Problem 94 - Algebra and Trigonometry

Question

Speed of a Boat Two fishing boats depart a harbor at the same time, one traveling east, the other south. (See the following figure.) The eastbound boat travels at a speed 3 \(\mathrm{mi} / \mathrm{h}\) faster than the southbound boat. After two hours the boats are 30 \(\mathrm{mi}\) apart. Find the speed of the southbound boat.

Step-by-Step Solution

Answer

The speed of the southbound boat is 9 miles per hour.

Step 1: Define Variables

Let the speed of the southbound boat be \( s \) miles per hour. Thus, the speed of the eastbound boat is \( s + 3 \) miles per hour, as it travels 3 miles per hour faster.

Step 2: Calculate Distance Traveled

After two hours, the southbound boat travels a distance of \( 2s \) miles, and the eastbound boat travels \( 2(s+3) \) miles.

Step 3: Use the Pythagorean Theorem

The distance between the two boats forms the hypotenuse of a right triangle with sides \( 2s \) and \( 2(s+3) \). According to the Pythagorean theorem, we have:\[(2s)^2 + [2(s+3)]^2 = 30^2\]

Step 4: Solve the Equation

Expanding the equation:\[4s^2 + 4(s+3)^2 = 900\]Simplify:\[4s^2 + 4(s^2 + 6s + 9) = 900\]\[4s^2 + 4s^2 + 24s + 36 = 900\]Combining terms:\[8s^2 + 24s + 36 = 900\]Subtract 900 from both sides:\[8s^2 + 24s - 864 = 0\]Divide by 8:\[s^2 + 3s - 108 = 0\]

Step 5: Factor the Quadratic Equation

We need to find two numbers that multiply to \(-108\) and add to \(3\). These numbers are \(12\) and \(-9\). Therefore, we factor the equation:\[(s + 12)(s - 9) = 0\].

Step 6: Solve for s

The solutions are \(s + 12 = 0\) or \(s - 9 = 0\). This gives us the solutions \(s = -12\) or \(s = 9\). Since speed cannot be negative, we have \(s = 9\).

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that relates the sides of a right triangle. It states that in any right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be expressed with the formula:

• \[a^2 + b^2 = c^2\]

In our boat exercise, the route taken by the two boats forms a right triangle, where their paths represent the two perpendicular sides of the triangle, and the distance apart after two hours is the hypotenuse.
To apply the Pythagorean theorem to the problem, we set the distances traveled by the eastbound and southbound boats as the legs of the triangle. Suppose the speed of the southbound boat is \(s\) miles per hour. After two hours, it would have traveled \(2s\) miles. The eastbound boat, traveling 3 miles per hour faster, would have covered \(2(s+3)\) miles.
Hence, the equation used to solve the problem is:

• \[(2s)^2 + [2(s+3)]^2 = 30^2\]

Using this method allows us to relate the speeds of the boats to their separation distance, effectively integrating algebraic problem-solving with geometric concepts.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They typically take the form:

• \[ax^2 + bx + c = 0\]

where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. Solving quadratic equations usually involves finding the values of \(x\) that satisfy the equation.
In our problem involving the boats, we derived a quadratic equation from the Pythagorean theorem:

• \[8s^2 + 24s - 864 = 0\]

Simplifying this by dividing all terms by 8, we focus on:

• \[s^2 + 3s - 108 = 0\]

The goal then is to find the values of \(s\) that will satisfy this quadratic equation. This process provides an algebraic way to solve problems that may not be directly solvable by simpler arithmetic means.

Factoring Quadratics

Factoring is a method used to solve quadratic equations. It involves rewriting the quadratic equation as a product of two binomials. This method is particularly useful when the quadratic is easily factorable, which means identifying two numbers that multiply to the constant term \(c\) and sum to the coefficient of the linear term \(b\).
In the context of our exercise, the equation to factor is:

• \[s^2 + 3s - 108 = 0\]

To factor it, we look for two numbers that multiply to \(-108\) and add up to \(3\). After examining different pairs of factors, we find that \(12\) and \(-9\) work:

• \[(s + 12)(s - 9) = 0\]

These factors correspond to the potential solutions for the speed of the southbound boat, which are \(s = -12\) and \(s = 9\). However, since speed cannot be negative, we conclude that \(s = 9\) miles per hour is the valid solution.