Orson rides his power boat up and down a canal. The water in the canal flows at 10 km/hr. Orson takes 5 hours longer to travel 360 km against the current than he does to travel 360 km with the current. What is the speed of Orson's boat in still water? (39.2 km/hr)
To find the speed of Orson's boat in still water, we can use the concept of relative speed.
Let's assume the speed of Orson's boat in still water is v km/hr, and the speed of the water current is 10 km/hr.
When Orson is traveling with the current, the effective speed of the boat will be v + 10 km/hr. Since he travels 360 km in that direction, we can write the equation:
Time taken downstream = Distance / Speed = 360 / (v + 10)
Similarly, when Orson is traveling against the current, the effective speed of the boat will be v - 10 km/hr. Since he takes 5 hours longer to travel the same distance, we can write the equation:
Time taken upstream = Distance / Speed = 360 / (v - 10)
According to the problem, the time taken upstream is 5 hours longer than the time taken downstream. So we can write the equation as:
360 / (v - 10) = 360 / (v + 10) + 5
To solve this equation, we can simplify it by canceling out the common terms:
360(v + 10) = 360(v - 10) + 5(v + 10)(v - 10)
Expanding and simplifying further:
360v + 3600 = 360v - 3600 + 5(v^2 - 100)
Canceling out the common terms:
3600 = 5v^2 - 500
Rearranging the terms and solving for v:
5v^2 = 4100
v^2 = 4100 / 5
v^2 = 820
Taking the square root of both sides:
v = √820
v ≈ 28.6356 km/hr (rounded to 4 decimal places)
So, the speed of Orson's boat in still water is approximately 28.6356 km/hr or 28.6 km/hr (rounded to 1 decimal place).