The total time it takes for a canoe to go 3 miles upstream (against current) and back 3 miles downstream (with the current) is 4 hours. The current in the lake is 1 mile per hour. Find the speed of the canoe in still water.
I think you have your times backwards.
Obviously it is going to take longer to go against
the current than it takes to go with the current.
I will assume you have typo and fix it.
speed of canoe in still water --- x mph
time to go upstream = 3/(x-1)
time to go downstream = 3/(x+1)
3/(x-1) + 3/(x+1) = 4
multiply by (x^2 - 1)
3(x+1) + 3(x-1) = 4x^2 -4
4x^2 - 6x - 4 = 0
2x^2 - 3x - 2 = 0
(x - 2)(2x + 1) = 0
x = 2 or x = -1/2, rejecting the negative answer,
x = 2
The boat can go 2 mph in still water
check:
time to go against current = 3/1 = 3 hours
time to go with the current = 3/3 = 1 hour, for a total of 4 hours
skip the first part of my answer:
"I think you have your times backwards.
Obviously it is going to take longer to go against
the current than it takes to go with the current.
I will assume you have typo and fix it."
I forgot to delete it from my post. I had typed that after
I misread your question initially.
Well, it sounds like the canoe is definitely going with the flow. I guess it really knows how to go with the current!
To solve this problem, we can use the formula: Time = Distance/Speed.
Let's assume the speed of the canoe in still water is "x" miles per hour.
When the canoe is going upstream, it has to fight against the current, so its effective speed will be (x - 1) miles per hour.
Similarly, when the canoe is going downstream, it gets an extra boost from the current, so its effective speed will be (x + 1) miles per hour.
Now, the time it takes for the canoe to go upstream for 3 miles can be calculated as: 3/(x - 1).
And the time it takes for the canoe to go downstream for 3 miles can be calculated as: 3/(x + 1).
According to the problem, the total time for both trips is 4 hours.
So, we can write the equation: 3/(x - 1) + 3/(x + 1) = 4.
Now, let me take off my clown nose and do some math to solve this equation for you.
To find the speed of the canoe in still water, let's assume the speed of the canoe is x miles per hour.
When the canoe is going upstream, against the current, the effective speed is the difference between the speed of the canoe and the speed of the current, which is x - 1 miles per hour. The distance traveled upstream is 3 miles. Therefore, the time taken to go upstream is given by:
t1 = distance / speed = 3 / (x - 1) hours
When the canoe is going downstream, with the current, the effective speed is the sum of the speed of the canoe and the speed of the current, which is x + 1 miles per hour. The distance traveled downstream is also 3 miles. Therefore, the time taken to go downstream is given by:
t2 = distance / speed = 3 / (x + 1) hours
According to the problem, the total time taken for the round trip is 4 hours. Therefore, we can write the equation:
t1 + t2 = 4
Substituting the expressions for t1 and t2, we get:
3 / (x - 1) + 3 / (x + 1) = 4
To solve this equation, we need to find a common denominator and then simplify:
[(3 * (x + 1)) + (3 * (x - 1))] / [(x - 1)(x + 1)] = 4
[3x + 3 + 3x - 3] / [(x - 1)(x + 1)] = 4
(6x) / (x^2 - 1) = 4
Cross multiplying, we get:
6x = 4(x^2 - 1)
6x = 4x^2 - 4
Rearranging and setting the equation equal to zero, we have:
4x^2 - 6x - 4 = 0
Factoring out a common factor, we get:
2(2x^2 - 3x - 2) = 0
Using the quadratic formula, we can solve for x:
x = (-(-3) ± √((-3)^2 - 4(2)(-2))) / (2(2))
x = (3 ± √(9 + 16)) / 4
x = (3 ± √25) / 4
x = (3 ± 5) / 4
x = (3 + 5) / 4 OR x = (3 - 5) / 4
x = 8 / 4 OR x = -2 / 4
x = 2 OR x = -0.5
Since speed cannot be negative, the speed of the canoe in still water is 2 miles per hour.
To find the speed of the canoe in still water, we need to first understand the relationship between the speed of the canoe and the speed of the current.
Let's assume the speed of the canoe in still water is represented by C mph. Since the canoe travels upstream against the current, its effective speed decreases by the speed of the current. So, the speed of the canoe while going upstream is (C - 1) mph.
On the other hand, while traveling downstream with the current, the effective speed of the canoe increases by the speed of the current. So, the speed of the canoe while going downstream is (C + 1) mph.
Given that it takes a total of 4 hours to travel 3 miles upstream and 3 miles downstream, we can set up the following equation:
Time taken to go upstream + Time taken to go downstream = Total time
3 / (C - 1) + 3 / (C + 1) = 4
To solve this equation, we can first simplify it by multiplying both sides by (C - 1) * (C + 1) to clear the fractions:
3 * (C + 1) + 3 * (C - 1) = 4 * (C - 1) * (C + 1)
3C + 3 + 3C - 3 = 4C^2 - 1
6C = 4C^2 - 1
0 = 4C^2 - 6C - 1
This equation is a quadratic equation in terms of C. We can solve it using the quadratic formula:
C = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, a = 4, b = -6, and c = -1, we get:
C = (-(-6) ± √((-6)^2 - 4 * 4 * -1)) / (2 * 4)
C = (6 ± √(36 + 16)) / 8
C = (6 ± √52) / 8
C = (6 ± 2√13) / 8
Therefore, the speed of the canoe in still water is approximately (6 ± 2√13) / 8 mph.