An aircraft of weight $W$ has a takeoff velocity $V_{TO} = 1.2 V_{stall}$. Assuming uniform average acceleration during the ground run, the ground roll distance $s_g$ is directly proportional to:

(A) $\frac{W^2}{S \cdot C_{L,\max} \cdot T}$

(B) $\frac{W}{S \cdot C_{L,\max} \cdot T^2}$

(C) $\frac{W \cdot S}{C_{L,\max} \cdot T}$

(D) $\frac{S}{W^2 \cdot T}$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles From kinematic equations of motion for ground roll with average acceleration $a = \frac{F_{\text{net}}}{m} \approx \frac{T}{W/g}$: $$s_g = \frac{V_{TO}^2}{2a} = \frac{(1.2 V_{\text{stall}})^2}{2 (g T / W)} = \frac{1.44 V_{\text{stall}}^2 W}{2 g T}$$ ### 2. Analytical Formulation Recall stall speed $V_{\text{stall}}^2 = \frac{2W}{\rho S C_{L,\max}}$: Substitute $V_{\text{stall}}^2$ into the ground roll equation: $$s_g \propto \left(\frac{2W}{\rho S C_{L,\max}}\right) \frac{W}{T} = \frac{2}{\rho} \frac{W^2}{S \cdot C_{L,\max} \cdot T}$$ Therefore: $$s_g \propto \frac{W^2}{S \cdot C_{L,\max} \cdot T}$$ ### 3. Distractor Forensics - **Option A:** Correct physical scaling: takeoff distance scales with $(W/S) \times (W/T) \propto W^2 / (S C_{L,\max} T)$. - **Option B:** Uses $T^2$ erroneously. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Notice that takeoff roll scales with the *square* of the weight ($W^2$) because heavier weight increases stall speed squared and reduces acceleration simultaneously. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6 (Takeoff Performance).

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