An aircraft flies through air at 30°C with a speed of 1260 km/h. How should its speed regime be classified (relative to the local speed of sound)?

Introduction / Context:
Aircraft speed regimes are classified by the Mach number, which is the ratio of true airspeed to the local speed of sound. Because the speed of sound varies primarily with temperature (and weakly with composition), the same numerical speed can be subsonic at cold altitudes and sonic or supersonic at warmer conditions near sea level. This problem asks you to categorize a given speed at 30°C.

Given Data / Assumptions:

• Ambient temperature T ≈ 30°C.
• Aircraft speed V = 1260 km/h.
• Speed of sound a ≈ 331 + 0.6*T (m/s) for air near sea level.

Concept / Approach:
Compute a from temperature, convert km/h to m/s, then find Mach M = V/a. If M ≈ 1, the regime is “sonic”. If M < 1 (significantly), subsonic; if M > 1, supersonic.

Step-by-Step Solution:
Compute a: a ≈ 331 + 0.6*30 = 331 + 18 = 349 m/s.Convert V: 1260 km/h ÷ 3.6 = 350 m/s (approx).Find Mach: M = V/a ≈ 350 / 349 ≈ 1.00.Classification: M ≈ 1 ⇒ sonic.

Verification / Alternative check:
Small rounding differences in a (due to humidity or lapse-rate assumptions) still leave M essentially unity at 30°C and 1260 km/h, so the sonic classification holds.

Why Other Options Are Wrong:

• Subsonic: would require M well below 1.
• Supersonic: would require clear margin above a (e.g., M ≥ 1.2 for practical supersonic flight).
• “Mach” alone is a unitless ratio name, not a regime classification.

Common Pitfalls:

• Forgetting to convert km/h to m/s (1 m/s = 3.6 km/h).
• Using a fixed a = 340 m/s regardless of temperature; here 30°C slightly raises a toward 349 m/s.

Final Answer:
Sonic