A canoe in still water travels at a rate of 12 miles per hour. The current today is traveling at a rate of 2 miles per hour. If it took an extra hour to travel upstream, how far was the trip one way?
since time = distance/speed,
d/(12-2) = d/(12+2) + 1
d = 35 miles
check:
35/10 = 3.5
35/14 = 2.5
To find the distance of the trip one way, we need to determine the speed of the canoe relative to the ground when it is traveling upstream and downstream.
Let's denote the speed of the canoe in still water as C, and the speed of the current as R. Given that the canoe travels at a rate of 12 miles per hour in still water (C = 12 mph) and the current is traveling at a rate of 2 miles per hour (R = 2 mph).
When the canoe is traveling upstream (against the current), its effective speed is reduced. So, the speed of the canoe relative to the ground when traveling upstream is (C - R) = (12 - 2) = 10 mph.
Since it took an extra hour to travel upstream, we can deduce that the time taken to return downstream is one hour less.
Let's denote the distance of the trip one way as D. The time taken to travel upstream is (D / (C - R)), and the time taken to return downstream is (D / (C + R)).
Given that the time taken to travel upstream is one hour more than the time taken to return downstream, we can write the equation:
D / (C - R) = D / (C + R) + 1
Substituting the values we know, we have:
D / (12 - 2) = D / (12 + 2) + 1
D / 10 = D / 14 + 1
To solve this equation, we can start by getting rid of the denominators by multiplying each term by the least common multiple (LCM) of 10 and 14, which is 70:
70 * (D / 10) = 70 * (D / 14 + 1)
Simplifying, we get:
7D = 5D + 70
Subtracting 5D from both sides, we have:
2D = 70
Dividing both sides by 2, we find:
D = 35
Therefore, the trip one way is 35 miles.
To determine the distance of a one-way trip, we first need to understand the concept of motion relative to a current. When a canoe is moving upstream (against the current), its effective speed is reduced because it has to overcome the current's opposing force. Conversely, when the canoe is moving downstream (with the current), its effective speed is increased as the current pushes it along.
Let's assume the distance of a one-way trip is represented by "d" miles.
When the canoe is moving upstream (against the current), it travels at a rate of 12 miles per hour (canoe's speed) minus 2 miles per hour (current's speed), which gives us an effective speed of 10 miles per hour.
When the canoe is moving downstream (with the current), it travels at a rate of 12 miles per hour (canoe's speed) plus 2 miles per hour (current's speed), resulting in an effective speed of 14 miles per hour.
Now, let's calculate the time it takes to travel upstream and downstream:
Time taken to travel upstream = d miles / 10 miles per hour = d/10 hours
Time taken to travel downstream = d miles / 14 miles per hour = d/14 hours
According to the given information, it took an extra hour to travel upstream, so we can set up the equation:
d/10 + 1 = d/14
To get rid of the fractions, we can cross-multiply:
14d + 140 = 10d
Subtracting 10d from both sides:
4d + 140 = 0
Subtracting 140 from both sides:
4d = -140
Dividing by 4:
d = -35
Since distance cannot be negative, there must have been an error in the calculations. Let's reexamine the equation:
14d + 140 = 10d
Simplifying further:
4d + 140 = 0
Subtracting 140 from both sides:
4d = -140
Dividing by 4:
d = -35
It appears that there was an error in the equation setup. Let's remove the extra 0 from the equation:
4d + 14 = 10d
Subtracting 4d from both sides:
14 = 6d
Dividing by 6:
d = 14/6
Simplifying:
d ≈ 2.33
Therefore, the distance of the one-way trip is approximately 2.33 miles.