Part (a): Minimum Coefficient of Friction

The problem specifies that the Center of Gravity (CG) is equidistant from all wheels. Thus, the normal reaction on each wheel is: $$N = \frac{Mg}{4}$$ Hint Analysis: The hint states that frictional forces providing centripetal force act only on the steered wheels. In a rear-wheel-drive car, the front wheels are steered.

Total Centripetal Force required: $$F_c = \frac{Mv^2}{R}$$ This force is provided by the static friction on the two front wheels: $$2 \cdot f_{front} \ge \frac{Mv^2}{R}$$ Substituting max friction $f_{front} = \mu N = \mu \frac{Mg}{4}$: $$2 \left( \mu \frac{Mg}{4} \right) \ge \frac{Mv^2}{R}$$ $$\frac{\mu Mg}{2} \ge \frac{Mv^2}{R} \implies \mu \ge \frac{2v^2}{Rg}$$

Substituting values ($v=20, R=200, g=10$): $$\mu \ge \frac{2(20)^2}{200(10)} = \frac{800}{2000} = 0.4$$

Part (b): Minimum Time to Increase Speed

New coefficient of friction $\mu’ = 2 \mu = 0.8$.
We need the minimum time to change speed from $20$ to $20.5$ m/s. This implies maximizing tangential acceleration ($a_t$).

Front Wheels (Steering): Provide Centripetal Force.
Check if they slip: Max capacity = $2(\mu’ \frac{Mg}{4}) = 0.4Mg$. Required = $\frac{M(20)^2}{200} = 0.2Mg$.
Capacity ($0.4Mg$) > Required ($0.2Mg$). Front wheels are stable.

Rear Wheels (Driving): Provide Tangential Force (Acceleration).
Based on the specific model implied by the hint, rear wheels do not contribute to centripetal force here, so their entire friction budget is available for tangential acceleration. $$F_{tangential} = 2 \cdot f_{rear_{max}} = 2 \left( \mu’ \frac{Mg}{4} \right) = \frac{\mu’ Mg}{2}$$ $$Ma_t = \frac{0.8 Mg}{2} = 0.4 Mg \implies a_t = 4 \text{ m/s}^2$$

Time Calculation: $$\Delta v = 20.5 – 20 = 0.5 \text{ m/s}$$ $$t = \frac{\Delta v}{a_t} = \frac{0.5}{4} = 0.125 \text{ s}$$