Let's define the variables:

• $p$ = airspeed of the plane (speed in still air) in miles per hour.
• $w$ = wind speed in miles per hour.

When flying with a tailwind, the plane’s effective speed is $p + w$ (airspeed plus wind speed). On the return flight against the wind, the plane’s effective speed is $p - w$ (airspeed minus wind speed).

Step 1: Convert time to hours

The plane flew 700 miles in 1 hour and 20 minutes with the tailwind. Converting 1 hour and 20 minutes to hours: $1 \text{ hour} + \frac{20}{60} \text{ hours} = \frac{80}{60} = \frac{4}{3} \text{ hours}.$ So, the time with the tailwind is $\frac{4}{3}$ hours.

Step 2: Write equations based on distance = speed × time

For the flight with the wind: $700 = (p + w) \times \frac{4}{3}$

For the return flight against the wind, where the plane took 2 hours: $700 = (p - w) \times 2$

Step 3: Solve the system of equations

Equation 1:

$700 = (p + w) \times \frac{4}{3}$ Multiplying both sides by 3: $2100 = 4(p + w)$ Dividing by 4: $p + w = 525 \quad \text{(Equation 1)}$

Equation 2:

$700 = (p - w) \times 2$ Dividing by 2: $p - w = 350 \quad \text{(Equation 2)}$

Step 4: Solve the system

Add Equation 1 and Equation 2: $(p + w) + (p - w) = 525 + 350$ $2p = 875$ $p = 437.5 \quad \text{(airspeed of the plane in still air)}$

Substitute $p = 437.5$ into Equation 1: $437.5 + w = 525$ $w = 525 - 437.5 = 87.5 \quad \text{(wind speed)}$

Final Answer:

• The airspeed of the plane is 437.5 miles per hour.
• The wind speed is 87.5 miles per hour.

Would you like further details or clarification?

Related Questions:

1. How does wind speed affect flight time in general?
2. What would happen if there was no wind during both flights?
3. How can you model this problem with other distances?
4. What other real-world factors could affect the plane's airspeed?
5. How can you use this model to predict fuel consumption?

Tip:

When solving problems involving wind or current, always break the situation into components with and against the force (wind or current).