Final answer:
Trey's swimming speed in still water is 4 km/h, calculated by setting up equations for the time taken to swim with and against the current, given that the rate of the current is 2 km/h.
Step-by-step explanation:
To solve the problem of how fast Trey would swim if there were no current, we must first understand that the time taken to swim a certain distance is the same whether he swims with the current or against it. Let's define Trey's swimming speed in still water as v (in kilometers per hour) and the current's speed as 2 km/h.
Swimming against the current, Trey's effective speed would be v - 2 km/h, and swimming with the current, his effective speed would be v + 2 km/h. Since he swam 5 km against the current and 15 km with the current in the same amount of time, we can set up the following equations:
Time against the current = Distance against the current / Speed against the current = 5 km / (v - 2 km/h)
Time with the current = Distance with the current / Speed with the current = 15 km / (v + 2 km/h)
Setting both times equal to each other:
5 km / (v - 2 km/h) = 15 km / (v + 2 km/h)
Cross-multiply and solve for v:
5(v + 2) = 15(v - 2)
5v + 10 = 15v - 30
10 + 30 = 15v - 5v
40 = 10v
v = 4 km/h
Therefore, Trey's swimming speed in still water is 4 km/h.