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When an airplane flies with the wind, it travels 800 miles in 4 hours. Against the wind, it takes 5 hours to cover the same distance. Find the plane’s rate in still air and the rate of the wind.

Short Answer

Expert verified
The plane's speed in still air is 180 mph, and the speed of the wind is 20 mph.

Step by step solution

01

Setting Up the Equations

Let \(a\) be the speed of the airplane in still air (in miles per hour), and \(w\) be the speed of the wind. When the plane flies with the wind, the speeds add up, so the equation becomes \(4(a+w) = 800\). When the plane flies against the wind, the wind speed subtracts from the airplane's, leading to \(5(a-w) = 800\).
02

Solving the First Equation

The first equation simplifies to \(a + w = 200\). This equation gives the combined speed of the airplane and the wind.
03

Solving the Second Equation

The second equation simplifies to \(a - w = 160\). This gives the speed of the airplane when there's resistance from the wind.
04

Solving the System of Equations

Add the two simplified equations together: \((a+w) + (a-w) = 200 + 160\), which simplifies to \(2a = 360\). Solving this for \(a\), we get \(a = 180\) mph.
05

Finding the Wind Speed

Substitute \(a = 180\) in either of the simplified equations. Putting it into \(a+w = 200\), we get \(180 + w = 200\). Solving for \(w\), we get \(w = 20\) mph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Equations
Algebraic equations are the foundation of algebra and represent a statement of equality between two expressions. These equations often include variables, which are symbols that stand in for unknown values. In problem-solving, algebraic equations are skillfully crafted to model real-world scenarios.

For example, if we let 'a' denote the speed of an airplane, and 'w' the wind speed, we can use algebraic equations to express the scenario provided in our exercise. The speed of the plane with the wind is the sum of its speed in still air and the wind's speed, yielding the equation, \[ a + w \], while against the wind it's the speed of the plane minus the wind's speed, expressed as \[ a - w \]. These simple algebraic expressions play a crucial part in solving the problem.
System of Equations
A system of equations consists of two or more equations with a common set of variables. The goal is to find values for the variables that will satisfy all equations in the system simultaneously.

In our airplane and wind speed problem, we deal with a system made up of two equations: \[ 4(a+w) = 800 \] and \[ 5(a-w) = 800 \]. These equations are interdependent, and solving the system requires methods like substitution, elimination, or matrix operations. Here, we apply elimination by adding and subtracting the equations to find the value of the variables 'a' and 'w'.

Understanding how to manipulate and solve such a system is a critical skill in algebra that extends to various disciplines including economics, engineering, and even computer science.
Problem-solving with Algebra
Problem-solving with algebra involves identifying unknowns, forming equations, and then systematically working towards a solution. This logical process can be segmented into identifiable steps, as seen in the given problem.

First, we define our variables and set up equations based on the problem scenario. Next, we manipulate these equations to isolate the variables, allowing us to find their values. This process may involve simplifying equations, as we did by diving both sides of the initial equations by the number of hours to get a direct relationship between speed and distance.

Finally, we interpret our results in the context of the original problem, ensuring that we understand not only how to find the solution but also what the solution means in real-world terms. This approach enhances critical thinking and application skills.
Speed, Distance, and Time Relationships
The relationship between speed, distance, and time is a fundamental concept in physics and mathematics, encapsulated by the formula \[ \text{speed} = \frac{\text{distance}}{\text{time}} \]. This relationship is linear and can help solve various problems involving motion.

In the context of our airplane problem, we apply this concept to determine the speed of the airplane in still air and the wind speed affecting it. The distance remains constant (800 miles), but the time changes depending on whether the plane is flying with or against the wind. The speed is then calculated for both scenarios.

A strong grasp of these relationships not only supports solving textbook problems but also aids in understanding more complex situations in the real world, such as estimating travel times and planning logistics.

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