Q:Related rates problem

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Question: A person can swim from A to B (with current) in 40 minutes, and back (against current) in 45 minutes. How long does it take them to kayak from A to B if the return trip (B to A, against current) takes 15 minutes.

Assume the speed of the current, and swim and kayaking speeds relative to current are all constant.

I set up the following equations, and tried to solve:

x = distance (constant), y = swim speed, z = current, a = kayak speed

t = time to get from B to A kayaking

  1. $x = 45(y-z)$
  2. $x= 15(a-z)$
  3. $x = 40(y+z)$
  4. $x = s (a + z)$

I got y = 17z from 1 and 3, and tried to create a proportion from them and use that to get the kayaking time, but it that didn't work.

Any ideas? Any and all help is appreciated.

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Let x be 1.

The kayaking rate is unaffected by the swim speed, only the current, so you should solve for the current, z, with Equations 1 and 3.

As a hint, don't do 45(y-z)=40(y+z), as that will give you z in terms of y when you can solve for z itself.

Assuming s is the kayaking time from A to B, notice equations 2 and 4 are two equations with 2 unknowns after you find z, namely a and s.

Can you take it from here?