Two boat landings are 6.0 km apart on the same bank of a stream that flows at 2.8 km/h. A motorboat makes the round trip between the two landings in 50 minutes. What is the speed of the boat relative to the water?
To find the speed of the boat relative to the water, we need to break down the problem and calculate the boat's speed and the stream's speed separately.
Let's denote the speed of the boat as 'x' km/h and the stream's speed as 's' km/h.
Since the boat is making a round trip, it will be traveling upstream and downstream. The upstream speed of the boat is the difference between its actual speed and the stream's speed, and the downstream speed is the sum of the boat's speed and the stream's speed.
Given that the stream flows at 2.8 km/h, we can set up two equations:
Equation 1: Upstream Speed = Boat's Speed - Stream's Speed
Equation 2: Downstream Speed = Boat's Speed + Stream's Speed
Now, let's calculate the time it takes for the boat to travel upstream and downstream.
The distance between the boat landings is 6.0 km. When the boat is going upstream (against the stream), its effective speed is the difference between the boat's speed and the stream's speed (x - 2.8 km/h). So, the time taken to travel upstream is equal to the distance divided by the effective speed:
Time Upstream = 6.0 km / (x - 2.8 km/h)
Similarly, when going downstream (with the stream), the effective speed is the sum of the boat's speed and the stream's speed (x + 2.8 km/h). So, the time taken to travel downstream is:
Time Downstream = 6.0 km / (x + 2.8 km/h)
According to the problem statement, the boat takes a total of 50 minutes (or 50/60 = 5/6 hours) for the round trip. This means that the total time taken for both the upstream and downstream journeys is equal to 5/6 hours:
Time Upstream + Time Downstream = 5/6 hours
Plugging in the values from the previous equations, we get:
6.0 km / (x - 2.8 km/h) + 6.0 km / (x + 2.8 km/h) = 5/6 hours
Now, we have an equation with one variable (x), the boat's speed. Let's solve it to find the speed of the boat relative to the water.
The next steps involve algebraic manipulation to solve for x. Multiply the entire equation by the least common multiple (LCM) of the denominators to eliminate the fractions:
[(x - 2.8 km/h) * (x + 2.8 km/h) * 6.0 km] / (x - 2.8 km/h) + [(x - 2.8 km/h) * (x + 2.8 km/h) * 6.0 km] / (x + 2.8 km/h) = 5/6 hours * LCM
Simplify the equation:
[(x + 2.8 km/h) * 6.0 km + (x - 2.8 km/h) * 6.0 km] / [(x - 2.8 km/h) * (x + 2.8 km/h)] = 5/6 hours * LCM
Multiply through by [(x - 2.8 km/h) * (x + 2.8 km/h)] to eliminate the denominators:
[(x + 2.8 km/h) * 6.0 km + (x - 2.8 km/h) * 6.0 km] = [(x - 2.8 km/h) * (x + 2.8 km/h)] * (5/6 hours * LCM)
Expand and simplify the equation:
6.0 km * x + 6.0 km * 2.8 km/h + 6.0 km * x - 6.0 km * 2.8 km/h = (x^2 - (2.8 km/h)^2) * (5/6 hours * LCM)
Combine like terms:
12.0 km * x = (x^2 - (2.8 km/h)^2) * (5/6 hours * LCM)
Expand and simplify:
12.0 km * x = (x^2 - 7.84 km^2) * (5/6 hours * LCM)
Distribute the multiplication:
12.0 km * x = (5/6 hours * LCM) * x^2 - (5/6 hours * LCM) * 7.84 km^2
Rearrange the equation to isolate the quadratic term:
(5/6 hours * LCM) * x^2 - 12.0 km * x - (5/6 hours * LCM) * 7.84 km^2 = 0
Now we have a quadratic equation of the form ax^2 + bx + c = 0, where a = (5/6 hours * LCM), b = -12.0 km, and c = -(5/6 hours * LCM) * 7.84 km^2. We can solve this quadratic equation to find the value(s) of x, which will give us the boat's speed relative to the water.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
The quadratic formula gives two possible solutions for x, which correspond to the boat's speed relative to the water.