Let's solve this problem step by step: --- ### **Given Data** - **Elevation of Station A:** \( +1000 \) m - **Elevation of Station B:** \( +125 \) m - **Tunnel Length:** \( 4 \) km - **Curve Length:** \( 6 \) km - **Radius of Curves:** \( 2800 \) m - **Tunnel Resistance:** \( 2 \) kg/t - **Curve Resistance:** \( \frac{700}{R} \) kg/t, \( R = 2800 \) m - **Train Speed:** \( 210 \) km/h - **Running Resistance (\( \omega_0 \))**: \( \omega_0 = 2.5 + \frac{v^2}{4500} \) kg/t - **No braking or traction in tunnel (coasting)** ### **Required** Find the **distance between the two stations**. --- ## **Step 1: Calculate the Total Resistance** ### **1.1. Running Resistance (\( \omega_0 \)):** Given: - \( v = 210 \) km/h \[ \omega_0 = 2.5 + \frac{v^2}{4500} \] \[ \omega_0 = 2.5 + \frac{(210)^2}{4500} \] \[ = 2.5 + \frac{44100}{4500} \] \[ = 2.5 + 9.8 = 12.3 \text{ kg/t} \] ### **1.2. Tunnel Resistance** Given: \( 2 \) kg/t (for 4 km) ### **1.3. Curve Resistance** \[ \text{Curve resistance} = \frac{700}{R} \] \[ = \frac{700}{2800} = 0.25 \text{ kg/t (for 6 km)} \] --- ## **Step 2: Calculate the Total Resistance Over Each Section** - **Tunnel (4 km):** \( 12.3 + 2 = 14.3 \) kg/t - **Curve (6 km):** \( 12.3 + 0.25 = 12.55 \) kg/t --- ## **Step 3: Calculate the Work Done Against Resistances** Let \( S \) be the **total distance** between the stations (in km). - Tunnel length: \( 4 \) km - Curve length: \( 6 \) km - Remaining (level/straight): \( S - 4 - 6 = S - 10 \) km ### **Total Resistance (kg/t):** - For tunnel: \( 14.3 \) (for 4 km) - For curves: \( 12.55 \) (for 6 km) - For straight: \( 12.3 \) (for \( S-10 \) km) --- ## **Step 4: Use the Principle of Work and Energy** **Energy lost = Change in Potential Energy** Since the train moves at a constant speed (no change in kinetic energy), the work done against resistances equals the loss in potential energy due to the elevation drop: \[ \text{Work Against Resistance} = \text{Loss in Potential Energy} \] \[ (W_{\text{tunnel}} \times 4) + (W_{\text{curve}} \times 6) + (W_{\text{straight}} \times (S-10)) = g \times (\text{Height A} - \text{Height B}) \] Here, \( g = 9.81 \) m/s², but since both sides involve kg/t, we can omit it for unit balance as per railway resistance convention. \[ (14.3 \times 4) + (12.55 \times 6) + (12.3 \times (S-10)) = 1000 - 125 = 875 \text{ m} \] Convert meters to kg/t-km: - 1 m elevation drop = 1 kg/t resistance per km So, the total resistance multiplied by distance equals the elevation drop in meters. --- ## **Step 5: Plug in the Values and Solve for \( S \)** \[ (14.3 \times 4) + (12.55 \times 6) + (12.3 \times (S-10)) = 875 \] \[ (57.2) + (75.3) + (12.3S - 123) = 875 \] \[ 132.5 + 12.3S - 123 = 875 \] \[ 12.3S + 9.5 = 875 \] \[ 12.3S = 875 - 9.5 = 865.5 \] \[ S = \frac{865.5}{12.3} \] \[ S \approx 70.4 \text{ km} \] --- ## **Final Answer** \[ \boxed{70.4 \ \text{km}} \] **is the distance between the two stations.** --- ### **Summary Table** | Section | Length (km) | Resistance (kg/t) | Product (kg/t-km) | |-----------------|-------------|-------------------|-------------------| | Tunnel | 4 | 14.3 | 57.2 | | Curves | 6 | 12.55 | 75.3 | | Straight | S - 10 | 12.3 | 12.3(S - 10) | | **Total** | S | | 875 | Let me know if you need further clarification!