Problem 127 - Elementary and Intermediate Algebra

Question

Wind Speeds. A plane that can fly 200 mph in still air makes a 330 -mile flight with a tail wind and returns, flying into the same wind. Find the speed of the wind if the total flying time is \(3 \frac{1}{3}\) hours.

Step-by-Step Solution

Answer

The wind speed is approximately 142 mph.

Step 1: Define Variables

Let's denote the speed of the wind as \(w\) in mph. Therefore, the effective speed of the plane flying with the tailwind is \(200 + w\), and against the wind, it is \(200 - w\).

Step 2: Setup Equations for Time

To find the total flying time, we need the times for both flights. The time for the flight with the wind is \( \frac{330}{200+w} \), and the time for the flight against the wind is \( \frac{330}{200-w} \). The total flying time is given as \(3 \frac{1}{3}\) hours or \(\frac{10}{3}\) hours.

Step 3: Formulate the Total Time Equation

The equation for the total flight time is: \[ \frac{330}{200+w} + \frac{330}{200-w} = \frac{10}{3} \]

Step 4: Simplify and Solve for w

Multiply every term by \( (200+w)(200-w) \) to eliminate the fractions: \[ 330(200-w) + 330(200+w) = \frac{10}{3}(200+w)(200-w) \]This simplifies to: \[ 66000 = \frac{10}{3}(40000 - w^2) \]Now multiply through by 3: \[ 198000 = 10(40000 - w^2) \]Divide by 10: \[ 19800 = 40000 - w^2 \]Rearrange to find \(w^2\): \[ w^2 = 40000 - 19800 \]\[ w^2 = 20200 \]Take the square root: \[ w = \sqrt{20200} \approx 142.14 \] mph.

Step 5: Validate the Solution

Now check the solution to ensure it satisfies the problem's condition. Calculate the flight times: - With the wind: \( \frac{330}{200+142.14} \approx 0.99 \) hours.- Against the wind: \( \frac{330}{200-142.14} \approx 2.34 \) hours.The total is approximately \(3.33\) hours, which is consistent with \(3 \frac{1}{3}\) hours.

Key Concepts

Understanding Wind Speed and Its Impact on Flight

Wind plays a significant role in aviation, affecting both flight time and fuel consumption. When a plane flies with a tailwind, the wind adds to its airspeed, which can reduce the travel time. Conversely, a headwind diminishes the airspeed, increasing the time it takes to reach the destination.
In this exercise, we define the wind speed as a variable, typically denoted by \( w \). When the plane flies with a tailwind, the effective speed becomes the sum of the plane's speed in still air and the wind speed, \( 200 + w \) mph. However, against the wind, the effective speed is reduced to \( 200 - w \) mph.

Key implications include:

• The wind speed can dramatically change the performance and efficiency of a flight.
• Understanding and computing the effective speeds helps in scheduling and fuel management.

By setting up equations based on these principles, we can solve for wind speed and better understand its effect on flight dynamics.

Quadratic Equations in Algebra Word Problems

Quadratic equations often arise in word problems related to speed and distance, such as this one. These equations are defined by the general form \( ax^2 + bx + c = 0 \). In this exercise, we simplify the time equation and arrive at a quadratic form, which allows us to solve for the unknown variable, the wind speed.
The equation development involves:

• Combining the time equations for with and against the wind.
• Eliminating the fractions through multiplication by the common denominator.
• Simplifying the resulting equation to identify the coefficient values \( a \), \( b \), and \( c \).

In solving our quadratic equation, we transition from linear to quadratic calculus, illustrating how quadratic equations facilitate solutions where two factors (like distance and speed) are in play. By solving \( w^2 = 20200 \), we apply the square root to find \( w \).

Flight Time Calculation with Wind Influences

Calculating flight time is crucial for both airlines and passengers. In the context of this word problem, understanding how to compute flight time when wind affects the speed is essential. Flight time is the distance traveled divided by the effective speed. Here, the plane travels 330 miles with and without the wind.
To determine flight time:

• Calculate the time with the wind using: \( \frac{330}{200+w} \) hours.
• Calculate the time against the wind using: \( \frac{330}{200-w} \) hours.
• Add both times to ensure they equate to the total time given: \( 3 \frac{1}{3} \) hours.

This problem exemplifies how effective speed can increase or decrease travel time, and reaffirms the importance of accurate calculations in ensuring reliable and efficient flight scheduling.