Problem 40 - College Algebra and Calculus: An Applied Approach

Question

Airplane Speed Two planes start from the same airport and fly in opposite directions. The second plane starts \(\frac{1}{2}\) hour after the first plane, but its speed is 50 miles per hour faster. Find the air speed of each plane if, 2 hours after the first plane departs, the planes are 2000 miles apart.

Step-by-Step Solution

Answer

The air speed of the first plane is approximately 550 mph and the air speed of the second plane is 600 mph.

Step 1: Define the Variables

Let's define \(x\) as the speed of the first plane in miles per hour, therefore the speed of the second plane will be \(x + 50\). As the second plane flies half an hour less than the first one, the first plane flies for 2 hours and the second for 1.5 hours.

Step 2: Set up the equation

According to the problem, 2 hours after Plane 1 departs, they are 2000 miles apart. So the equation will be: distance covered by Plane 1 + distance covered by Plane 2 = total distance between the planes, which will result in: \(2x + 1.5(x + 50) = 2000\).

Step 3: Solve the Equation

First, expand and simplify the equation: \((2x + 1.5x + 75) = 2000\). Combine the x terms to get \(3.5x + 75 = 2000\). Subtract 75 from both sides to get \(3.5x = 1925\). Then, divide both sides by 3.5 to solve for x which will be approximately \(550\).

Step 4: Find the Speed of the Second Plane

Since the second plane is 50 miles per hour faster than the first, its speed will be \(550 + 50 = 600\) mph.

Key Concepts

System of Equations

Understanding how to solve a system of equations can be incredibly useful in algebra.To solve problems like the one about airplanes moving in opposite directions, you often need to set up and solve a system.In this particular exercise, defining the variables is the first important step.

• Let \( x \) represent the speed of the first plane in miles per hour.
• Consequently, the speed of the second plane becomes \( x + 50 \) because it is 50 mph faster.

After defining the variables, the next step is to set up an equation based on the problem's conditions.Here, the key equation is:

• The sum of the distances covered by both planes equals 2000 miles.
• This can be expressed as: \( 2x + 1.5(x + 50) = 2000 \)

Once you've set up the equation, you can solve it by combining like terms and simplifying, giving you the speeds you need.

Distance-Rate-Time Problems

Distance-rate-time problems often involve understanding how distance, speed (rate), and time are related.The basic formula connecting these three is:

• \( \text{Distance} = \text{Rate} \times \text{Time} \)

In the airplane problem, you need to apply this formula to both planes.For the first plane:

• Distance covered is \( 2x \).
• This is because it travels for 2 hours and its speed is \( x \).

For the second plane:

• Distance covered is \( 1.5(x+50) \).
• It travels for 1.5 hours at a speed of \( x+50 \).

Adding together these distances gives us the total distance of 2000 miles covered by both planes.Breaking down a complex word problem like this into segments where you apply the distance-rate-time formula is crucial in simplifying and solving these types of questions.

Linear Equations

Solving linear equations is a fundamental skill in algebra that involves finding the value of variables that make the equation true.In this case, the equation formed is a linear equation which eventually looks like:

• \( 3.5x + 75 = 2000 \)

Linear equations can be simplified and solved by performing basic operations on both sides.

• You first simplify the equation by combining like terms.
• Next, isolate the variable \( x \) by subtracting 75 from both sides resulting in: \( 3.5x = 1925 \).
• Finally, divide both sides by 3.5 to solve for \( x \), arriving at: \( x = 550 \).

This gives the speed of the first plane.Since the second plane is 50 mph faster, we find its speed to be 600 mph.By breaking down the process into clear steps, solving linear equations becomes manageable and logical.