I am rowing on a river between two cities that are 19 miles apart. When going downstream, I make the journey in .76
hour(s). When I return upstream, it takes me 2.11
hour(s).
What is the current of the river? what is the speed I would row in still water?
If my speed is s, and the current is c, then since
distance = speed*time,
.76(s+c) = 19
2.11(s-c) = 19
Now just solve for s and c.
To find the current of the river and the speed you would row in still water, we can use the concept of relative velocity.
Let's say the speed of your rowing boat in still water is 'v' mph, and the speed of the current of the river is 'c' mph.
When rowing downstream with the current, your effective speed is increased by the speed of the current. So, the effective speed downstream is (v + c) mph.
Likewise, when rowing upstream against the current, your effective speed is reduced by the speed of the current. So, the effective speed upstream is (v - c) mph.
Given that the distance between the two cities is 19 miles and the time taken for downstream is 0.76 hours and for upstream is 2.11 hours, we can write the following equations:
Downstream: distance = speed * time
19 = (v + c) * 0.76
Upstream: distance = speed * time
19 = (v - c) * 2.11
Now, we have two equations with two variables. We can solve for 'v' (speed in still water) and 'c' (speed of the current).
Let's rearrange the first equation to solve for v:
v + c = 19 / 0.76
v + c = 25
And the second equation:
v - c = 19 / 2.11
v - c = 9
Adding the two equations together eliminates 'c':
(v + c) + (v - c) = 25 + 9
2v = 34
v = 17
Now, substitute the value of 'v' back into one of the equations to solve for 'c':
v + c = 25
17 + c = 25
c = 25 - 17
c = 8
Therefore, the current of the river is 8 mph, and your speed rowing in still water is 17 mph.