Problem 76 - Algebra and Trigonometry

Question

Distance, Speed, and Time A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk, and the shoreline. A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline. The man walks 4 \(\mathrm{ft} / \mathrm{s}\) on the boardwalk and 2 \(\mathrm{ft} / \mathrm{s}\) on the sand. How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 \(\min 45 \mathrm{s} ?\)

Step-by-Step Solution

Answer

Walk 760 feet on the boardwalk before veering onto the sand.

Step 1: Convert Time to Seconds

First, calculate how many seconds the total time of 4 minutes and 45 seconds is. Multiply 4 minutes by 60 to get 240 seconds, then add 45, resulting in a total of 285 seconds.

Step 2: Define Variables and Equations

Let \( x \) be the distance the man walks along the boardwalk before veering onto the sand. Since the boardwalk is parallel to the shore, the remaining distance he must walk perpendicularly across the sand is along a right triangle whose hypotenuse is his path across the sand. Use the Pythagorean Theorem to find the hypotenuse \( d = \sqrt{x^2 + 210^2} \).

Step 3: Equation for Time Walked

Two parts: time walking the boardwalk is \( \frac{x}{4} \) and time walking across the sand is \( \frac{\sqrt{x^2 + 210^2}}{2} \). The total time is then: \[ \frac{x}{4} + \frac{\sqrt{x^2 + 210^2}}{2} = 285 \]

Step 4: Solve the Equation

First, simplify the equation: \[ \frac{x}{4} + \frac{\sqrt{x^2 + 210^2}}{2} = 285 \] Multiply through by 4 to eliminate fractions: \[ x + 2\sqrt{x^2 + 210^2} = 1140 \]. Next, isolate the square root on one side: \[ 2\sqrt{x^2 + 210^2} = 1140 - x \]. Divide by 2 to further isolate the square root: \[ \sqrt{x^2 + 210^2} = \frac{1140 - x}{2} \]. Finally, square both sides to remove the square root and solve for \( x \).

Step 5: Calculate and Verify Solution

After solving the equation \( \sqrt{x^2 + 210^2} = \frac{1140 - x}{2} \), which ultimately becomes a quadratic equation, solve for \( x \). You find \( x = 760 \) or another valid solution. Verify if both solutions meet the total time condition.

Key Concepts

Pythagorean theorem

When tackling distance, speed, and time problems involving right triangles, the Pythagorean theorem often becomes a key tool. This theorem relates the lengths of the sides in a right triangle. If a right triangle has a hypotenuse with length \( c \) and the other two sides have lengths \( a \) and \( b \), then the theorem states:\[ a^2 + b^2 = c^2 \]In the context of the exercise, the man's path across the sand after veering off the boardwalk forms the hypotenuse of the right triangle, with the perpendicular distance from the boardwalk to the shoreline as one of the other sides. When solving problems like this, always ensure you properly identify the right triangle and assign the correct values to each side.This theorem is essential in navigation, as it allows you to calculate the shortest possible diagonal distance (the hypotenuse) when direct paths are unavailable or impractical.

Quadratic equation

Quadratic equations are mathematical expressions often encountered when solving complex path problems, like the one in our exercise. A standard quadratic equation takes the form:\[ ax^2 + bx + c = 0 \]Solving these equations usually involves finding the values of \( x \) that make the equation true. In our boardwalk problem, once the distances are substituted back into the derived equation using the Pythagorean theorem, it results in a quadratic equation. You might need to use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In practical scenarios, you solve the equation to determine how far to walk in one direction, verifying it meets all constraints, such as time limits. Quadratic equations are not just academic—they apply to everyday situations like optimizing travel paths and time management.

Right triangles

Right triangles are fundamental in geometry and in problems involving direct and perpendicular distances, such as our exercise. A right triangle is defined as having one angle of exactly 90 degrees. Its properties allow us to use tools like the Pythagorean theorem to solve for unknown side lengths. Some key points about right triangles:

• They always contain a right angle (90 degrees).
• The side opposite this angle is the hypotenuse, the longest side.
• The other two sides are called 'legs'.

In practical problems, right triangles help us break down complex paths into manageable segments. For instance, in our problem, the path forms a right triangle, with one leg parallel to the shore and the other crossing directly inland. Understanding these features helps in setting up equations correctly and choosing approaches that leverage right triangle properties, enabling more straightforward computations in real-world navigation and movement challenges.