8. The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for his business. Every day, his route takes him 22.5 miles against the current and back to his dock, and he needs to make this trip in a total of 9 hours. He has a boat in mind, but he can only test it on a lake where there is no current. How fast must the boat go on the lake in order for it to serve the ferry operator’s needs? *
Speed of boat he needs --- x mph
Speed with the current = x + 6 mph
speed against the current = x - 6 mph
22.5/(x+6) + 22.5/(x-6) ≤ 9
let's use the equality...
multiply each term by (x-6)(x+6)
22.5(x-6) + 22.5(x+6) = 9(x-6)(x+6)
22.5x - 135 + 22.5x + 135 = 9x^2 - 324
9x^2 - 45x - 324 = 0
x^2 - 5x + 36 = 0
(x - 9)(x + 4) = 0
x = 9 or x = -4, but we obviously reject the x = -4
His new boat must be able to go at least 9 mph, (slow boat?)
tysm
Thank you Mathhelper!
Well, it seems like the ferry operator is in a bit of a dilemma. Let me see what I can come up with!
If the ferry operator is traveling against the current for 22.5 miles, and the speed of the current is 6 mph, it means that the boat's speed relative to the ground is 6 mph slower than its speed on the lake.
To make the round trip in 9 hours, the ferry operator would need to spend half of that time going against the current, and the other half going with the current. So, each leg of the journey would take 4.5 hours.
Now, let's calculate the boat's required speed on the lake. Since it takes 4.5 hours to go 22.5 miles against the current, the boat's speed (relative to the ground) would need to be:
22.5 miles / 4.5 hours = 5 mph
So, the boat would need to go at least 5 mph on the lake for the ferry operator to meet his timing requirements.
To determine the speed the boat must go on the lake in order to serve the ferry operator's needs, we need to consider the effect of the current on the boat's speed.
Let's start by understanding the time it takes for the ferry operator to travel against the current and back to the dock. The total distance covered in the round trip is 22.5 miles. Since the current is 6 mph, it effectively slows down the boat's speed.
Let's assume the boat's speed on the lake is represented by "x" mph.
When the ferry operator travels against the current, the boat's effective speed is reduced by the speed of the current. So, the time it takes to travel against the current is:
Time against current = distance / (speed of boat - speed of current)
Time against current = 22.5 miles / (x mph - 6 mph)
Similarly, when the ferry operator travels with the current, the current adds to the boat's speed. So, the time it takes to travel with the current is:
Time with current = distance / (speed of boat + speed of current)
Time with current = 22.5 miles / (x mph + 6 mph)
Given that the total time for the round trip is 9 hours, we can set up the following equation:
Time against current + Time with current = Total time
22.5/(x - 6) + 22.5/(x + 6) = 9
To solve this equation, we can multiply through by (x - 6)(x + 6) to eliminate the denominators:
22.5(x + 6) + 22.5(x - 6) = 9(x - 6)(x + 6)
Simplifying further, we get:
22.5x + 135 + 22.5x - 135 = 9(x^2 - 36)
45x = 9x^2 - 324
9x^2 - 45x - 324 = 0
Now we can solve this quadratic equation using the quadratic formula. The formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our equation, we get:
x = (-(-45) ± √((-45)^2 - 4*9*(-324))) / (2*9)
x = (45 ± √(2025 + 11664)) / 18
x = (45 ± √(13689)) / 18
Simplifying further:
x = (45 ± 117) / 18
x = (45 + 117) / 18 or x = (45 - 117) / 18
x = 162 / 18 or x = -72 / 18
x = 9 or x = -4
Since speed cannot be negative, the speed of the boat on the lake would need to be 9 mph in order for it to serve the ferry operator's needs.
Therefore, the boat must go at a speed of 9 mph on the lake to compensate for the current and meet the ferry operator's requirements.