Jim paddles a canoe upstream at a rate o 3mi/hr. Traveling downstream, he travels at 8 mi/hr. What is his rate in still water and what is the rate of the current?
PLEASE SHOW THE SOLUTION EQUATION
Eq1: Vc - Vw = 3
Eq2: Vc + Vw = 8
Add the Eqs:
2Vc = 11
Vc = 5.5 mi/h = Velocity of canoe in
stil water.
In Eq2, substitute 5.5 for Vc:
5.5 + Vw = 8
Vw = 8-5.5 = 2.5 mi/h. = Velocity of
the water.
To solve this problem, we can use the concept of relative velocity. Let's denote the rate of the canoe in still water as 'V' and the rate of the current as 'C'.
When Jim paddles upstream, his effective velocity is the difference between the rate of the canoe in still water and the rate of the current. So, his effective velocity upstream is (V - C) mi/hr.
Similarly, when Jim paddles downstream, his effective velocity is the sum of the rate of the canoe in still water and the rate of the current. So, his effective velocity downstream is (V + C) mi/hr.
Given that Jim's effective velocity upstream is 3 mi/hr and his effective velocity downstream is 8 mi/hr, we can set up the following equations:
Upstream velocity: V - C = 3
Downstream velocity: V + C = 8
To solve this system of equations, we can use a method called substitution. Rearrange the first equation to get V = 3 + C.
Substitute this expression for V in the second equation:
(3 + C) + C = 8
3C + 3 = 8
3C = 5
C = 5/3
Now substitute the value of C back into the first equation to find V:
V - (5/3) = 3
V = 3 + 5/3
V = 14/3
Therefore, Jim's rate in still water, V, is 14/3 mi/hr, and the rate of the current, C, is 5/3 mi/hr.