Question

Imagine you are canoeing on the Colorado River. In still water, your average canoeingrate is 3 miles per hour. It takes you the same amount of time to travel 10 milesdownstream, with the current, as 2 miles upstream, against the current. What is the rateof the water’s current?

Answer 1

The rate of the water's current is 2 mph

Step-by-step explanation

Let r represents the rate of water's currrent. (In miles per hour)

Downstream

Distance = 10 miles Rate = 3+r

Using d = rt,

t = d/r

`t=(10)/(3+r)`

Upstream

Distance = 2 miles Rate = 3- r

`t=(2)/(3-r)`

Times are the same

Hence, equate the two ts

`(10)/(3+r)=(2)/(3-r)`

Cross-multiply

10(3-r) = 2(3+r)

Open the parenthesis.

30 - 10r = 6 + 2r

collect like term

30 - 6 = 2r + 10r

24 = 12r

Divide both-side by 12

2 = r

r = 2

Therefore, the rate of the water's current is 2 mph