Problem 18 - Contemporary Precalculus

Question

In the remaining exercises, solve the applied problems. A student leaves the university at noon, bicycling south at a constant rate. At 12: 30 P.M., a second student leaves the same point and heads west, bicycling 7 mph faster than the first student. At 2: 00 P.M., they are 30 miles apart. How fast is each one going?

Step-by-Step Solution

Answer

Answer: The first student is going at a constant speed of approximately 7.92 mph, and the second student is going at a constant speed of approximately 14.92 mph.

Step 1: Define variables

Let x be the speed of the first student (in mph) who is going south and y be the speed of the second student who is going west. Since the second student is 7 mph faster than the first one, we can write y = x + 7.

Step 2: Calculate the time each student traveled

The first student started at noon and by 2:00 P.M.,i.e, for 2 hours. The second student started at 12:30 P.M. and by 2:00 P.M., i.e, for 1.5 (1 hour and 30 minutes) hours.

Step 3: Apply the distance formula for both students

Distance = Speed × Time For the first student, the distance traveled is d1 = x * 2. For the second student, the distance traveled is d2 = y * 1.5.

Step 4: Use the Pythagorean theorem

Since the students' paths create a right triangle, we use the Pythagorean theorem: d1^2 + d2^2 = (distance between them)^2 (x * 2)^2 + (y * 1.5)^2 = 30^2 Remember that y = x + 7, and we substitute it into the equation: (x * 2)^2 + ((x + 7) * 1.5)^2 = 30^2

Step 5: Solve for x

Expand and simplify the equation: (2x)^2 + (1.5x + 10.5)^2 = 900 4x^2 + 2.25x^2 + 31.5x + 110.25 = 900 6.25x^2 + 31.5x - 789.75 = 0 Now, solve the quadratic equation for x using the quadratic formula: x = (-B ± √(B²-4AC)) / 2A x = (-31.5 ± √(31.5²-4(6.25)(-789.75))) / (2(6.25)) Since the speed cannot be negative, we discard the negative value, and we get x ≈ 7.92 mph.

Step 6: Solve for y

Substitute the value of x found in step 5 into the equation y = x + 7: y = 7.92 + 7 y ≈ 14.92 mph

Step 7: State the results

The first student is going at a constant speed of approximately 7.92 mph, and the second student is going at a constant speed of approximately 14.92 mph.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that describes the relationship between the sides of a right-angled triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In mathematical terms, it is written as:
\[ c^2 = a^2 + b^2 \]
For the problem at hand, where students are moving in perpendicular directions, their paths form a right-angled triangle. Here, the actual distance they are apart (30 miles) becomes the hypotenuse. The horizontal distance covered by the student traveling west and the vertical distance by the student going south are the other two sides. Applying the Pythagorean Theorem helps in calculating these individual distances given a composite distance.
We set up the equation:

• \( d1^2 + d2^2 = 30^2 \)
• Substitute the expressions for \(d1\) and \(d2\) using speeds and times into the equation.

Quadratic Equation

A quadratic equation is any equation that can be rearranged in standard form as: \[ ax^2 + bx + c = 0 \]In this problem, we derive a quadratic equation by applying the Pythagorean Theorem and substituting the speeds. Our aim is to solve for the speed of the first student, \(x\), resulting in an equation:

• \[ 6.25x^2 + 31.5x - 789.75 = 0 \]

This form allows us to employ the quadratic formula to find \(x\), which is:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The formula gives two solutions. We discard the negative one, as negative speed doesn't make sense in this context.

Variable Definition

Understanding variables is crucial when dealing with algebraic problems. A variable is essentially a symbol that represents an unknown value in mathematical expressions or equations. Here, two primary variables are defined:

• \(x\): the speed of the first student heading south.
• \(y\): the speed of the second student heading west.

Additionally, a relationship between \(x\) and \(y\) is established based on the problem statement: \[ y = x + 7 \]Defining these variables clearly from the start helps in setting up equations that simplify solving the problem.

Distance Formula

The distance formula in physics relates speed, time, and distance. Specifically, it expresses that distance is the product of speed and time: \( Distance = Speed \times Time.\)For each student in the problem, we apply this formula separately to find their respective distances:

• First Student: \( d1 = x \times 2 \), assuming \(x\) is the speed, and the student travels for 2 hours.
• Second Student: \( d2 = (x + 7) \times 1.5 \), as this student travels at a speed 7 mph faster, for 1.5 hours.

We integrate these distances in applying the Pythagorean Theorem to ultimately find their relative speeds.