Power to overcome drag scales with v³

Steps:

• Drag force F ∝ v², so F at 40 m/s = 800 N × (40/20)² = 800 × 4 = 3200 N.
• For steady speed, engine power P equals drag force times velocity: P = F × v.
• P = 3200 N × 40 m/s = 128,000 W.
• Convert to kW: 128,000 W = 128 kW.

Why C is correct:

• Power P = F v, with F ∝ v², so P ∝ v³; original P at 20 m/s is 800 × 20 = 16 kW, thus at 40 m/s: 16 kW × (2)³ = 128 kW.

Why the others are wrong:

• A: Doubles velocity but ignores force quadrupling, yielding only 32 kW.
• B: Quadruples force but forgets doubling velocity, yielding 64 kW.
• D: Overestimates scaling (e.g., as v⁴), yielding 512 kW.

Final answer: C