Subtopic: Thrust Equations & Specific Fuel Consumption
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. 𝛼VLO (𝛼 > 0)
denotes the airplane speed at 0.5 SLO. Neglecting changes in the airplane mass
during the ground roll segment, the value of 𝛼 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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