Conceptual Analysis

The motion consists of two distinct phases:

1. Slipping Phase: The wheels slip on the ground. The car accelerates under a constant kinetic friction force. The power delivered by the engine exceeds the rate of work done by friction, so the wheels spin up rapidly. This phase ends when the velocity $v$ reaches a critical value $v_c$ such that power $P = F \cdot v_c$.
2. Rolling Phase: Once pure rolling begins, static friction acts. We assume full power $P$ is now converted into kinetic energy (neglecting rotational KE of light wheels).

Case 1: Four-Wheel Drive (4WD)

Since all 4 wheels drive and support the total mass $m$ equally (symmetry), the total normal force is $mg$. Kinetic friction acts on all wheels.

Phase 1 (Slipping):
Total friction force $F = \mu mg$.
Acceleration $a = \frac{F}{m} = \mu g$.
This phase continues until the power required to maintain this acceleration matches the motor’s output $P$: $$ P = F \cdot v_c \Rightarrow v_c = \frac{P}{\mu mg} $$ Time for this phase $t_1$: $$ t_1 = \frac{v_c}{a} = \frac{P / (\mu mg)}{\mu g} = \frac{P}{m(\mu g)^2} $$

Phase 2 (Rolling):
Power $P$ increases kinetic energy directly. Using the work-energy theorem from $v_c$ to $v$: $$ \int_{t_1}^{t} P \, dt = \frac{1}{2}mv^2 – \frac{1}{2}mv_c^2 $$ $$ P(t – t_1) = \frac{1}{2}mv^2 – \frac{1}{2}m\left(\frac{P}{\mu mg}\right)^2 $$ $$ t = t_1 + \frac{mv^2}{2P} – \frac{m P^2}{2P m^2 \mu^2 g^2} = \frac{P}{m \mu^2 g^2} + \frac{mv^2}{2P} – \frac{P}{2 m \mu^2 g^2} $$ $$ t_{4WD} = \frac{mv^2}{2P} + \frac{P}{2m(\mu g)^2} $$

Calculation:
Substituting $m=1.0, v=20, P=50, \mu=0.5, g=10$: $$ t_{4WD} = \frac{1(20)^2}{2(50)} + \frac{50}{2(1)(0.5 \times 10)^2} = \frac{400}{100} + \frac{50}{2(25)} = 4 + 1 = 5.0 \text{ s} $$

Case 2: Rear-Wheel Drive (RWD)

Only rear wheels are powered. Assuming CoM is central, rear wheels support $N_R = mg/2$.

Phase 1 (Slipping):
Friction force $F = \mu N_R = \frac{\mu mg}{2}$.
Acceleration $a = \frac{F}{m} = \frac{\mu g}{2}$.
Critical velocity $v_c$: $$ P = F \cdot v_c \Rightarrow v_c = \frac{P}{\mu mg / 2} = \frac{2P}{\mu mg} $$ Time $t_1$: $$ t_1 = \frac{v_c}{a} = \frac{2P / (\mu mg)}{\mu g / 2} = \frac{4P}{m(\mu g)^2} $$

Phase 2 (Rolling):
Using the energy equation again: $$ t – t_1 = \frac{m}{2P}(v^2 – v_c^2) $$ $$ t = \frac{4P}{m(\mu g)^2} + \frac{mv^2}{2P} – \frac{m}{2P}\left(\frac{2P}{\mu mg}\right)^2 $$ $$ t = \frac{4P}{m(\mu g)^2} + \frac{mv^2}{2P} – \frac{2P}{m(\mu g)^2} $$ $$ t_{RWD} = \frac{mv^2}{2P} + \frac{2P}{m(\mu g)^2} $$

Calculation:
$$ t_{RWD} = \frac{1(20)^2}{2(50)} + \frac{2(50)}{1(0.5 \times 10)^2} = 4 + \frac{100}{25} = 4 + 4 = 8.0 \text{ s} $$