algebra 2 help!?

Question

Here's the problem:
There is a kayaker that goes upstream 15 miles and back again in 32 minutes. If the current is 5 m/hr then what is the paddlers speed? Please help!

Answer 1

That kayaker sure is fast.

Let
s = kayaker's speed in still water
d = oneway distance
t = round trip time
r = average rate roundtrip

d = 15 miles
t = 32 minutes = 32/60 hour
___________

r = 2d/t = (2*15)/(32/60) = 30/(32/60)
= (30*60)/32 = 1800/32 = 225/4

r = 2d/t = 2d / [d/(s - 5) + d/(s + 5)]
225/4 = 2 / [1/(s - 5) + 1/(s + 5)]

225 [1/(s - 5) + 1/(s + 5)] = 2*4
225 [(s + 5) + (s - 5)] = 8(s + 5)(s - 5)
225(2s) = 8(s² - 25)
450s = 8s² - 200
8s² - 450s - 200 = 0
4s² - 225s - 100 = 0

s = {225 ± √[225² - 4*4*(-100)]} / (2*4)
s = {225 ± √52225} / 8
s = {225 ± 5√2089} / 8
s ≈ -0.4409872, 56.690987

Since the answer must be positive we are left with:

s = {225 + 5√2089} / 8 ≈ 56.690987 mph in still water