Abby Tanenbaum paddled her canoe 10 miles upstream, and then paddled back to her starting point. If the rate of the current was 3 mph,and the entire trip took 3.5 hours, what was Abby's rate (in still water)?
To solve this problem, we can set up a few equations.
Let's denote Abby's rate in still water as "r" (in mph).
When Abby paddles upstream (against the current), her effective speed is the difference between her paddling rate and the rate of the current, which is (r - 3) mph.
When Abby paddles downstream (with the current), her effective speed is the sum of her paddling rate and the rate of the current, which is (r + 3) mph.
We know that Abby paddled 10 miles upstream and then paddled back to her starting point, so the total distance is 2 * 10 = 20 miles.
We also know that the entire trip took 3.5 hours.
Now, using the formula Time = Distance / Speed, we can create two equations:
1. Upstream: 10 / (r - 3)
2. Downstream: 10 / (r + 3)
The total time for both trips is 3.5 hours, so we can set up the equation:
10 / (r - 3) + 10 / (r + 3) = 3.5
Now, let's solve this equation to find Abby's rate (r) in still water.
Multiply the whole equation by (r - 3)(r + 3) to eliminate the denominators:
10(r + 3) + 10(r - 3) = 3.5(r - 3)(r + 3)
Simplify the equation:
10r + 30 + 10r - 30 = 3.5(r^2 - 9)
20r = 3.5(r^2 - 9)
20r = 3.5r^2 - 31.5
3.5r^2 - 20r - 31.5 = 0
Now, we have a quadratic equation. We can solve it using the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = 3.5, b = -20, and c = -31.5.
Plugging in the values and simplifying, we get two solutions for r:
r ≈ 6.53 or r ≈ -1.43
We can disregard the negative value since rate is always positive.
Therefore, Abby's rate in still water is approximately 6.53 mph.