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Step 1: Draw a figure from the given information

Radius of the circular race track R = 300 m

Angle of banking \theta = 15 \degree

Coefficient of friction \mu_s = 0.2

Acceleration due to gravity g = 9.8 m/s^2

N \cos \theta = mg + f_r \sin \theta

N \sin \theta + f_r \cos \theta = \frac{mv^2}{r}(centripetal force)

Required formulas to solve this question

On a banked road, the horizontal component of normal force and frictional force combine to generate centripetal force, which keeps the vehicle driving in a circular turn without skidding.

Maximum sped of the car on a banked road. $v_{max} = [Rg \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} ]^{\frac{1}{2}}$

The frictional force is not required at the optimum speed because the normal reaction component provides enough centripetal force. \mu_s = 0

Optimum speed of the car v_o = (Rg \tan \theta)^{\frac{1}{2}}

Step 2: Calculating the optimum speed of the vehicle

v_o = (Rg \tan \theta)^{\frac{1}{2}}

$v_o = [(300 m )(9.8 m/s^2) \tan 15\degree]^{\frac{1}{2}}$

v_o = 28.1 m/s

a) optimum speed of the race- car to avoid wear and tear on its tyres, v_o = 28.1 m/s

Step 3: Calculating the maximum speed of the vehicle

$v_{max} = [Rg \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} ]^{\frac{1}{2}}$

$v_{max} = [(300m)(9.8m/s^2) \frac{0.2 + \tan 15\degree}{1 - 0.2 \tan 15\degree}]^{\frac{1}{2}}$

v_{max} = \sqrt{1458 .7}

v_{max} = 38.1 m/s

b) maximum possible speed to avoid slipping, v_{max} = 38.1 m/s