Problem 70 - Algebra A Combined Function

Question

A pilot can fly a DC-10 1365 miles against the wind in the same time he can fly 1575 miles with the wind. If the speed of the plane in still air is 490 miles per hour, find the speed of the wind. (Source: Air Transport Association of America)

Step-by-Step Solution

Answer

The speed of the wind is 35 mph.

Step 1: Define Variables

Let the speed of the wind be \( w \) mph. The effective speed of the plane against the wind is \( 490 - w \) mph, and with the wind is \( 490 + w \) mph.

Step 2: Set Up the Time Equation

Time is given as \( \text{time} = \frac{\text{distance}}{\text{speed}} \). For the journey against the wind, the time is \( \frac{1365}{490 - w} \). For the journey with the wind, the time is \( \frac{1575}{490 + w} \). Since both times are equal, we set up the equation: \[ \frac{1365}{490 - w} = \frac{1575}{490 + w} \]

Step 3: Cross-Multiply

Cross-multiply to eliminate the fractions: \[ 1365(490 + w) = 1575(490 - w) \]

Step 4: Distribute and Simplify

Expand both sides to get: \[ 1365 \times 490 + 1365w = 1575 \times 490 - 1575w \]Simplifying, it becomes:\[ 668850 + 1365w = 771750 - 1575w \]

Step 5: Solve for \( w \)

Add \( 1575w \) to both sides and subtract \( 668850 \) from both sides: \[ 1365w + 1575w = 771750 - 668850 \]Combine like terms: \[ 2940w = 102900 \]Divide by 2940: \[ w = \frac{102900}{2940} \]Solving this gives \( w = 35 \).

Step 6: Confirm the Solution

Verify by checking if the times match for wind speed 35 mph:- Against the wind: \( \frac{1365}{455} = 3 \) hours- With the wind: \( \frac{1575}{525} = 3 \) hoursBoth times are equal, confirming the solution is correct.

Key Concepts

Distance, Speed, and Time

Distance, speed, and time form a fundamental trio in algebra word problems. These parameters are interrelated through the formula: \[ \text{time} = \frac{\text{distance}}{\text{speed}} \]This means that time is equal to distance divided by speed.
When dealing with motion problems like flying, driving, or running, it's essential to understand how these variables interact. The speed of any moving object, such as a plane or car, changes depending on external factors, like wind or road conditions.

• Distance: This is the length of the path traveled by an object. In the exercise, distances of 1365 miles and 1575 miles are given for flight paths.
• Speed: This refers to how quickly an object moves. The airplane's speed in still air is 490 miles per hour, but the presence of wind alters this speed.
• Time: Time is how long the journey takes, calculated using distance and speed.

In problems similar to our example, recognizing how the movement against or with wind affects speed is crucial. The effective speed against the wind is slower, while speed with the wind is faster. This difference in speed leads to the fact that time remains the same for both scenarios.

Variables and Equations

Variables and equations are key to solving algebra word problems. In our context, a variable is an unknown quantity we need to find, often represented by letters like \( x \), \( y \), or \( w \). In the given exercise, we define the speed of the wind as \( w \).
By establishing variables, we can translate word problems into mathematical equations, which we can then solve step by step.

• Variables: These are placeholders for unknown values. Here, \( 490 - w \) represents the speed against the wind, and \( 490 + w \) represents the speed with the wind.
• Equations: Equations are mathematical statements that assert the equality of two expressions. The equation in this problem is set by equating the two journey times: \[ \frac{1365}{490 - w} = \frac{1575}{490 + w} \]

To solve algebraic equations, operations such as addition, subtraction, cross-multiplying, and simplification are applied. Understanding how to manipulate equations correctly allows us to isolate and solve for the unknown variable.

Problem Solving Steps

Successfully solving algebra problems requires a systematic approach. This involves breaking down the problem into manageable steps, applying logical reasoning, and validating the solution.

• Define Variables: Identify what you need to find and represent it with a variable. In our exercise, the speed of the wind is \( w \).
• Set Up Equations: Use the problem's context to formulate an equation that captures the described relations. The given times against and with the wind form our equation.
• Manipulate the Equation: Apply algebraic operations like cross-multiplying and simplifying to make the variable more accessible.
• Solve and Verify: Find the value of the variable and check your work. Consistency between both scenarios supports the correctness of the solution.

Following these methodical steps not only helps in solving the particular problem but also strengthens overall problem-solving skills. It's all about connecting words and numbers through logical operations to uncover the unknowns.