Step 1: Understanding the Question:
The problem asks us to calculate the exact distance of the plain section of a road connecting two points A and B.
The total length of the road is given as 11.5 km.
The journey from A to B takes 2 hours and 54 minutes, and comprises uphill, plain, and downhill sections.
The return journey from B to A takes 3 hours and 6 minutes.
The person walks at 3 km/h uphill, 4 km/h on plain ground, and 5 km/h downhill.
Step 2: Key Formula or Approach:
The fundamental relationship is $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
Let the distance of the uphill part be $U$, the plain part be $P$, and the downhill part be $D$.
We know that the total distance $U + P + D = 11.5$ km.
On the return journey, the downhill section becomes uphill, and the uphill section becomes downhill.
We will set up two time equations for the forward and return journeys and add them together to eliminate variables.
Step 3: Detailed Explanation:
Let the length of the uphill section be $U$, the plain section be $P$, and the downhill section be $D$.
The total length of the road is given by the equation: \[ U + P + D = 11.5 \]
The time taken for the forward journey from A to B is 2 hours and 54 minutes.
We must convert 54 minutes into hours by dividing by 60, giving $\frac{54}{60} = \frac{9}{10} = 0.9$ hours.
So, the total forward time is $2.9$ hours or $\frac{29}{10}$ hours.
The equation for the forward journey is: \[ \frac{U}{3} + \frac{P}{4} + \frac{D}{5} = \frac{29}{10} \]
For the return journey, the person travels back over the same route.
The old downhill part $D$ is now traveled uphill at 3 km/h, and the old uphill part $U$ is traveled downhill at 5 km/h.
The plain part $P$ is still traveled at 4 km/h.
The time taken for the return journey is 3 hours and 6 minutes.
Converting 6 minutes to hours gives $\frac{6}{60} = \frac{1}{10} = 0.1$ hours.
So, the total return time is $3.1$ hours or $\frac{31}{10}$ hours.
The equation for the return journey is: \[ \frac{D}{3} + \frac{P}{4} + \frac{U}{5} = \frac{31}{10} \]
Now, we add the forward journey equation and the return journey equation together.
\[ \left( \frac{U}{3} + \frac{U}{5} \right) + \left( \frac{D}{3} + \frac{D}{5} \right) + \left( \frac{P}{4} + \frac{P}{4} \right) = \frac{29}{10} + \frac{31}{10} \]
\[ U\left( \frac{1}{3} + \frac{1}{5} \right) + D\left( \frac{1}{3} + \frac{1}{5} \right) + \frac{2P}{4} = \frac{60}{10} \]
\[ (U + D)\left( \frac{8}{15} \right) + \frac{P}{2} = 6 \]
From our total distance equation, we know that $U + D = 11.5 - P$.
We substitute this expression into our combined time equation.
\[ (11.5 - P)\left( \frac{8}{15} \right) + \frac{P}{2} = 6 \]
Now we expand the terms to solve for $P$.
\[ 11.5 \times \frac{8}{15} - P\left(\frac{8}{15}\right) + \frac{P}{2} = 6 \]
Since $11.5 = \frac{23}{2}$, the first term is $\frac{23}{2} \times \frac{8}{15} = \frac{23 \times 4}{15} = \frac{92}{15}$.
We isolate the terms containing $P$.
\[ P \left( \frac{1}{2} - \frac{8}{15} \right) = 6 - \frac{92}{15} \]
To simplify the parentheses, we use the common denominator 30.
\[ P \left( \frac{15 - 16}{30} \right) = \frac{90 - 92}{15} \]
\[ P \left( \frac{-1}{30} \right) = \frac{-2}{15} \]
Multiplying both sides by -30 gives: \[ P = \frac{2}{15} \times 30 = 2 \times 2 = 4 \]
The length of the plain part of the journey is precisely 4 km.
Step 4: Final Answer:
The length of the plain part of the journey is 4km.