A motorboat is capable of traveling at a speed of 15 miles per hour in still water. On a particular day, it took 30 minutes longer to travel a distance of 10 miles upstream than it took to travel the same distance downstream. What was the rate of current in the stream on that day?

Answer 1

Let's say the rate of the current is x miles per hour.
When the motorboat travels upstream, it has to overcome the current, so its effective speed is decreased by x mph.
Therefore, the time it takes to travel 10 miles upstream is 10 / (15 - x) hours.
When the motorboat travels downstream, it benefits from the current, so its effective speed is increased by x mph.
Therefore, the time it takes to travel 10 miles downstream is 10 / (15 + x) hours.
The problem states that it took 30 minutes longer to travel upstream than downstream, which can be expressed mathematically as:
10 / (15 - x) = 10 / (15 + x) + 0.5
Multiplying both sides of the equation by (15 - x)(15 + x), we get:
10(15 + x) = 10(15 - x) + 0.5(15 - x)(15 + x)
150 + 10x = 150 - 10x + 112.5 - 0.5x^2
0.5x^2 + 20x - 112.5 = 0
Dividing the equation by 0.5, we get:
x^2 + 40x - 225 = 0
Using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = 1, b = 40, and c = -225, we get:
x = (-40 ± sqrt(40^2 - 4(1)(-225))) / (2(1))
x = (-40 ± sqrt(1600 + 900)) / 2
x = (-40 ± sqrt(2500)) / 2
x = (-40 ± 50) / 2
x = (50 - 40) / 2 or x = (-50 - 40) / 2
x = 10 / 2 or x = -90 / 2
x = <<10/2=5>>5 or x = -45
Since the rate of the current cannot be negative, the rate of the current in the stream was 5 miles per hour. Answer: \boxed{5}.