wo 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 5 h the boats are 25 mi apart. Find the speed of the southbound boat.
mi/h
Let's assume the speed of the southbound boat is x mi/h.
Since the eastbound boat is traveling 1 mi/h faster, its speed will be x + 1 mi/h.
Using the formula for distance (distance = speed × time), we can calculate the distance traveled by each boat after 5 hours:
Distance traveled by the southbound boat = x × 5 = 5x mi
Distance traveled by the eastbound boat = (x + 1) × 5 = 5(x + 1) mi
Given that the boats are 25 mi apart, we can set up the following equation:
5x^2 + 25 = (5(x + 1))^2
Expanding the equation:
5x^2 + 25 = 25(x^2 + 2x + 1)
Simplifying:
5x^2 + 25 = 25x^2 + 50x + 25
Moving all terms to one side:
20x^2 + 50x = 0
Dividing both sides by 10x to isolate x:
2x + 5 = 0
Subtracting 5 from both sides:
2x = -5
Dividing both sides by 2:
x = -5/2
Therefore, the speed of the southbound boat is -5/2 mi/h. However, speed cannot be negative, so the speed of the southbound boat must be positive.
Thus, the speed of the southbound boat is 5/2 mi/h or 2.5 mi/h.
Let's assume the speed of the southbound boat is "x" mi/h.
Since the eastbound boat travels at a speed 1 mi/h faster, its speed would be "x + 1" mi/h.
In 5 hours, the eastbound boat would have traveled a distance of (x + 1) * 5 = 5x + 5 miles.
The southbound boat would have traveled a distance of x * 5 = 5x miles.
Since they are 25 miles apart, we can set up the following equation:
5x + 5 + 5x = 25
Simplifying the equation:
10x + 5 = 25
10x = 20
x = 2
Therefore, the speed of the southbound boat is 2 mi/h.