Delta s(t) \) is also continuous and differentiable because it is the difference of two continuous and differentiable functions.

   - At the start and finish, \( \Delta s(a) = 0 \) and \( \Delta s(b) = 0 \).

4. **Apply the MVT to \( \Delta s(t) \)**:
   By the Mean Value Theorem, there exists some \( c \in (a, b) \) such that:
   \[
   \Delta s'(c) = \frac{\Delta s(b) - \Delta s(a)}{b - a}.
   \]

   - Since \( \Delta s(b) = \Delta s(a) = 0 \), we have:
     \[
     \Delta s'(c) = 0.
     \]

5. **Interpret \( \Delta s'(c) = 0 \)**:
   The derivative of \( \Delta s(t) \) is:
   \[
   \Delta s'(t) = s_1'(t) - s_2'(t) = v_1(t) - v_2(t).
   \]

   Therefore, \( \Delta s'(c) = 0 \) implies:
   \[
   v_1(c) = v_2(c).
   \]

### Conclusion
At some point \( c \) during the race, the two runners must be running at exactly the same speed.

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Would you like a deeper explanation, an example, or clarification of any step? 

Here are five related questions to explore further:
1. How does the Mean Value Theorem apply to real-world scenarios beyond this race problem?
2. Can you explain why differentiability is required for the Mean Value Theorem to hold?
3. How would the conclusion change if the runners finished at different times?
4. What is the significance of continuity in ensuring the theorem applies?
5. Can the Mean Value Theorem be extended to functions of multiple variables?

**Tip**: The Mean Value Theorem is a fundamental result in calculus that often helps demonstrate equality or intermediate behavior over an interval—keep it in mind when analyzing motion or rates of change!

Consider two athletes running at variable speeds v1(t) and v2(t). The runners start and finish a race at exactly the same time. Explain why the two runners must be going the same speed at some point.

Understand why two runners starting and finishing a race at the same time must run at the same speed at some point. This explanation uses the Mean Value Theorem, exploring continuity, differentiability, and key calculus concepts. Suitable for advanced high school or undergraduate students.

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