Rayburn's boat has a speed of 14 mph in still water.The total time required for Rayburn to make a round trip from his home on the river to his favorite fishing hole is 7 hrs. If the speed of the current is 2 mph, what is the distance from Rayburn's home to the fishing hole?
[(14 - 2) / d] + [(14 + 2) / d] = 7
Hmmm. time = distance/speed ...
To find the distance from Rayburn's home to the fishing hole, we can start by determining the time it takes him to travel in each direction.
Let's assume that the distance from Rayburn's home to the fishing hole is "d" miles.
When Rayburn is going with the current, his effective speed will be the sum of his boat's speed and the speed of the current:
Speed with current = Boat speed + Current speed = 14 mph + 2 mph = 16 mph.
When Rayburn is going against the current, his effective speed will be the difference between his boat's speed and the speed of the current:
Speed against current = Boat speed - Current speed = 14 mph - 2 mph = 12 mph.
Let's suppose it takes Rayburn "t1" hours to reach the fishing hole while going with the current and "t2" hours to return home against the current.
Using the formula: distance = speed × time, we can create two equations:
Equation 1: d = 16t1 (when going with the current)
Equation 2: d = 12t2 (when going against the current)
We know that the total time required for the round trip is 7 hours, so:
t1 + t2 = 7 (Equation 3)
Now we have a system of equations: Equation 1, Equation 2, and Equation 3.
To solve this system, we'll use a method called substitution.
From Equation 1, we can rearrange it to find t1 in terms of d:
t1 = d/16
Substituting t1 in Equation 3, we have:
d/16 + t2 = 7
Rearranging this equation to get t2 in terms of d:
t2 = 7 - d/16
Now, substituting t2 in Equation 2:
d = 12(7 - d/16)
d = 84 - 3d/4
d + 3d/4 = 84
(4d + 3d)/4 = 84
7d/4 = 84
7d = 336
d = 336/7
d = 48 miles
Therefore, the distance from Rayburn's home to the fishing hole is 48 miles.