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Let's break down and solve this step-by-step.

### **Problem Setup:**
A plane travels on a 2780-mile flight and arrives at its destination in 5.7 hours. We are tasked with using the **Mean Value Theorem (MVT)** and **Intermediate Value Theorem (IVT)** to show that there are at least two times during the flight when the speed of the plane is **300 miles per hour**.

### **Step 1:**
We are given that the plane begins its takeoff at **2:00 p.m.** and the total flight distance is 2780 miles.

The position function \( S(t) \) represents the distance traveled by the plane at time \( t \). The time \( t = 0 \) corresponds to **2:00 p.m.**, and the total flight takes **5.7 hours**.

We fill in the following information:

- \( S(0) = 0 \): At time \( t = 0 \), the plane hasn't traveled any distance.
- \( S(5.7) = 2780 \): After 5.7 hours, the plane reaches its destination, 2780 miles away.

Thus:
\[
S(0) = 0 \quad \text{and} \quad S(5.7) = 2780
\]

### **Step 2:**
The **Mean Value Theorem (MVT)** says that there exists a time \( t_0 \), where \( 0 < t_0 < 5.7 \), such that the instantaneous velocity at that time equals the average velocity over the entire flight. Mathematically, this is expressed as:

\[
S'(t_0) = \frac{S(5.7) - S(0)}{5.7 - 0} = \frac{2780 - 0}{5.7} = 487.72 \text{ miles per hour}
\]

Thus, the average speed over the entire flight is **487.72 miles per hour**, and there exists at least one time \( t_0 \) during the flight when the plane's speed is exactly 487.72 miles per hour.

### **Step 3:**
We now analyze the speed using the **Intermediate Value Theorem (IVT)**.

We know:
- \( v(0) = 0 \): The plane's speed at takeoff is 0 miles per hour.
- \( v(5.7) = 0 \): The plane's speed at landing is also 0 miles per hour.
- \( v(t_0) = 487.72 \): The plane's speed reaches 487.72 miles per hour at some time \( t_0 \).

Using the **Intermediate Value Theorem**, since the speed goes from 0 at takeoff to 487.72 at some point and then back to 0 at landing, the speed must cross **300 miles per hour** at least **twice** — once while accelerating and once while decelerating.

Therefore, by the IVT, there are at least two times during the flight when the speed of the plane was exactly **300 miles per hour**.

---

### Final Answers:
- Step 1: \( S(0) = 0 \) and \( S(5.7) = 2780 \)
- Step 2: Average velocity = 487.72 miles per hour
- Step 3: By the Intermediate Value Theorem, the speed must be 300 miles per hour at least twice during the flight.

---

Would you like to explore more details on how these theorems are applied?

### **Related Questions:**
1. What is the formal statement of the Mean Value Theorem and how is it applied to velocity?
2. Can the Mean Value Theorem guarantee more than two instances where the speed is 300 miles per hour?
3. How does the Intermediate Value Theorem apply in contexts other than velocity?
4. If the plane’s speed varies non-linearly, how would the Mean Value Theorem's application change?
5. What is the significance of knowing the average speed in real-world flight scenarios?

#### **Tip:** The Mean Value Theorem is widely used to prove that a function has a certain value at some point, making it a valuable tool in understanding real-world rates of change.

A plane begins its takeoff at 2:00 p.m. on a 2780-mile flight. After 5.7 hours, the plane arrives at its destination. Explain why there are at least two times during the flight when the speed of the plane is 300 miles per hour.

Learn how to apply the Mean Value Theorem and Intermediate Value Theorem to explain why a plane's speed must be 300 miles per hour at least twice during a 2780-mile flight completed in 5.7 hours. This problem covers calculus concepts such as velocity and the application of key theorems.

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