someone rides 1 mile upstream ane then 1 mile back in 1 hour and the speed of the river is 2 mph, what's the

Question

rate of speed of the ride in still water?
Years later I still have trouble with this one

Answer 1

the distance travelled both ways is the same (one mile)
and d = rt
let x be rate of speed in still water
then upstream is (x-2) and downstream is (x+2)
so rewrite d=rt as t = d/r
time upstream is 1/(x-2)
time downstream is 1/(x+2)
since the round trip took 1 hour total......
1/(x-2) + 1/(x+2) = 1
solve this

LCD will be (x-2)(x+2)
multiply both sides by this and get
(x+2) + (x-2) = 1(x-2)(x+2) == x² - 4
2x = x² - 4
0 = x² - 2x - 4 use quadratic formula to solve this and get 1 ± √5
of course you only want 1 + √5 because the other answer would be negative and you will not have negative rate of speed.

Answer 2

Let S be the speed in still water.
Let T be time it takes to go upstream. Then (1-T) is the amount of time it takes to get back.

You can go at a rate of (S-2) upstream and (S+2) downstream.

Therefore:

(S-2)*T = 1
(S+2)*(1-T) = 1

So

T = 1/(S-2)

That means:

(S+2)*(1-1/(S-2)) = 1

If you multiply both sides of this equation by (S-2) you will get a quadratic equation. Only one of the roots of that equation should work.

Answer 3

uhh try the d=rt formula

distance=rate times time

Answer 4

idk why are they traveling by water its cold out.