Problem 42 - Intermediate Algebra

Question

Crop Dusting. \(\quad\) A helicopter spraying fertilizer over a field can fly 0.5 mile downwind in the same time as it can fly 0.4 mile upwind. Find the speed of the wind if the helicopter travels 45 mph in still air when dusting crops.

Step-by-Step Solution

Answer

The wind speed is 5 mph.

Step 1: Identify Variables

Let \( v \) be the speed of the wind in mph. The speed of the helicopter flying downwind is \( 45 + v \) mph, and flying upwind is \( 45 - v \) mph.

Step 2: Set Up the Equation

The time it takes to fly a certain distance is given by distance/speed. For downwind, time \( t_1 = \frac{0.5}{45+v} \) and for upwind, time \( t_2 = \frac{0.4}{45-v} \). Since the times are equal, set \( t_1 = t_2 \).

Step 3: Simplify the Equation

Set up the equation: \( \frac{0.5}{45+v} = \frac{0.4}{45-v} \). Cross-multiply to get: \( 0.5(45-v) = 0.4(45+v) \).

Step 4: Solve for v

Simplify to \( 22.5 - 0.5v = 18 + 0.4v \). Combine like terms to obtain \( 4.5 = 0.9v \). Divide both sides by 0.9 to find \( v = 5 \).

Key Concepts

Speed and Distance Problems

Speed and distance problems are a common type of question found in intermediate algebra that involve calculating distance, rate, or time.
They typically revolve around the formula:

• Distance = Speed × Time

In these problems, the key is to identify what you know and what you need to find.
The question here involves a helicopter flying downwind and upwind, with known distances both ways and a speed in still air.
The core task is to find the wind's speed.

The concept of downwind means the helicopter travels with the wind, effectively increasing its speed.
Conversely, upwind implies traveling against the wind, reducing its speed.
In problem-solving, it's crucial to understand these directions' effects on the effective speed of the helicopter.

Algebraic Equations

Algebraic equations are mathematical statements that use symbols and numbers to express relationships between different quantities.
They are usually solved by finding the value of unknown variables.
In this case, identifying proper equations involves using the formula for time:

• Time = Distance / Speed

The exercise utilizes two such equations to describe the helicopter's journey.
One equation is for the time taken to travel downwind, and the other is for upwind:

• Downwind time: \( \frac{0.5}{45+v} \)
• Upwind time: \( \frac{0.4}{45-v} \)

Both times are equal because the helicopter spends the same amount of time flying both distances.
Establishing an equation where these times are equal allows us to cross-multiply and solve for the variable \( v \), which represents the wind speed.

Problem Solving Steps

Approaching speed and distance problems involves systematic problem-solving steps to arrive at a solution.
Let's walk through these steps:

• **Identify Variables**: Determine what each variable represents in the context of the problem.
  Here, \( v \) denotes the wind's speed, and the helicopter's speeds are expressed as \( 45 + v \) and \( 45 - v \).
• **Set Up Equations**: Use the problem's information to create equations based on known formulas—like the time formula here.
• **Simplify and Solve**: After setting the equation, use algebra to simplify it.
  In this scenario, cross-multiplying leads to a solvable equation for \( v \).

Finally, by simplifying and rearranging the terms, you isolate \( v \) and solve to find the wind speed is 5 mph.
This process is a practical way to tackle algebraic problems in a structured, logical sequence.