$A$ circular racetrack of radius $300 \; m$ is banked at an angle of $15^{\circ}$. If the coefficient of friction between the wheels of a race-car and the road is $0.2$,what is the
$(a)$ optimum speed of the racecar to avoid wear and tear on its tyres,and
$(b)$ maximum permissible speed to avoid slipping?

(A) On a banked road,the horizontal component of the normal force and the frictional force contribute to provide the centripetal force required to keep the car moving on a circular turn without slipping.
$(a)$ At the optimum speed,the horizontal component of the normal reaction is sufficient to provide the necessary centripetal force,and the frictional force is not required. The optimum speed $v_{o}$ is given by the formula:
$v_{o} = \sqrt{R g \tan \theta}$
Given $R = 300 \; m$,$\theta = 15^{\circ}$,and $g = 9.8 \; m/s^{2}$:
$v_{o} = \sqrt{300 \times 9.8 \times \tan(15^{\circ})} = \sqrt{2940 \times 0.2679} \approx \sqrt{787.6} \approx 28.1 \; m/s$.
$(b)$ The maximum permissible speed $v_{\max}$ to avoid slipping is given by the formula:
$v_{\max} = \sqrt{R g \left( \frac{\mu_{s} + \tan \theta}{1 - \mu_{s} \tan \theta} \right)}$
Given $\mu_{s} = 0.2$:
$v_{\max} = \sqrt{300 \times 9.8 \times \left( \frac{0.2 + 0.2679}{1 - (0.2 \times 0.2679)} \right)} = \sqrt{2940 \times \left( \frac{0.4679}{0.9464} \right)} \approx \sqrt{2940 \times 0.4944} \approx \sqrt{1453.5} \approx 38.1 \; m/s$.