The mad river flows at a rate of 2 KM/H. In order for a boat to travel 79.2 KM of River and then return in a total of eight hours how fast must about travel in Still water?
You should be able to get 20 km/h as your answer.
To determine the speed of the boat in still water, we need to consider the effects of the river's current. Let's break down the problem:
Let B be the speed of the boat in still water (in KM/H).
The current of the river is given as 2 KM/H.
When the boat travels downstream (with the current):
The effective speed of the boat is increased by the speed of the current. Therefore, the boat's speed downstream is B + 2 KM/H.
When the boat travels upstream (against the current):
The effective speed of the boat is reduced by the speed of the current. Therefore, the boat's speed upstream is B - 2 KM/H.
Now, let's calculate the time it takes for the boat to travel both downstream and upstream:
1. Traveling downstream:
The total distance traveled downstream is 79.2 KM.
Using the formula: time = distance / speed, the time taken downstream is 79.2 / (B + 2) hours.
2. Traveling upstream:
The total distance traveled upstream is also 79.2 KM (since the boat travels back in the same distance).
Using the same formula, the time taken upstream is 79.2 / (B - 2) hours.
Given that the total time taken for the round trip is 8 hours, we can write the equation:
time downstream + time upstream = 8
Substituting the values, we have the equation:
79.2 / (B + 2) + 79.2 / (B - 2) = 8
To solve this equation, we can multiply both sides by (B + 2)(B - 2) to eliminate the denominators.
After simplification and rearrangement, we get a quadratic equation:
79.2(B - 2) + 79.2(B + 2) = 8(B + 2)(B - 2)
Solving this equation will give us the value of B, representing the boat's speed in still water.
time upriver=distance/(S-2)
time downriver= distance/(s+2)
8hours=79.2 ( 1/(s-2) + 1/(s+2))
a bit of algebra, but get a common denominator on the right (s^2-4),
8hours=79.2 ( s+2 + s-2)/s^2-4
then combine, put it in quadratic form, and solve for s.