Two fishing boats depart a harbor at the same time, one traveling east, the other south. The eastbound boat travels at a speed 1 mi/h faster than the southbound boat. After nine hours the boats are x = 45 mi apart. Find the speed of the southbound boat.
To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Let's represent the speed of the southbound boat as "s" mi/h. Since the eastbound boat travels 1 mi/h faster, its speed can be represented as "s + 1" mi/h.
After 9 hours, we can calculate the distance traveled by each boat. The distance for the southbound boat is 9s miles (since speed x time = distance). The distance for the eastbound boat is 9(s + 1) miles.
According to the Pythagorean theorem, the square of the distance between the two boats (x^2) is equal to the sum of the squares of the distances traveled by each boat. In this case, x^2 = (9s)^2 + (9(s + 1))^2.
Simplifying the equation, we can rewrite it as x^2 = 81s^2 + 81(s^2 + 2s + 1).
Expanding further, we have x^2 = 81s^2 + 81s^2 + 162s + 81.
Combining like terms, we get x^2 = 162s^2 + 162s + 81.
Since we know that x = 45 miles, we can substitute this value into the equation: 45^2 = 162s^2 + 162s + 81.
Simplifying further, we have 2025 = 162s^2 + 162s + 81.
Rearranging the equation, we get 324s^2 + 162s + 81 - 2025 = 0.
Simplifying the equation, we have 324s^2 + 162s - 1944 = 0.
Now we can solve this quadratic equation for "s". Using the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = 324, b = 162, and c = -1944. Substituting these values into the quadratic formula, we have s = (-162 ± √(162^2 - 4 * 324 * -1944)) / (2 * 324).
Simplifying further, we get s = (-162 ± √(26244 + 251136)) / 648.
Continuing to simplify, we have s = (-162 ± √277380) / 648.
By evaluating the square root, we have s = (-162 ± 526.612) / 648.
Now, we can calculate both possible values for "s":
s1 = (-162 + 526.612) / 648 ≈ 0.541 mi/h
s2 = (-162 - 526.612) / 648 ≈ -12.60 mi/h
Since speed cannot be negative, we can ignore s2. Therefore, the speed of the southbound boat is approximately 0.541 mi/h.