A kayaker paddled 2 hours with a 6 mph current in a river. The return trip against the same current took 3 hours. Find the speed the kayaker would make in still water.
Let x = kayaker paddle speed.
Then going down the river his speed is 6+x and since distance = rate*time then distance = (6+x)*2 hrs or d = 12 + 2x.
On the return trip his speed is 6-x and the distance is (6-x)*3 hrs or d = 18-3x
The two distances are equal; therefore, set the two equations equal and solve for x. Check my thinking. Check my work.
To find the speed the kayaker would make in still water, we can use the concept of relative velocity. Let's assume the speed of the kayaker in still water is represented by 'x' mph.
During the downstream trip, the kayaker is paddling with the current, so the effective speed is the sum of the kayaker's speed in still water (x mph) and the speed of the current (6 mph). Therefore, the effective speed is (x + 6) mph.
We are given that the kayaker paddled downstream for 2 hours, so the distance covered is (x + 6) * 2.
During the upstream trip, the kayaker is paddling against the current, so the effective speed is the difference between the kayaker's speed in still water (x mph) and the speed of the current (6 mph). Therefore, the effective speed is (x - 6) mph.
We are given that the kayaker paddled upstream for 3 hours, so the distance covered is (x - 6) * 3.
Since the downstream trip and the upstream trip cover the same distance (since they are going from the same starting point to the same ending point), we can equate the distances:
(x + 6) * 2 = (x - 6) * 3
Simplifying this equation, we get:
2x + 12 = 3x - 18
Bringing all the 'x' terms to the left side and the constant terms to the right side, we get:
2x - 3x = -18 - 12
-x = -30
Solving for 'x', we get:
x = 30
Therefore, the speed the kayaker would make in still water is 30 mph.