RBD BYU 37

Step 1: Normal Reaction Calculation

The problem states that the center of gravity (C.G.) is equidistant from all four wheels. Therefore, the weight $mg$ is distributed equally between the front and rear axles.

Total normal reaction on rear wheels ($N_R$):

$$ N_R = \frac{mg}{2} $$

Similarly, the normal reaction on the front wheels is $N_F = \frac{mg}{2}$.

Step 2: Phase 1 – Motion Limited by Friction

Initially, as the car starts from rest, the engine power $P$ would theoretically imply infinite force at zero velocity ($F = P/v$). However, the force is physically limited by the static friction between the tires and the road. The car is rear-wheel drive, so the driving force comes from friction on the rear wheels.

Maximum acceleration occurs when friction is limiting ($\mu N_R$):

$$ ma = f_{\text{max}} = \mu N_R = \mu \left( \frac{mg}{2} \right) $$ $$ a = \frac{\mu g}{2} $$

Since acceleration is constant in this phase, the velocity as a function of time is:

$$ v(t) = a t = \frac{\mu g}{2} t \quad \text{for } 0 \le t \le t_0 $$

Step 3: Determining the Transition Time ($t_0$)

This phase of constant acceleration continues until the power required to maintain this acceleration equals the engine’s maximum constant power $P$.

$$ P = F \cdot v(t_0) $$ $$ P = \left( \frac{\mu mg}{2} \right) \left( \frac{\mu g}{2} t_0 \right) $$ $$ P = \frac{\mu^2 m g^2}{4} t_0 $$

Solving for the transition time $t_0$:

$$ t_0 = \frac{4P}{\mu^2 m g^2} $$

The velocity at this transition moment ($v_0$) is:

$$ v_0 = \frac{\mu g}{2} t_0 = \frac{\mu g}{2} \left( \frac{4P}{\mu^2 m g^2} \right) = \frac{2P}{\mu m g} $$

Step 4: Phase 2 – Motion Limited by Power ($t > t_0$)

Once $t > t_0$, the engine power $P$ becomes the limiting factor. The friction force adjusts to match the power output: $F = P/v$.

Using Newton’s Second Law:

$$ m \frac{dv}{dt} = \frac{P}{v} $$ $$ v \, dv = \frac{P}{m} dt $$

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

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

Step 5: Final Substitution

Substitute $v_0 = \frac{2P}{\mu m g}$ and simplify:

$$ v = \sqrt{ \left( \frac{2P}{\mu m g} \right)^2 + \frac{2P}{m}t – \frac{2P}{m}t_0 } $$

We know that at $t_0$, the kinetic energy term $\frac{1}{2}mv_0^2$ was built up by power $P$ over time $t_0/2$ (average velocity was $v_0/2$), or we can just simplify algebraically using $t_0 = \frac{4P}{\mu^2 m g^2}$:

$$ \frac{2P}{m} t_0 = \frac{2P}{m} \left( \frac{4P}{\mu^2 m g^2} \right) = \frac{8P^2}{\mu^2 m^2 g^2} = 2 v_0^2 $$

Substituting this back into the velocity equation:

$$ v^2 = v_0^2 + \frac{2P}{m}t – 2v_0^2 $$ $$ v^2 = \frac{2P}{m}t – v_0^2 $$ $$ v = \sqrt{ \frac{2P}{m}t – \frac{4P^2}{\mu^2 m^2 g^2} } $$

Rearranging to match the format of the provided solution:

$$ v = \sqrt{ \frac{2P}{m} \left( t – \frac{2P}{\mu^2 m g^2} \right) } $$