Problem 233 - Elementary Algebra

Question

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Step-by-Step Solution

Answer

The speed of the jet in still air is 415 mph, and the speed of the wind is 19 mph.

Step 1 - Define the Variables

Let the speed of the jet in still air be denoted as \(j\) (in miles per hour, mph) and the speed of the wind as \(w\) (also in mph).

Step 2 - Set Up Equations for Tailwind

With a tailwind, the effective speed of the jet is \(j + w\). The jet travels 868 miles in 2 hours, so we set up the equation: \[ (j + w) \times 2 = 868 \]Solving for \(j + w\): \[ j + w = 434 \]

Step 3 - Set Up Equations for Headwind

With a headwind, the effective speed of the jet is \(j - w\). The jet travels 792 miles in 2 hours, so we set up the equation: \[ (j - w) \times 2 = 792 \]Solving for \(j - w\): \[ j - w = 396 \]

Step 4 - Add the Equations

Add the equations from Steps 2 and 3 to eliminate \(w\): \[ (j + w) + (j - w) = 434 + 396 \]This simplifies to: \[ 2j = 830 \]Solving for \(j\): \[ j = 415 \]

Step 5 - Substitute \(j\) Back into One Equation

Use the value of \(j\) found in Step 4 and substitute it into the equation \(j + w = 434\): \[ 415 + w = 434 \]Solving for \(w\): \[ w = 19 \]

Key Concepts

speed problems

Speed problems often involve calculating distances traveled over time at certain speeds. These types of problems can seem tricky because they involve variables and algebraic manipulation. By breaking down the problem and addressing it step-by-step, you can make the calculations much more manageable. The key is to define your variables clearly and set up equations that accurately represent the problem. This approach helps in understanding the problem and finding the solution systematically.

algebraic equations

Algebraic equations are mathematical statements that establish the equality between two expressions. They involve variables, constants, and arithmetic operators. In speed problems, these equations are essential for solving unknowns. For instance, in the given exercise, we set up two equations based on the conditions with a tailwind and a headwind:

• With a tailwind: \( (j + w) \times 2 = 868 \)
• With a headwind: \( (j - w) \times 2 = 792 \)

These equations help us express the relationships mathematically, making it easier to find the speeds of the jet and wind.

tailwind and headwind effects

Tailwind and headwind significantly impact an aircraft's effective speed. A tailwind blows in the same direction as the aircraft, increasing its speed. For instance, if a jet's speed is \( j \) and the wind speed is \( w \), then with a tailwind: \( j + w \). Conversely, a headwind blows against the direction of travel, reducing the aircraft's speed. In this case, the effective speed is: \( j - w \). Understanding these effects is crucial for setting up the correct algebraic equations when solving such problems.

distance-rate-time relationship

The distance-rate-time (DRT) relationship is fundamental in solving speed problems. It is mathematically expressed as: \( \text{Distance} = \text{Rate} \times \text{Time} \). In the given problem, we used this relationship to express the distances traveled by the jet in terms of its speed and the wind speed. For instance:

• With a tailwind, the jet's effective speed is \( j + w \), and it travels 868 miles in 2 hours, so \( (j + w) \times 2 = 868 \).
• With a headwind, the effective speed is \( j - w \), and it travels 792 miles in 2 hours, so \( (j - w) \times 2 = 792 \).

By using this core relationship, we can set up and solve systems of equations to find the unknowns.