Respuesta :
Answer:2.55
Step-by-step explanation:Let's denote the speed of the current as \( C \) (in miles per hour).
When the tugboat travels downstream (with the current), its effective speed is the sum of its speed in still water and the speed of the current. So, the speed downstream is \( 14 + C \) mph.
When the tugboat travels upstream (against the current), its effective speed is the difference between its speed in still water and the speed of the current. So, the speed upstream is \( 14 - C \) mph.
We know that the distance traveled downstream and upstream is the same, which is 125 miles.
We can use the formula:
\[
\text{time} = \frac{\text{distance}}{\text{speed}}
\]
For the downstream trip, the time taken is \( \frac{125}{14 + C} \) hours.
For the upstream trip, the time taken is \( \frac{125}{14 - C} \) hours.
Since the total time taken for the round trip is 19 hours, we have:
\[
\frac{125}{14 + C} + \frac{125}{14 - C} = 19
\]
Now, we can solve this equation to find the value of \( C \), the speed of the current.
\[ \frac{125}{14 + C} + \frac{125}{14 - C} = 19 \]
\[ \frac{125(14 - C) + 125(14 + C)}{(14 + C)(14 - C)} = 19 \]
\[ \frac{125(14 - C) + 125(14 + C)}{196 - C^2} = 19 \]
\[ \frac{125 \cdot 14 - 125C + 125 \cdot 14 + 125C}{196 - C^2} = 19 \]
\[ \frac{125 \cdot 28}{196 - C^2} = 19 \]
\[ \frac{125 \cdot 28}{19} = 196 - C^2 \]
\[ \frac{125 \cdot 28}{19} = 196 - C^2 \]
\[ \frac{3500}{19} = 196 - C^2 \]
\[ 196 - \frac{3500}{19} = C^2 \]
\[ C^2 = 196 - \frac{3500}{19} \]
\[ C^2 = 196 - \frac{3500}{19} \]
\[ C^2 = \frac{19 \cdot 196 - 3500}{19} \]
\[ C^2 = \frac{3624 - 3500}{19} \]
\[ C^2 = \frac{124}{19} \]
\[ C^2 = \frac{124}{19} \]
\[ C = \sqrt{\frac{124}{19}} \]
\[ C \approx \sqrt{6.5263} \]
\[ C \approx 2.55 \text{ mph (approximately)} \]
So, the speed of the current is approximately 2.55 miles per hour.
I assume that the 19 hours is the time for the round trip.
Let the speed of the current equal x and the time it took to get down hill y.
Then it takes y=125/(14+x) hours to go down hill.
Also it takes 19-y= 125/(14-x) hours to get up hill.
Substituting equation one into two and then solving for x we get approximately 3.4 mph.
