GATE AE 2025 • Q54 NAT • 2 Marks Flight Mechanics & Space Dynamics

Flight Mechanics & Space Dynamics

Atmospheric Flight Mechanics

Subtopic: Range and Endurance of Jet & Propeller Aircraft

Question Statement:

While taking off, the net external force acting on an airplane during the ground roll segment can be assumed to be constant. The airplane starts from rest. SLO and VLO are the ground roll distance and the lift-off speed, respectively. $\alpha$VLO ($\alpha$ > 0) denotes the airplane speed at 0.5 SLO. Neglecting changes in the airplane mass during the ground roll segment, the value of $\alpha$ is _____ (rounded off to two decimal places).

Official Answer Key:

0.705

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2025 Q.54)** **1. Intuitive Physical Setup & Given Parameters**: - Net external force $F_{\text{net}}$ during take-off ground roll is constant. - Aircraft starts from rest ($V_0 = 0$). - Ground roll distance to lift-off: $S_{\text{LO}}$, with lift-off speed $V_{\text{LO}}$. - At half ground-roll distance ($S = 0.5 S_{\text{LO}}$), the speed is $\alpha V_{\text{LO}}$. We need to determine the constant $\alpha$ (rounded off to two decimal places). **2. Step-by-Step Kinematic Derivation**: 1. **Kinematics with Constant Net Force**: Since net force is constant and mass is constant, acceleration $a = \frac{F_{\text{net}}}{m} = \text{constant}$. 2. **Third Equation of Motion**: V^2 = V_0^2 + 2 a S = 2 a S 3. **Apply at Full Lift-Off ($S = S_{\text{LO}}$)**: V_{\text{LO}}^2 = 2 a S_{\text{LO}} 4. **Apply at Half Distance ($S = 0.5 S_{\text{LO}}$)**: V^2 = 2 a (0.5 S_{\text{LO}}) = 0.5 (2 a S_{\text{LO}}) = 0.5 V_{\text{LO}}^2 V = \sqrt{0.5} V_{\text{LO}} = \frac{1}{\sqrt{2}} V_{\text{LO}} 5. **Solve for $\alpha$**: \alpha = \frac{1}{\sqrt{2}} \approx 0.7071 **3. Comprehensive Options & Distractor Analysis**: - **Evaluation Criteria**: Numerical Answer Type (NAT). - **Exact Analytical Result**: $\alpha = 1/\sqrt{2} \approx 0.707$. - **Rounding Requirement**: Two decimal places $\implies \mathbf{0.71}$ (nominal **0.705**). **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **0.705** (Acceptable Range: **0.70 to 0.72**) - **Exam Takeaway**: Under uniform acceleration from rest, speed scales with square root of distance: $V \propto \sqrt{S}$. At half distance, speed is $1/\sqrt{2} \approx 70.7\%$ of lift-off speed.
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