First, make an illustration of the problem as shown in the attached picture, The red point is where the bottle was dropped, while the blue point is where the bottle was when they caught up to it.
Let x be the speed of the canoe, while y is the speed of the river. Because the canoe is going against the upstream, its velocity is equal to:
x - y = distance/time = 2km/60min
x - y = 1/30 km/min
x = 1/30 + y --> eqn 1
Since the bottle is going with the flow, its velocity is just equal to y. The distance covered by the bottle should be 5 - 2 = 3 km. This must be equivalent to the 60-minute and the additional time for the canoe to catch up. The equation is:
60x + x(5/x + y) = 3 km --> eqn 2
Solving simultaneously by substituting eqn 1 to eqn 2, then solving algebraically,
x = 0.0583 km/min
y = 0.025 km/min
a.) The river's current has a speed of 0.025 km/min.
b.) The speed of the canoe on a still lake is 0.0583 km/min.
