1. Conceptual Analysis

The car accelerates due to the kinetic friction force on the rear driving wheels. There are two distinct phases of motion:

• Phase 1 (Constant Force): Initially, the engine power is sufficient to spin the wheels, but the car’s acceleration is limited by the kinetic friction coefficient (same as static) $\mu$. The car accelerates at a constant rate.
• Phase 2 (Constant Power): As speed increases, the driving force required to maintain that maximum acceleration would require power $P > F_{max}v$. Since power is limited to constant $P$, the driving force decreases ($F = P/v$), and the car moves under constant power.

2. Calculating the Transition Point

Since the center of gravity is equidistant from the wheels, the normal force on the rear axle (two wheels) is half the weight:

$$N_{rear} = \frac{mg}{2}$$

The maximum frictional force (driving force) is:

$$F_{max} = \mu N_{rear} = \frac{\mu mg}{2}$$

The car accelerates with constant acceleration $a$ until the power required ($P = F v$) exceeds the engine’s constant power $P$. Let $v_0$ be the velocity where this transition happens:

$$P = F_{max} \cdot v_0 \implies v_0 = \frac{P}{F_{max}} = \frac{P}{(\mu mg / 2)} = \frac{2P}{\mu mg}$$

3. Solving for Phase 1 ($0 \le v \le v_0$)

During this phase, force is constant at $F_{max}$.

$$a = \frac{F_{max}}{m} = \frac{\mu mg / 2}{m} = \frac{\mu g}{2}$$

Velocity as a function of time:

$$v(t) = at = \frac{\mu g t}{2}$$

The time $t_0$ when the transition occurs is:

$$t_0 = \frac{v_0}{a} = \frac{(2P / \mu mg)}{(\mu g / 2)} = \frac{4P}{m \mu^2 g^2}$$

4. Solving for Phase 2 ($t > t_0$)

Now, the power is the limiting factor. $P = \text{constant}$.

$$P = Fv = (ma)v = m v \frac{dv}{dt}$$

Integrating from the transition point $(t_0, v_0)$ to an arbitrary point $(t, v)$:

$$\int_{v_0}^{v} v \, dv = \int_{t_0}^{t} \frac{P}{m} \, dt$$ $$\left[ \frac{v^2}{2} \right]_{v_0}^{v} = \frac{P}{m} (t – t_0)$$ $$\frac{v^2}{2} – \frac{v_0^2}{2} = \frac{P}{m} \left( t – \frac{4P}{m \mu^2 g^2} \right)$$

Substitute $v_0 = \frac{2P}{\mu mg}$:

$$\frac{v^2}{2} = \frac{P}{m}t – \frac{P}{m}\left(\frac{4P}{m \mu^2 g^2}\right) + \frac{1}{2}\left(\frac{2P}{\mu mg}\right)^2$$ $$\frac{v^2}{2} = \frac{P}{m}t – \frac{4P^2}{m^2 \mu^2 g^2} + \frac{2P^2}{m^2 \mu^2 g^2}$$ $$\frac{v^2}{2} = \frac{P}{m}t – \frac{2P^2}{m^2 \mu^2 g^2}$$ $$v^2 = \frac{2P}{m}t – \frac{4P^2}{m^2 \mu^2 g^2} = \frac{2P}{m} \left( t – \frac{2P}{m \mu^2 g^2} \right)$$