Joseph was kayaking on the Hudson River. While looking at the Breakneck Ridge trail-head, he lost a whole bag of donuts. Joseph didn’t realize he had lost it for fifteen minutes. That’s when he turned back and started going in the opposite direction. When he found the bag, which was going at the speed of the Hudson’s current, it was two miles from the Breakneck Ridge trail-head. What is the speed of the current in the Hudson River?

Guest May 23, 2023

#1**0 **

Let x be the speed of Joseph's kayak and y be the speed of the current.

We know that Joseph was kayaking for 15 minutes and that the bag was 2 miles away from the Breakneck Ridge trail-head when he found it.

This means that Joseph traveled 15x miles and the bag traveled 15y miles.

We also know that the bag was carried downstream by the current, so the distance it traveled is equal to the distance it would have traveled if Joseph had been kayaking in the opposite direction for 15 minutes.

This means that 15y = 15x - 2

Solving for y, we get y = x - 2/3

Therefore, the speed of the current in the Hudson River is x - 2/3 = 2 - 2/3 = 4/3 miles per hour.

So the answer is 4/3

Guest May 23, 2023

#2**0 **

let his speed upstream=S

Let the speed of the current=C

Let the distance he covered paddling upstream=D

D / [S - C] = 15/60,

[D + 2] / [S + C]==[D+2] /C,

[2 - D] / C ==15/60, solve for C, D, S

**C==4 mph - speed of the current**

S ==8 mph - his speed in still water

D==1 mile - upstream from the Ridge when he realized he had lost his donuts.

Guest May 24, 2023