A-Level Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency: worked solution

Question

A top-selling electric car measures 1.62 m high by 1.88 m wide, has drag coefficient 0.29 and peaks at 100 kW. With drag F_D = ½ρv²CA (air 1.29 kg m⁻³, A the frontal area facing travel), obtain its speed at full motor power. [3]

Principle: at top speed the car is no longer accelerating, so by Newton's first law the resultant force on it is zero. That means the forward driving force from the motor exactly balances the backward drag force. Also, the power delivered by a force moving at constant speed is P = F v. Step 1 - write the power in terms of speed alone. At top speed, driving force F = drag force F_D = ½ ρ v² C A. Substituting this into P = F v: P = (½ ρ v² C A) × v P = ½ ρ C A v³ So the power needed grows with the CUBE of the speed. Step 2 - work out the frontal area A. The car presents a rectangle of height 1.62 m by width 1.88 m to the oncoming air: A = 1.62 m × 1.88 m A = 3.05 m² Step 3 - rearrange P = ½ ρ C A v³ for v. Multiply both sides by 2: 2P = ρ C A v³ Divide both sides by ρ C A: v³ = 2P / (ρ C A) Take the cube root: v = [2P / (ρ C A)]^(1/3) Step 4 - substitute the values, in SI units. P = 100 kW = 100 × 10³ W ρ = 1.29 kg m⁻³ C = 0.29 (no units) A = 3.05 m² Top line: 2P = 2 × 100 × 10³ = 2.00 × 10⁵ W Bottom line: ρ C A = 1.29 × 0.29 × 3.05 = 1.14 kg m⁻¹ Step 5 - divide, then take the cube root. v³ = 2.00 × 10⁵ / 1.14 v³ = 1.76 × 10⁵ m³ s⁻³ v = (1.76 × 10⁵)^(1/3) v = 56.0 m s⁻¹ The top speed is about 56 m s⁻¹ (roughly 200 km h⁻¹).