Problem 71 - College Algebra and Calculus: An Applied Approach

Question

Two planes leave simultaneously from the same airport, one flying due east and the other due south (see figure). The eastbound plane is flying 50 miles per hour faster than the southbound plane. After 3 hours the planes are 2440 miles apart. Find the speed of each plane.

Step-by-Step Solution

Answer

The speed of the southbound plane is 660 mph and the eastbound plane is 710 mph.

Step 1: Setup the Pythagorean theorem

Because the planes travel due east and due south, they form two legs of a right triangle, with their resultant distance apart providing the hypotenuse. Therefore, if we denote the speed of the Southbound airplane as \(x\) mph, then the speed of the Eastbound airplane is \(x + 50\) mph. After 3 hours, the distances they traveled form the two legs of a right triangle. This gives us the ratios: \((3x)^2 + (3(x+50))^2 = 2440^2\).

Step 2: Simplifying the equation

Simplify the equation to solve for \(x\). It becomes \(9x^2 + 9(x^2 + 100x + 2500) = 2440^2\), and further simplifies to \(18x^2 + 900x - 2440^2 = 0\). This now resembles a quadratic equation in the form \(ax^2 + bx + c = 0\).

Step 3: Solving quadratic equation

We solve the quadratic equation by using the quadratic formula \(x= -b \pm \sqrt{b^2 - 4ac} / 2a\) for \(x\), where \(a = 18\), \(b=900\), and \(c=-2440^2\). After computation, one of the solutions of \(x\) turns out to be negative which is not reasonable for this case as speed cannot be negative. Thus, \(x\) is approximately 660 mph.

Step 4: Find the speed of both planes

With \(x\) being the speed of the southbound plane, we have 660 mph, and for the eastbound plane (which is has a speed 50 mph more than the southbound plane), we get a speed of 660 + 50 = 710 mph.

Key Concepts

Right Triangles

In the scenario of two planes leaving an airport, the paths they travel form a right triangle. A right triangle is a triangle with one angle measuring 90 degrees. This shape allows us to use the Pythagorean theorem, which is a crucial tool in solving problems involving right triangles. The theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the two other sides.

For the exercise, the two planes fly in perpendicular directions, due east and due south. This setup creates a right triangle with:

• The eastbound plane’s path as one leg.
• The southbound plane’s path as another leg.
• The straight-line distance between them as the hypotenuse.

Thus, by applying the Pythagorean theorem, we can connect these distances and arrival times. It helps us find the unknown side, which in this case, are the speeds of the planes.

Quadratic Equations

Quadratic equations are algebraic expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. This kind of equation often appears in problems involving areas, projectile motion, and in this case, the paths of two airplanes.

In solving this problem, simplifying the attributes of the right triangle led to a quadratic equation. Initially, by expressing the planar distances in terms of time and speed, a quadratic form emerged: \(18x^2 + 900x - 2440^2 = 0\). The task was to find \(x\), which represents the unknown speed of the southbound plane.

• You can solve quadratic equations using various methods, with the quadratic formula being common: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Choosing the positive solution is essential since speed can't be negative.

This step is vital because accurate calculation leads us to understand the dynamics between the planes’ speeds.

Distance-Speed-Time Relationships

The relations between distance, speed, and time form the basis for solving many practical problems, like this one involving two planes. The fundamental formula connecting these three variables is:

• Distance = Speed × Time

This equation is extraordinarily helpful.
For instance, when calculating how far each plane travels in the given time (3 hours), we use:

• For the southbound plane: Distance = \(3x\)
• For the eastbound plane: Distance = \(3(x + 50)\)

The problem used these distance expressions to set up a relationship as part of the right triangle. Understanding how these elements work together is key in deducing one variable when you have the other two. This foundational link shows us how these speeds interact over time to create the distances featured.