Let's solve the question step by step as requested: --- ### **Step 1: Find Pressure, Temperature, and Density Altitudes** #### **Given:** - Indicated Altitude = 14,500 ft - Outside Air Temperature (OAT) = 5°F - Indicated Airspeed (IAS) = 126 knots #### **a. Pressure Altitude** Pressure altitude is the altitude corrected for non-standard pressure; at standard setting (29.92 in Hg), it's the same as indicated altitude if the altimeter is set to 29.92. **Assume altimeter set to 29.92 in Hg (unless otherwise stated):** **Pressure Altitude = Indicated Altitude = 14,500 ft** **Explanation:** Pressure altitude is used as a baseline for other calculations and assumes standard pressure. --- #### **b. Temperature Altitude** Convert OAT from °F to °C: \[ T_{(°C)} = (T_{(°F)} - 32) \times \frac{5}{9} = (5 - 32) \times \frac{5}{9} = -27 \times \frac{5}{9} \approx -15°C \] **Explanation:** The temperature at the aircraft's altitude is -15°C. This is required for further calculations. --- #### **c. Density Altitude** Density altitude is the pressure altitude corrected for non-standard temperature: \[ \text{Density Altitude} = \text{Pressure Altitude} + [120 \times (\text{OAT} - \text{ISA Temp})] \] ISA Temperature at 14,500 ft: - Standard lapse rate: 2°C per 1000 ft - Sea level: 15°C \[ \text{ISA Temp at 14,500 ft} = 15°C - (2 \times 14.5) = 15°C - 29°C = -14°C \] So, \[ \text{Density Altitude} = 14,500 + [120 \times (-15 - (-14))] \] \[ \text{Density Altitude} = 14,500 + [120 \times (-1)] = 14,500 - 120 = 14,380 \text{ ft} \] **Explanation:** Density altitude is slightly lower than pressure altitude because the temperature is 1°C colder than standard at this altitude. --- ### **Step 2: Convert IAS to m/s** IAS (Indicated Airspeed) = 126 knots 1 knot = 0.51444 m/s \[ \text{IAS (m/s)} = 126 \times 0.51444 = 64.82 \text{ m/s} \] **Explanation:** IAS is simply converted from knots to meters per second using the standard conversion. --- ### **Step 3: Calculate TAS (True Airspeed) in m/s** The formula for TAS is: \[ \text{TAS} = \text{IAS} \times \sqrt{\frac{\rho_0}{\rho}} \] Alternatively, a quick estimate: \[ \text{TAS} \approx \text{IAS} \times \left(1 + 0.02 \times \frac{\text{Altitude (ft)}}{1000}\right) \] But let's use the more accurate method: - At 14,500 ft, standard pressure = 58.3 kPa, temperature -14°C - Standard sea level density (\(\rho_0\)) = 1.225 kg/m³ First, calculate actual air density (\(\rho\)) using: \[ \rho = \frac{p}{R T} \] Where: - \(p\) = pressure (Pa) - \(R\) = 287 J/kg·K (specific gas constant for dry air) - \(T\) = temperature (K) Pressure at 14,500 ft ≈ 58.3 kPa = 58,300 Pa Temperature = -15°C = 258 K \[ \rho = \frac{58300}{287 \times 258} \approx \frac{58300}{73946} \approx 0.788 \text{ kg/m}^3 \] Now, \[ \frac{\rho_0}{\rho} = \frac{1.225}{0.788} \approx 1.55 \] \[ \sqrt{1.55} \approx 1.245 \] \[ \text{TAS} = 64.82 \times 1.245 \approx 80.7 \text{ m/s} \] **Explanation:** TAS accounts for the lower air density at altitude and is higher than IAS. --- ## **Final Answers** ### **a. Altitudes:** - **Pressure Altitude:** **14,500 ft** - **Temperature at Altitude:** **-15°C** - **Density Altitude:** **14,380 ft** ### **b. Speeds:** - **IAS:** **64.8 m/s** - **TAS:** **80.7 m/s**